View Full Version : [SOLVED] Astronomy Q&A Game
Nicool002
Mar8-03, 10:56 PM
Hi guys! Most of you know how this works but for the newcomers:
The rules are this: someone will ask a question and if the question is not answered correctly within 3 days then a new question will be posted. If an answer to a question is posted and the person that posted the question does not respond to the answer within 2 to 3 days, then the first person to have answered the question will then be able to post their own question. HAVE FUN AND LEARN!
I will start:
Question: What is the brightest star in the Northern Sky? (excluding the sun)
I'm making an educated guess here:
Alpha Canis Majoris (Sirius)?
Techincally, I think Deneb is the "brightest" but only in absolute magnitude. It's so far away that it really doesn't make much of a difference since its apparent magnitude is not as bright as Sirius.
Hey it's only a guess.
oh, man. I better lock this now before you guys swamp our server resources again.
-jk [:D]
Nicool002
Mar9-03, 03:23 PM
haha *hurt look* us? Phobos why ever would you say that? hehe
Yes sting you are correct, your go.
My go? My turn to ask a question (I never participated in the original thread so I have no idea how this works)?
Okay, my question: What is the name of the large multiringed basin located on Mercury at a longitude of 180 degrees?
Nicool002
Mar9-03, 03:57 PM
I believe that it is the Caloris Basin. Am I correct?
$#%^! I sure pick easy questions don't I?
Yes, you got it right. I guess, it's "your go"
Nicool002
Mar9-03, 04:09 PM
[:D] I knew that from when I studied Astronomy for Science Olympiad (By the way guys I got a gold medal in Astronomy for the Competition [:D] )
Ok. True Or False
Do we havea supermassive black hole at the center of our galaxy?
Shoot, that was quick...
Uhhh truefalse...
Actually, I don't know if it was ever "proven" but I'm going with "true"
Nicool002
Mar9-03, 04:23 PM
Correct, it is true. On Discover, they lookd at something like 37 galaxies and in each one, they found a supermassive black hole. Then, they looked at our galaxy and guesss what they found... Oh wait, sting already said it.
Yippee!
Okay, I guess it's my turn:
What is the name given to the energy generation in which a stellar object initiates helium burning by the triple-alpha process?
Nicool002
Mar9-03, 04:47 PM
OK I don't think I know this one but I'm gonna go out on a limb and say the Nuclear Fusion that a star relies on to live.... but I am not sure if that's right.
for how many years does pluto enter into neptune's orbit, and how many years does it take pluto to make one revolution around the sun?
no cheating...
Nicool002
Mar9-03, 05:15 PM
Kerrie I'm gonna answer your question but first The way I put it in the rules was that one question had to be answered before the next was asked that way you dont have a bunch of questions going on at once [:)]
Answer: Pluto takes 247 years that is the first answer. The second is... well I am not sure but I think it is around 100 years although that is basicaly just a guess.
OK I don't think I know this one but I'm gonna go out on a limb and say the Nuclear Fusion that a star relies on to live.... but I am not sure if that's right.
Generally, yes, but I was thinking more of a specific name but I'll give you a clue (two words)
What is the 'helium flash'?
how many years does it take pluto to make one revolution around the sun
Well if I was being tricky I would say exactly 1. Plutonian year that is [;)]
Raavin
axeeonn
Mar17-03, 07:44 PM
Darn, I do believe damgo got it.
RuroumiKenshin
Mar17-03, 07:51 PM
Originally posted by Raavin
Well if I was being tricky I would say exactly 1. Plutonian year that is [;)]
Raavin
It takes pluto 250 earth years to make one revolution around the sun.[:D]
I think Damgo got it too. The 'initiation' would seem to be the helium flash which continues into helium fusion. I think that's right, as the helium fuses to carbon 12, then you then have carbon flash and I assume you could have oxygen and neon flash but, for reasons that elude me, it would seem that it is naturally unlikely do to the changes/reduction in pressure until it gets to iron.
Raavin [:D]
Yep, he's right.
Helium flash was the answer I was looking for.
:) Hmm.... okay, what does /\-CDM stand for, and what is it?
jjalexand
Mar18-03, 07:06 AM
This may sound a bit sour but... now we are starting a new leaf, how about you starting a separate topic called "Astronomy Questions Game" so we can use the word "Astronomy" for a general topic of Astronomy without turning it into a sort of "Who wants to be a Millionaire" (an Aussie TV program :)
By the way, what happened to all the old posts, have they been classified due to the impending war or something?
[ome]
...cold dark matter...
...cold dark matter is composed of objects massive enough to move at sub-relativistic velocities...
cragwolf
Mar18-03, 10:49 AM
Originally posted by damgo
:) Hmm.... okay, what does /\-CDM stand for, and what is it?
As the previous poster pointed out CDM stands for cold dark matter, but the /\ stands for the dark energy or cosmological constant. I couldn't begin to tell you what cold dark matter is or what dark energy is, because no one knows! [:)]
I have turned to the darkside...
For I have tasted the truth...
(CEO Morgan's Recycling Tanks of Humour)
Yep, cragwolf got it.... it's the cosmological model with large lamda, omega_matter mostly dark.
Nicool003
Mar18-03, 03:26 PM
Well first off JJalexand were you on the old PF? and second off it's because this was the name on the old PF and I didn't want to change it and this thread IS about astronomy. I made it a game so it would be more fun.
(as you can see I had to change my name having login troubles seeing what we can do about it)
cragwolf
Mar18-03, 09:13 PM
Originally posted by damgo
Yep, cragwolf got it.... it's the cosmological model with large lamda, omega_matter mostly dark.
Hi damgo, can you award the prize to Bogdan, since he got most of the question right, and I'm feeling generous? [:)]
Nicool003
Mar18-03, 09:27 PM
Well right now I don't know the answer to your question. I will try to find the answer but I cant really guarantee anything. Does anyone know the answer?
Nicool003
Mar19-03, 03:27 PM
I believe it is CragWolf's question now am I right? SO go ahead and ask away crag.
cragwolf
Mar19-03, 06:05 PM
No, it's bogdan's question. Go ahead, bogdan, ask any astronomy question you like.
Thx...in what constellation is the star named Alkalurops ?
Originally posted by bogdan
Thx...in what constellation is the star named Alkalurops ?
The Plowman, I think his name is Buotus, although that's probably not the spelling!
cragwolf
Mar20-03, 05:01 PM
Boötes, I'm just correcting LURCH's spelling.
Nicool003
Mar20-03, 07:50 PM
That's correct but I'm sure you'd rather wait for bogdans confirmation... right?
By the way welcome to everyone from the old forums and new!
The Philosopher
Mar20-03, 10:09 PM
Alkalurops or Inkalunis or Icalurus or Clava or Venabulum... I guess you mean this star, well, it is in Bootes, in fact, it's Mu 1 Bootis. Alkalurops means the "shepherd's crook".
Bootes is the correct answer...
So...Lurch...it's your turn...thx to cragwolf...
I'm sorry Philosopher...maybe next time...
The Philosopher
Mar21-03, 11:12 AM
awwww..........shoot! I'll try next time too then.
M.
Nicool003
Mar21-03, 01:45 PM
Good we always like more people coming to join this therad welcome to PF[:D]
Ok Lurch ask away [6)]
Okey-Dokey,
Fortunately for astronomers,our solar system is currently moving into a rather large, empty portion of the galaxy. Sort-of like a "clearing" in the Milky Way Forest.
To what cause do astronomers attribute this clearing?
Nicool003
Mar22-03, 11:18 AM
Cosmic Acceleration
http://www.anzwers.org/free/universe/2000lys.html
Lurch, you say solar system is currently entering a clearing.
I found this map of the Orion Arm (which is the small arm we are in)
It is a projection on the galactic plane with the the sun in the center and all the Hipparchos cataloged stars out to 2000 ly.
the sun's motion is to the right of the map towards the stars marked cygnus, I believe.
there is also a link there to a largerscale map of the Orion Arm, everything out to 5000 ly.
maybe these maps will help someone answer the question.
What you say is interesting and makes me wonder what the size and distance of this clearing is and what is the timescale of our entering it.
Fortunately for astronomers,our solar system is currently moving into a rather large, empty portion of the galaxy. Sort-of like a "clearing" in the Milky Way Forest. I will take a stab from memory without any internet site to point to.
The distances mentioned at the site above about the Orion Arm are very small distances compared to the size of our galaxy. The Sun, with us, happens to be located at the "inside edge" of the Orion arm with relatively little between us and the Sagittarius arm, the next one in toward the galactic center. We are ~33,000 light years from the galactic core with this "clear space" allowing us to see across to the Sagittarius arm and the galactic core behind, with radio and infrared.
However, contrary to popular belief, a spiral arm does not make a full rotation of a galaxy intact as an arm with its general size and shape. Our Sun, will rotate a bit faster than the "rest" of the Orion arm as the stars farther from the core of the galaxy start to lag behind. The sun will make one full revolution in about 250 million years, and in that time the Orion arm, and all others, will dissapate as new spiral arms form from newly-formed stars and re-distribution of gas and dust. In other words, we will be moving more and more into an "empty space" between spiral arms as a few (million) years pass.
Nicool003
Mar22-03, 03:42 PM
Isn't it odd how in science a long time is a short time? Like on the radio (it was on a regular music station actually. I was quite surprised) They were saying that in only a few million years the Earth will begin to get colder because the sun will be changing into a Red Giant. ONLY A FEW MILLION YEARS FOLKS! PUT IT ON YOUR CALENDER!
(psst lurch did I get it right?)
Nope, yer thinkin' too big; occupational hazard in astronomy. I'm talking about the small (on the cosmic scale) clearing in our neighborhood of this galaxy.
Nicool003
Mar22-03, 03:53 PM
Oh I see. Or I will. I will think this over again. Cosmic acceleration would be the whole galaxy so I need something smaller...hmm
Nope, yer thinkin' too big; occupational hazard in astronomy. I'm talking about the small (on the cosmic scale) clearing in our neighborhood of this galaxy. Ok, another guess. If you are talking on a small scale, we happenen to be located in a large area of nearby star-forming regions. As the stars form and the dust and gas condense, we will be in, and looking through, a "clearer" area of space to whatever is beyond.
cragwolf
Mar23-03, 02:20 AM
A supernova. Just a guess, though.
Nicool003
Mar23-03, 09:04 AM
Hmmm I don't think it is a supernova but I got it wrong so I guess I wouldn't know...
So if cosmic acceleration is too big and that is the whole galaxy then....
There is something similar too cosmic acceleration occuring in our own galaxy, which is pushing all the solar systems away.
(which could be caused by the supermassive black hole at the center of our galaxy)
Originally posted by cragwolf
A supernova. Just a guess, though.
Darn good guess, Crag. As among the many "lucky accidents" of our kind of life developing on Earth, this supernova is believed to have taken place at sufficient distance so as not to bombard us with so much radiation that it sterrilized the surface, yet close enough to clear away the area into which ur system was about to pass, thus elliminating much of the debris that many exogeologists believe might prevent life from forming.
So I guess astronomers doubly benifit from that blast; it provides them with a clearer view, but it also permitted them to come into existance.
Your go Cragwolf.
There have been many "nearby" supernovae in recent times cosmologicaly speaking. Did you have something particular in mind, or just a generic event of many that could have been the answer to your question?
Questions on this thread should probably be more specific than just "something I had in mind", I think.
Nicool003
Mar23-03, 03:12 PM
Hmmm I don't think it is a supernova but I got it wrong so I guess I wouldn't know...
Good thing I said that! Wow I got something wrong twice in one question!
cragwolf
Mar23-03, 09:36 PM
OK, my turn. The globular cluster system of the Milky Way (and indeed, of many galaxies) is comprised of two distinct populations. What is the main observational parameter which distinguishes these two populations? The answer isn't size or luminosity, but that should give you an idea of what I mean by observational parameter.
I'm making a wild guess.... mass?
axeeonn
Mar23-03, 10:16 PM
Open and globular clusters. Where open clusters are typically young metal rich pop I stars and globular are of older low-metal pop II stars.
Color. Either observational or spectral. Older globulars are more orange-red, indicating population II stars, and any younger globulars would contain second generation stars of population I colored more toward the blue-white end of the spectrum. This would also show in the different types of Cepheids found in older verses newer globulars.
Nicool003
Mar24-03, 03:47 PM
Crag are you here? can you confirm the answers...?
cragwolf
Mar24-03, 05:48 PM
labguy's answer is more or less correct. Mostly, astronomers speak of a bimodality in the metallicity distribution of globular clusters in our Galaxy (metallicity tends to scale roughly linearly (*) with colour, so the two terms tend to be interchangeably used). There appears to be two populations of globulars within the Milky Way: a metal-poor population ([Fe/H ~ -1.5] associated with the galactic halo, and a metal-rich population ([Fe/H] ~ -0.5) associated with either the galactic disk or bulge. The metal-rich globulars seem to be roughly 2 billion years younger (on average) than the metal-poor globulars (but note, this is roughly the same as the uncertainty in the ages of globular clusters). One theory of globular cluster formation is that the younger, metal-rich clusters were formed during a merging episode (i.e. when the host galaxy merged with another galaxy), and the older, metal-poor globulars formed during the formation of the host galaxy itself. Observations of mergers show very young globulars.
(*) But in recent years this assumption of linearity has been challenged.
axeeonn
Mar24-03, 06:49 PM
Erm, I said metal.
Erm, I said metal. Yes, you did. But, you were also making the distinction between open and globular clusters, while the question was only about globulars.
I had a question, before the PF format changed, that I don't think was ever answered.(?)
QUESTION:
In the Crab Nebula (Supernova Remnant),
(1) What is particularly rare about the light that we can see?
(2) What causes this rare property?
(3) What is the name given to this type of light (EM radiation)?
axeeonn
Mar24-03, 08:20 PM
The pulsar produces synchrotron radiation, which is what partially makes the nebula glow. This is caused when charged particles, I think deutrons, sprial around a magnetic field. The photons are then emitted from the poles of the mag field.
The pulsar produces synchrotron radiation, which is what partially makes the nebula glow. This is caused when charged particles, I think deutrons, sprial around a magnetic field. The photons are then emitted from the poles of the mag field. Synchrotron radiation is right for the name, but I need two more specific answers for the rest. Right track, though.
Labguy asked:
QUESTION:
In the Crab Nebula (Supernova Remnant),
(1) What is particularly rare about the light that we can see?
(2) What causes this rare property?
(3) What is the name given to this type of light (EM radiation)?
(1) We see light in all known forms/frequencies and it does not follow the blackbody radiation curve, increasing toward lower frequencies rather than toward high. (It is said to be "nonthermal".)
(2) The fastest known Pulsar, near the middle of the nebula.
(3) Synchrotron Radiation.
1) It is polarized; different regions have different polarizations.
2) Electrons spiraling around regions of uniform magnetic field.
3) Synchrotron radiation.
- Warren
Originally posted by chroot
1) It is polarized; different regions have different polarizations.
2) Electrons spiraling around regions of uniform magnetic field.
3) Synchrotron radiation.
- Warren This one is correct. The main "property" I was looking for was that this type of EMR is polarized.
Your question, Chroot.
Hmmm... my turn... I gotta think of a good one. :)
- Warren
1) What major astronomical event occurred in 1987? (Hint: Large Magellanic Cloud)
2) The event was observed by more than just optical means. A particularly weakly-interacting type of particle was implicated. What were the particles?
3) How were these particles thought to be formed?
4) How many were detected, and when? How does that evidence help us understand what happened?
- Warren
Originally posted by chroot
1) What major astronomical event occurred in 1987? (Hint: Large Magellanic Cloud)
2) The event was observed by more than just optical means. A particularly weakly-interacting type of particle was implicated. What were the particles?
3) How were these particles thought to be formed?
4) How many were detected, and when? How does that evidence help us understand what happened?
- Warren
(1) Supernova 1987A, Type II Supernova.
(2) Neutrinos.
(3) A significant portion of the energy from the "rebound explosion" in a Type II supernova is in the form of Neutrinos.
http://www.solstation.com/x-objects/sn1987a.htm
(4) Detected 02/23/1987. Approximately 10^17 hit the detector and a total of 10 were detected, of ~10^58 estimated to be emitted. They were detected before the visable light reached the first sensors/cameras. The neutrino:light delay was ~2 to 3 hours. The amount (number) of neutrinos from a SN will tell us (a) the energy effenciency of different types of supernovae explosions, (b) which type of supernova was seen, (c) the different "rebound" energy effeciencies in supernovae from stars of different original chemical compositions (this "rebound" took ~2 hours to emit light in the visable after the neutrinos were produced in the explosion), and (d) lots of other neat stuff filling the book right beside my computer....[:)]
Labguy: You get a gold star. :) Your question?
- Warren
Originally posted by chroot
Labguy: You get a gold star. :) Your question?
- Warren
Ok, thanks. We might as well stay on the supernova subject, in general. In SN 1987A (last question), we happened to be lucky enough to get the early neutrino detections, and source identification, about 2-3 hours before the visable light was detected. We also know that most of any supernova energy is released as neutrinos, the rest as various wavelengths of EM radiation. Assuming these facts:
QUESTION:
(A) For SN 1987A in particular, what could have, and did, make the visable light delay as long as 2-3 hours instead of a shorter time period?
(B) In any supernova, what would be the most likely reason (cause) for a longer or shorter delay between neutrinos reaching us versus the EM radiation reaching us? This is not the same, specific reason as for SN 1987A as above.
The question (A) above might be a tough one, but I know the answer is out on the web, and in books, somewhere. Leeway will be given for "close enough"..[:))]
Labguy asked:
QUESTION:
(A) For SN 1987A in particular, what could have, and did, make the visable light delay as long as 2-3 hours instead of a shorter time period?
(B) In any supernova, what would be the most likely reason (cause) for a longer or shorter delay between neutrinos reaching us versus the EM radiation reaching us? This is not the same, specific reason as for SN 1987A as above.
(A) Calculations suggest that in SN 1987A the initial shock wave did not make it out of the core on its own. The core needed to contract even more before it could become a true neutron star. It did so by vast neutrino losses. The neutrinos were produced by the annihilation of electron-positron pairs made by the gamma rays. The total energy emitted in the 10-second neutrino burst was enormous, about 250 times the energy of the material explosion. It is believed that a small fraction of these neutrinos revived the stalled shock and powered the great explosion of the star. By heating and expanding the star and triggering a new flurry of nuclear reactions in its layered interior, the revived shock was responsible for the supernova's optical display. The effect was delayed by about 2 to 3 hours however: the shock had to traverse the entire star before any light made it out. The neutrinos from the collapsing core easily outraced the shock. Passing through the rest of the star very close to the speed of light, and so they were the first signal to leave the supernova.
(B) It depends on the progenitor star, but I'm not sure which characteristics specifically it depends on. All my googling turned up "since the neutrinos reach us before the visible light, we can use this as a detection system for SNs..." not one of them said why.
If you will take it, I will call your answer correct, since on question "(A)" it was so accurate and complete. I had thought that the second question would be easier, and since it mentioned all supernovae in general, you are also correct.
I mentioned no specifics about a star in the second question, so the only reasonable answer left would have to be the MASS of the "progenitor" star. We all know that the neutrinos radiate first and unempeded, so they would always reach us first. In a star of large mass, the propogation of the reactions creating the EM radiation would take longer to complete and "reach the edge" from where we could see the radiation. The reverse would be the case for a star of smaller mass. SO, a long "neutrino-light" delay means a larger star exploding than one with a shorter "neutrino-light" delay.
Very imperssive answer!
Your Question:
Excellent. (Sorry it took so long to get back. Spring fever...)
Today's topic is binary stars. Several questions, mostly in a multiple choice format.
1) Which of these is the best description of the orbital motion of a binary star system?
A) The less massive star orbits around the more massive one.
B) Both stars orbit around a common center of mass.
C) The dimmer star orbits around the brighter one.
D) Each star orbits around the other.
2) For visual binaries, which of these stars is usually designated the 'primary' component?
A) the heaviest
B) the most iron-rich
C) the dimmest
D) the brightest
3) Of the four main types of binary stars, which is detected by 'wobbles' in a point of light's motion across the sky?
A) eclipsing
B) visual
C) astrometric
D) spectroscopic
4) Of the four main types of binary stars, which can be 'resolved' into its separate components through a telescope?
A) astrometric
B) eclipsing
C) spectroscopic
D) visual
5) True or false: Planets can exist in stable orbits in a binary star system.
6) Observations of a visual double must record two characteristics: the angle orientation of the secondary with respect to the primary, and ... what?
A) angular separation between the two stars
B) parallax of each star
C) linear separation between the two stars
D) mean differences in the solar spectra
7) What astronomer coined the phrase 'binary star'?
A) Sir Edmond Halley
B) Sir William Herschel
C) Clyde Tombaugh
D) Asaph Hall
E) J-Man
Originally posted by J-Man
1) Which of these is the best description of the orbital motion of a binary star system?
A) The less massive star orbits around the more massive one.
B) Both stars orbit around a common center of mass.
C) The dimmer star orbits around the brighter one.
D) Each star orbits around the other.
B
2) For visual binaries, which of these stars is usually designated the 'primary' component?
A) the heaviest
B) the most iron-rich
C) the dimmest
D) the brightest
D
3) Of the four main types of binary stars, which is detected by 'wobbles' in a point of light's motion across the sky?
A) eclipsing
B) visual
C) astrometric
D) spectroscopic
C
4) Of the four main types of binary stars, which can be 'resolved' into its separate components through a telescope?
A) astrometric
B) eclipsing
C) spectroscopic
D) visual
D
5) True or false: Planets can exist in stable orbits in a binary star system.
A tentative true. In certain systems, resonances exist in which a planet could conceivably exist in a stable orbit. The vast majority of orbits in most binary systems, however, throw the planet into a star, or out into space.
6) Observations of a visual double must record two characteristics: the angle orientation of the secondary with respect to the primary, and ... what?
A) angular separation between the two stars
B) parallax of each star
C) linear separation between the two stars
D) mean differences in the solar spectra
A
7) What astronomer coined the phrase 'binary star'?
A) Sir Edmond Halley
B) Sir William Herschel
C) Clyde Tombaugh
D) Asaph Hall
E) J-Man
B
- Warren
chroot answered:
1=B
2=D
3=C
4=D
5=true
6=A
7=B
Very good!
I would also have accepted "E" for question 7 even though it is completely wrong. I have a soft spot for brown-nosers. [;)]
It is now your turn for a question.
Since it's Messier marathon season, why don't we do a set of questions on the Messier catalogue? These questions can all be answered easily with the right references -- I have no idea if the information is on the web, though.
1) List the Messier catalogue objects which are thought to have been recorded erroneously. Extra points for describing which nearby objects are thought were intended.
2) List the Messier objects that were actually added to the catalogue by third parties long after both Messier and his assistant died.
- Warren
Originally posted by chroot
1) List the Messier catalogue objects which are thought to have been recorded erroneously. Extra points for describing which nearby objects are thought were intended.
2) List the Messier objects that were actually added to the catalogue by third parties long after both Messier and his assistant died.
1) Most Messier objects are nebulae, star clusters or galaxies. There are 3 that are not.
i) M24 - A Milky Way star cloud which contains an 11th mag open cluster (NGC 6603). The NGC erroneously takes this cluster for M24, although Messier without doubt described the star cloud.
ii) M40 - Binary star system that Messier found and logged when looking for a (non-existant) nebula reported by 17th century observer Jan Hevelius.
iii) M73 - a group or an asterism of four 10th to 12th magnitude stars, which Messier measured at the same time when he determined M72's position.
Some versions of the Messier list omit some or all of these objects, though they are without doubt real objects, and their appearance was correctly described by Messier. However, these objects can be hardly classified as deep sky objects at all: M40 and M73 are multiple stars (or asterisms), while M24 is perhaps no object at all, but a "window in the dust" obscurring the Milky Way, and/or a larger portion of a spiral arm.
Missing objects: Of the 103 objects in the full printed version of Messier's catalog, only 99 show up as described at their position, while four objects are missing: M47, M48, M91, and M102. For at least three of these entries, the described objects exist, but Messier gave a wrong position, only the case of M102 is still controversially discussed.
-------------
2) Additional Messier objects:
7 objects were added to the Messier list by others; objects M104 to M110.
See this page for more info on these objects: http://www.maa.agleia.de/Messier/addition.html
J-Man,
Correct on all counts. Good job!
- Warren
OK, let's think about asteroids for a bit. I'll use the multiple choice format again for several (loosely) related questions.
Question 1:
The first asteroid, 1 Ceres, was discovered on Jan 1, 1801, who discovered it?
A) WIlliam Hershel
B) Giuseppe Piazzi
C) Edwin Hubble
D) Issac Newton
------------------------------------
Question 2:
Asteroids that come within the orbit of the Earth at their perihelions (closest approach to the Sun) are known as ___________ asteroids.
A) Aten
B) Apollo
C) Trojan
D) Amors
------------------------------------
Question 3:
Asteroids that are always closer to the Sun than the Earth are called __________
asteroids.
A) Apollo
B) Amor
C) Trojan
D) Aten
------------------------------------
Question 4:
If all of the asteroids were lumped together, they would make a planet bigger than
Jupiter.
True or False
------------------------------------
Question 5:
What asteroid did the NEAR Shoemaker mission recently land on?
A) 243 Ida
B) 3 Juno
C) 433 Eros
D) 951 Gaspra
------------------------------------
Question 6:
The Galileo spacecraft had encounters with two asteroids while enroute to Jupiter, what are the names of the asteroids?
A) 2062 Aten and 4 Vesta
B) 243 Ida and 951 Gaspra
C) 2212 Hephaistos and 511 Davida
D) 52 Europa and 911 Agamemnon
------------------------------------
Question 7:
Which asteroid was discovered to have its own satellite?
A) 951 Gaspra
B) 243 Ida
C) 433 Eros
D) 3 Juno
------------------------------------
Question 8:
Between the main concentrations of asteroids in the main belt are relatively empty regions known as the _______________.
A) Kuiper belt
B) Cassini's division
C) Kirkwood gaps
D) Van Allen belts
Nicool003
Apr3-03, 07:51 PM
J-man can we erm try to keep it 3 questions or below next time? [;)]
I'm sorry. I guess I got carried away.
How about the 1st person to get half of em right gets the next question?
Question 1:
B) Giuseppe Piazzi
Question 2:
B) Apollo
Question 3:
D) Aten
Question 4:
False
Question 5:
C) 433 Eros
Question 6:
B) 243 Ida and 951 Gaspra
Question 7:
C) 433 Eros
Question 8:
B) Cassini's division
axeeonn answered:
Question 1:
B) Giuseppe Piazzi
Question 2:
B) Apollo
Question 3:
D) Aten
Question 4:
False
Question 5:
C) 433 Eros
Question 6:
B) 243 Ida and 951 Gaspra
Question 7:
C) 433 Eros
Question 8:
B) Cassini's division
Numbers 1 through 6 are correct; numbers 7 & 8 were incorrect
The answer list is:
1=B
2=B
3=D
4=false
5=C
6=B
7=B
8=C
You got 6 out of 8, but you only needed 4 so your turn axeeonn.
Use Hubble's law to determine the age of the universe (assuming Ho is actually costant).
If you use Ho = 70km/s/MPc, you get... ?
The Hubble time (tH, age of the universe) is defined as 1/H0. In term of years,
tH [yr] = 9.78 x 1011 [yr km / s Mpc] / H0 [km/s/Mpc]
If H0 = 70 km/s/Mpc, then tH = 13.97 Gyr.
- Warren
cragwolf
Apr7-03, 07:53 PM
The Hubble time is not necessarily the age of the universe. Indeed, it's unlikely to be so. But it's close. The age of the universe, at least in the most successful model, depends on the mass/energy density, curvature and size of the cosmological constant. The actual formula for the age of the universe, as derived by the Friedman-Lemaitre model is given by:
Ht = [inte] dy / y(Ω_m y^3 + Ω_R y^2 + Ω_λ)^1/2
where the integral goes from y=1 to y=[oo], and
y = 1+z (where z is the redshift)
Ω_m is the mass/energy density term
Ω_R is the curvature term
Ω_λ) is the cosmological constant term
H is the current Hubble constant
t is the age of the universe
Current data seem to suggest:
Ω_m = 0.27
Ω_R = 0
Ω_λ = 0.73
H = 71 km/s/Mpc
Anyone care to try out the integral? [:D]
Hmmm, I think I will pass on that integral. ;)
chroot, you're up.
cragwolf
Apr11-03, 11:39 PM
[zz)]
Hey chroot!
CHROOT!
[s(]
Sorry to wake you up, but it's your turn to ask a question.
[8)]
[:D]
Originally posted by cragwolf
The Hubble time is not necessarily the age of the universe. Indeed, it's unlikely to be so. But it's close. The age of the universe, at least in the most successful model, depends on the mass/energy density, curvature and size of the cosmological constant. The actual formula for the age of the universe, as derived by the Friedman-Lemaitre model is given by:
Ht = [inte] dy / y(Ω_m y^3 + Ω_R y^2 + Ω_λ)^1/2
where the integral goes from y=1 to y=[oo], and
y = 1+z (where z is the redshift)
Ω_m is the mass/energy density term
Ω_R is the curvature term
Ω_λ) is the cosmological constant term
H is the current Hubble constant
t is the age of the universe
Current data seem to suggest:
Ω_m = 0.27
Ω_R = 0
Ω_λ = 0.73
H = 71 km/s/Mpc
Anyone care to try out the integral? [:D]
This is out of order---a question concerning the integral. There is a related formula for the age of the universe. Does anyone have an online source for it? The formula might result from cragwolf's integral.
To make writing it easier let S = Ω_λ^1/2
so that if Ω_λ = 0.73, S will be 0.84.
age = (2/3H) (1/2S) ln [(1+S)/(1-S)]
I came across this online and have lost the URL. Can anyone supply a reference? Don't want to interupt the game, but would appreciate any help.
Originally posted by cragwolf
[zz)]
Hey chroot!
CHROOT!
[s(]
Sorry to wake you up, but it's your turn to ask a question.
[8)]
[:D]
Sorry folks. I have been asked to leave this forum by Greg Bernhardt, the owner.
- Warren
Originally posted by chroot
Sorry folks. I have been asked to leave this forum by Greg Bernhardt, the owner.
- Warren
Why?
Greg Bernhardt
Apr12-03, 06:13 PM
Please use the PM system for personal and off-topic questions.
Thanks
Nicool003
Apr13-03, 12:26 AM
Hey guys sorry I havent been around this thread much! I will be though! To get back on topic how about the person that asked the question to ask another one...
Originally posted by Nicool003
Hey guys sorry I havent been around this thread much! I will be though! To get back on topic how about the person that asked the question to ask another one...
That would be axeeonn who asked, on 5April:
" Use Hubble's law to determine the age of the universe (assuming Ho is actually constant).
If you use Ho = 70km/s/MPc, you get... ?"
This was the most recent question asked, so you are inviting
axeeonn to pose another.
[Long pause...]
I hope that, one way or another, you get the game started back up because from what I've read it is phenomenal----really interesting stuff
axeeonn
Apr14-03, 08:04 PM
Ok, let's try to get this rolling agian...
How many kilograms of hydrogen is converted to He in our sun per second?
Originally posted by axeeonn
Ok, let's try to get this rolling agian...
How many kilograms of hydrogen is converted to He in our sun per second? ~700 million tons of hydrogen.
http://www.seds.org/billa/tnp/sol.html
Should be ~6.35e11 Kg. H converted to He.
axeeonn
Apr14-03, 10:01 PM
Damgo's turn, go go go!
Cool. Okay, the LIGO project is designed to detect gravitational waves from space by measuring the change in distance between two objects -- a laser and a far-away mirror -- as the gravitational wave passes by.
1) For an expected incoming gravitational wave, by what percentage will it alter distances as it passes by? (order of magnitude)
2) How long are the LIGO interferometer arms? (distance from laser to mirror)
3) So what amount of change in distance must LIGO be sensitive to, in order to detect gravitation waves?
Originally posted by damgo
Cool. Okay, the LIGO project is designed to detect gravitational waves from space by measuring the change in distance between two objects -- a laser and a far-away mirror -- as the gravitational wave passes by.
1) For an expected incoming gravitational wave, by what percentage will it alter distances as it passes by? (order of magnitude)
*An E-18 meter change over a length of 4000 meters. The fraction is 0.25 x 10^-21. This is 0.25 x 10^-19 percent.*
2) How long are the LIGO interferometer arms? (distance from laser to mirror)
*4000 meters*
3) So what amount of change in distance must LIGO be sensitive to, in order to detect gravitation waves?
E-18 meter
http://www.nature.com/nsu/nsu_pf/991111/991111-3.html
I am just copying what the guy in Nature Science Update said.
He said the expected change is E-15 millimeter over a length of 4000 meters, approximately. One arm stretches out while the other arm perpendicular to it gets shrunk by the same amount.
****added later****
Here's another web-reference, an excellent article inPhysics Today
from October 1999 by the director of LIGO Barry Barish
http://www.aip.org/web2/aiphome/pt/vol-55/iss-5/pdf/vol52no10p44-50.pdf
He confirms that the fractional change in length that they expect to be sensitive to is on the order of 10^-21.
This would be 4E-18 meter in the arm length of 4000 meter.
^^^ yup... it's pretty amazing, no? Your go!
Originally posted by damgo
^^^ yup... it's pretty amazing, no? Your go!
measuring an attometer (E-18 meter) change in distance is pretty amazing, couldn't believe it at first and thought there was some mistake.
Here is a question. Think of a generic black hole that has the same mass as the earth, and imagine that you want to give a young person some round object---a baseball or pingpong ball or marble or whatever---to show them the size.
What is the diameter of the thing you are looking for?
had a second question but erased it, another time maybe
worried by lack of response. I'm a relative newcomer and
may be off-base somehow. the diameter (twice Schwarzschild
radius) of a non-rotating black hole with mass the same as Earth's?
Answer is 2GM/c^2 for radius.
0.4 inches; = 1cm is radius.
Double that for your diameter question; 2cm diameter.
Labguy
Right you are! Your go.
Originally posted by marcus
Labguy
Right you are! Your go. Cool. Ok, more Black Hole stuff.
The "original" Schwarzchild Radius calculated by Schwartzchild, Oppenheimer and many others was based on MASS, gravity, c and not much else. Later, Kerr and Newman added other properties to be considered. So now we have the "accepted" Kerr-Newman Black Hole.
QUESTION:
(A) What is the difference in the formula (therefore size) of the Event Horizon in a Schwarzchild Black Hole vs. a Kerr-Newman Black Hole??
(B) What "property" did Kerr add to the static Black Hole?
(C) What "property" did Newman add to the static Black Hole?
Originally posted by Labguy
Cool. Ok, more Black Hole stuff.
The "original" Schwarzchild Radius calculated by Schwartzchild, Oppenheimer and many others was based on MASS, gravity, c and not much else. Later, Kerr and Newman added other properties to be considered. So now we have the "accepted" Kerr-Newman Black Hole.
QUESTION:
(A) What is the difference in the formula (therefore size) of the Event Horizon in a Schwarzchild Black Hole vs. a Kerr-Newman Black Hole??
(B) What "property" did Kerr add to the static Black Hole?
(C) What "property" did Newman add to the static Black Hole?
(B) and (C)-----Kerr added spin and Newman added charge
(and that does it because mass spin and charge are the only properties the thing can have)
(A)----I think I have seen that the event horizon radius r+ is given by a formula like this:
r+ = (r/2) + sqrt[ (r/2)^2 - a^2 - Q^2)
where r is 2GM/c^2, the usual Schw. radius. Also Q is the charge and a is an angular momentum term J/Mc called the specific angular momentum normalized by c. This a clearly has the dimensions of length. Hmmm must think about Q. I will try to improve on this answer and clear up some confusion I have about it.
Anyway at least this reduces to the usual r when there is no spin and charge! That is, when a and Q are zero, then r+ is just the ordinary r.
(B) and (C)-----Kerr added spin and Newman added charge
(and that does it because mass spin and charge are the only properties the thing can have) B and C are correct. There is one more property that a BH can have, and that is a magnetic field. But, this is only true with an accreting BH, so your answer is still correct.
(A)----I think I have seen that the event horizon radius r+ is given by a formula like this:
r+ = (r/2) + sqrt[ (r/2)^2 - a^2 - Q^2)
where r is 2GM/c^2, the usual Schw. radius. Also Q is the charge and a is an angular momentum term J/Mc called the specific angular momentum normalized by c. The key words here are "where r is 2GM/c^2, the usual Schw. radius. Correct again. The event horizon radius is the same for a static or rotating Black Hole!
Q and J come into play only when the smart guys calculate any suspected properties of the Ring Singularity and one other "area of influence" that hasn't been mentioned yet here. This would be the Ergosphere necessary in any model of a Kerr-Newman Black Hole. So now, we will usually be dealing with (a) a Singularity, (b) an Event Horizon and (c) an Ergosphere.
You already knew all the answers here! Too quick for me; at least let one of my questions last more than an hour or two...[:)]
[i]
You already knew all the answers here! Too quick for me; at least let one of my questions last more than an hour or two...[:)] [/B]
You can rely on it! I think it's my turn so here is one about the Gamma Ray Burst of March 29 of this year. It was a big one apparently so there was a lot about it on the web. I am not quite sure how they rate those things---either by the gamma ray burst itself, which only satellite sensors can detect---or by the optical afterglow. In this case the optical afterglow was visible to the naked eye during the first minute or so I think. Anyway it was considered notable for whatever reason. My question is:
How far away was it and in what constellation did it appear?
screwball
Apr17-03, 10:51 AM
Originally posted by marcus
You can rely on it! I think it's my turn so here is one about the Gamma Ray Burst of March 29 of this year. It was a big one apparently so there was a lot about it on the web. I am not quite sure how they rate those things---either by the gamma ray burst itself, which only satellite sensors can detect---or by the optical afterglow. In this case the optical afterglow was visible to the naked eye during the first minute or so I think. Anyway it was considered notable for whatever reason. My question is:
How far away was it and in what constellation did it appear?
-it was a supernova in the constellation Leo
-i looked it up and it was about 2 billion light years away
very cool, they said the gamma ray burst out did the entire universe for about 30 sec in gamma rays
Originally posted by screwball
-it was a supernova in the constellation Leo
-i looked it up and it was about 2 billion light years away
very cool, they said the gamma ray burst out did the entire universe for about 30 sec in gamma rays
Leo was where, and so they did! Your go.
screwball
Apr17-03, 03:04 PM
Originally posted by marcus
Leo was where, and so they did! Your go.
ok thanks[:)]
i dont have any good questions rite now so ill just put up a simple one to pass my turn off
Pleiades a popular cluster is also know as "the Seven Sisters". But when viewed with the nakid eye on a fairly good night you can only count six. With a telescope or binoculars of coarse you can see many more than seven.
the question is
~If only six stars are viewable with the nakid eye why would people of ancient times call it "the Seven Sisters"? (there are 2 reasonable explinations for this, either one is fine)
cragwolf
Apr17-03, 07:14 PM
Here are my guesses.
1. One of the stars has decreased in brightness over the years.
2. Light pollution was much less significant back then.
Originally posted by screwball
ok thanks[:)]
i dont have any good questions rite now so ill just put up a simple one to pass my turn off
Pleiades a popular cluster is also know as "the Seven Sisters". But when viewed with the nakid eye on a fairly good night you can only count six. With a telescope or binoculars of coarse you can see many more than seven.
the question is
~If only six stars are viewable with the nakid eye why would people of ancient times call it "the Seven Sisters"? (there are 2 reasonable explinations for this, either one is fine)
in case anyone's interested,
Gibson at U Calgary posted this list of the brightest Pleiades:
Name Designation Visual Magnitude
Alcyone 25 Tau 2.90
Atlas 27 Tau 3.62
Electra 17 Tau 3.70
Maia 20 Tau 3.87
Merope 23 Tau 4.18
Taygeta 19 Tau 4.30
Pleione 28 Tau 5.09
-- HD 23985 5.23
Asterope 1+2 21+22 Tau 5.31 (combined)
-- HD 23753 5.44
http://www.ras.ucalgary.ca/~gibson/pleiades/pleiades_see.html
The modern names do not necessarily correspond to the Greek names, if the Greeks had consistent names for individuals in the group. Gibson says the Greeks had several ways of explaining why only six are commonly visible. They made up excuses like Elektra was saddened after Troy (which her son founded) fell and faded out etc. Here is an exerpt from his page:
http://www.ras.ucalgary.ca/~gibson/pleiades/pleiades_myth.html
''Lost Pleiad
The `lost Pleiad' legend came about to explain why only six are easily visible to the unaided eye (I have my own thoughts on this). This sister is variously said to be Electra, who veiled her face at the burning of Troy, appearing to mortals afterwards only as a comet; or Merope, who was shamed for marrying a mortal; or Celæno, who was struck by a thunderbolt. Missing Pleiad myths also appear in other cultures, prompting Burnham to speculate stellar variability (Pleione?) as a physical basis. It is difficult to know if the modern naming pays attention to any of this...."
My guess is that the Greeks had seven sisters in the story because seven is an appealing number to use in folktales. Then when they got the idea of applying it to that cluster of stars they just adapted it by various made-up explanations why only six were visible (to most people most of the time anyway.) It was a kludge.
Cragwolf's idea makes sense too. And he hasn't asked a question in a while.
screwball
Apr17-03, 09:35 PM
Originally posted by cragwolf
Here are my guesses.
1. One of the stars has decreased in brightness over the years.
2. Light pollution was much less significant back then.
this is one possible ansewer, the other was nailed by marcus [:D]
but cragwolf buzzed in first so its your turn buddy
cragwolf
Apr18-03, 03:16 AM
OK, my question is the following: what is meant by the term relaxation time, as applied to agglomerations of stars, e.g. open clusters, globular clusters, galaxies?
Originally posted by cragwolf
OK, my question is the following: what is meant by the term relaxation time, as applied to agglomerations of stars, e.g. open clusters, globular clusters, galaxies?
I think it is the time an (N-body) system takes to forget the initial conditions and get thoroughly scrambled by random interactions so that the velocities are distributed according to some probability curve.
Like a bunch of air molecules in a box. You put them in with whatever artificial distribution of velocities. Maybe half going 200 m/s and half going 300 m/s. Then after a while, by random collisions, they trade energy around and come to a "Maxwellian" distribution of velocities---a one-sided bell curve.
In the case of a globular cluster, relaxation time could be the time you expect it to take for the cluster to become spherical. By random gravitational interactions (analogous to collisions) between pairs and triples of stars.
I believe that at least in globular clusters this time might be proportional to N/logN and to the average crossing time----the time an average star takes to get from one side of the heap to the other.
My idea about this is pretty vague and if someone wants to be precise I would gladly defer.
With the solar system the relaxation time may involve getting sorted out into roughly a plane and getting the orbits circularized. I do not know how to correctly define relaxation time in all these different cases.
Maybe same as above, but "relaxation time" in a stellar cluster is the time that it takes for the cluster stars to fall to a motion that "behaves like a gas".
http://www.pas.rochester.edu/~dmw/ast142/Lectures/Lect_16b.pdf
Originally posted by Labguy
Maybe same as above, but "relaxation time" in a stellar cluster is the time that it takes for the cluster stars to fall to a motion that "behaves like a gas".
http://www.pas.rochester.edu/~dmw/ast142/Lectures/Lect_16b.pdf
That is a much better and more concise answer. Mine can be ignored. That is a great set of lecture notes by Dave Watson.
cragwolf
Apr18-03, 06:44 PM
Oops, I didn't see this last page, Marcus. OK, Labguy is right, and can ask the next question.
Originally posted by cragwolf
Oops, I didn't see this last page, Marcus. OK, Labguy is right, and can ask the next question. Thanks.
QUESTION:
What type of object (objects) is (are) considered to be the source of X-Ray Bursters? Include at least a small description of the "process".
Originally posted by Labguy
Thanks.
QUESTION:
What type of object (objects) is (are) considered to be the source of X-Ray Bursters? Include at least a small description of the "process".
The object is a binary star pair, one partner being a neutron star. The other partner expands to the point where material from its outer layers can flow over and accrete onto the surface of the neutron star.
this process itself must release energy in various forms including xray, but this energy from the gradual buildup of material is not the "burst" people talk about.
The burst happens when a critical mass of hydrogen, in a thick enough layer, builds up on the surface of the neutron star. The hydrogen layer suddenly fuses into helium----a thermonuclear explosion involving the dense plasma "atmosphere" of the neutron star.
If I remember right, fusion at the sun's core makes
X-rays roughly 1000 eV and up because the temperature is 1000-plus eV. So that is what naked fusion yields (luckily for us, the energy gets softened as it percolates out to the sun's surface). So xray is to expected in this case, but with no surrounding material to buffer it.
What figure is used for the "peak absolute magnitude" of a Type Ia supernova?
Briefly sketch the process leading up to a supernova of that type.
Why do they all have about the same luminosity----allowing them to be used as a standard candle for estimating distance?
Originally posted by marcus
What figure is used for the "peak absolute magnitude" of a Type Ia supernova?
Briefly sketch the process leading up to a supernova of that type.
Why do they all have about the same luminosity----allowing them to be used as a standard candle for estimating distance?
Peak absolute magnitude for a type Ia supernava is at the top (brightest) point on the light curve after the explosion. By number, it is defined as MIa= -19.5 (+/-)0.2 Mag.
A Type Ia supernova is from a binary star system consisting of a white dwarf with a red giant companion, where the white dwarf accretes enough material to place it above the "Chandra's Limit". When this happens, the total mass of the white dwarf collapses and all material explodes in a huge nuclear fusion process and the Type Ia supernava occurs, leaving no "stellar remnant" behind. IOW, there is no Neutron Star or Black Hole remnant. About 99% of the energy is emitted as neutrinos, but that still leaves enough for the huge amount of visable light that can be detected; brighter than any other class of supernova.
The luminosity of Type Ia Supernovae is always nearly equivalent because the process of the explosion, and the chemical composition of the White Dwarf, must always meet nearly identical conditions. Most texts will say that the limit is the famous 1.44 Solar Masses (Sm), but in fact, the limit reached by the White Dwarf is somewhat less at 1.38 Sm. This is because the process (explosion) is the result of "Carbon deflagration" which must propogate throughout the star at a speed near, but not above, the speed of sound in that medium. Not all accreting white dwarfs, in fact very few, meet these conditions of size and composition, so Type Ia Supernovae are very rare, but very much a consistant "standard candle"
Pleased to say this told me more than I knew before about type Ia SN.
The figure of -19.5 was what I was looking for, plus the something about the process.
BTW since the sun is abs. magnitude 4.8 this must mean that the supernova at peak is 24-some steps of visual magnitude more luminous than the sun...
Anyway, your go, Labguy
Ok, an easy one.
QUESTION(S):
(1) About when was the redshift of galaxies discovered?
(2) Where (observatory) was this redshift discovered?
(3) Who discovered the redshift?
screwball
Apr21-03, 02:07 PM
Originally posted by Labguy
Ok, an easy one.
QUESTION(S):
(1) About when was the redshift of galaxies discovered?
(2) Where (observatory) was this redshift discovered?
(3) Who discovered the redshift?
In 1929 by E. Hubble. On the Mount Wilson Observatory? not sure...
Originally posted by screwball
In 1929 by E. Hubble. On the Mount Wilson Observatory? not sure...
(1) No.
(2) No.
(3) No.
Keep trying folks...[:)]
screwball
Apr21-03, 03:46 PM
Originally posted by Labguy
(1) No.
(2) No.
(3) No.
Keep trying folks...[:)]
Hubble was wrong? the date too? ok
Originally posted by screwball
Hubble was wrong? the date too? ok Yes, no Hubble, no 1929. No comment on the location...[:)]
Originally posted by Labguy
Ok, an easy one.
QUESTION(S):
(1) About when was the redshift of galaxies discovered?
(2) Where (observatory) was this redshift discovered?
(3) Who discovered the redshift?
I want to comment even though I have no guess about the answer. this is an interesting question because it does not mention the LINEAR RELATION between distance and redshift which Hubble discovered and is known for.
Hubble used cepheids (and luminosity of the whole galaxy) to estimate the distance. So he could plot an approximately linear relation. But someone else at some other observatory could have
previously discovered the redshift-----apparently according to Labguy they did----and simply not related it to distance in a pattern. Neat question because of the element of surprise.
In 1922 the astronomer Vesto Slipher, at Lowell Observatory, published his findings about the redshifts of galaxies.
He had found that they are mostly redshifted rather
than blueshifted.
The first redshift measurement he made was in 1912, I
believe, and he eventually compiled a list of 41 extragalactic
objects nearly all of which were redshifted.
Slipher (1875-1969) was born in Indiana. A history of Lowell Observatory (Flagstaff Arizona) says that Vesto's middle name was Melvin, and that his measurements of the 41 extragalactic redshifts was from 1912 to 1917.
He became director of the Observatory in 1916 and helped to find Pluto.
screwball
Apr21-03, 05:25 PM
ok i looked it up and found this in one of my old high school astronomy books:
"In 1913 V. M. Slipher at Lowell Observatory reported on the spectra of faint, nebulous objects in the sky. Their spectra seemed to be composed of a mixture of stellar spectra: some had Doppler shifts that suggested rotation, most had red shifts as if they were receding, and the faintest had the largest red shifts. Within two decades, astronomers concluded that the faint objects were galaxies similar to our own Milky Way and that the galaxies are indeed receding from us in a general expansion."
that was a good question very interesting to realize hubble wasnt the first to observe red shift but just the one to put it all together.
screwball
Apr21-03, 05:26 PM
lol u beat me to it
Originally posted by marcus
In 1922 the astronomer Vesto Slipher, at Lowell Observatory, published his findings about the redshifts of galaxies.
He had found that they are mostly redshifted rather
than blueshifted.
The first redshift measurement he made was in 1912, I
believe, and he eventually compiled a list of 41 extragalactic
objects nearly all of which were redshifted.
Slipher (1875-1969) was born in Indiana. A history of Lowell Observatory (Flagstaff Arizona) says that Vesto's middle name was Melvin, and that his measurements of the 41 extragalactic redshifts was from 1912 to 1917.
He became director of the Observatory in 1916 and helped to find Pluto. CORRECT. However, the first measurement made was of the Andromeda Galaxy which was blue shifted. Then came the redshift findings.
Your question.
A)What is the peak luminosity of a Type Ia supernova expressed as a wattage?
Don't count other ways the supernova might be releasing energy, like neutrinos, I guess that's implicit when one asks about the luminosity.
B)What is the luminosity of the sun expressed also expressed as a wattage?---the total output of light in all directions.
C)What is the approximate ratio of the two wattages? By what factor is a Ia SN brighter than the sun?
My handbook gives the sun's luminosity in ergs per second.
I'm asking for the answer in watts (one watt = 10^7 ergs per second) because that's more conventional although ergs still seem to be current in a good deal of astronomical writing.
Originally posted by marcus
A)What is the peak luminosity of a Type Ia supernova expressed as a wattage?
Don't count other ways the supernova might be releasing energy, like neutrinos, I guess that's implicit when one asks about the luminosity.
B)What is the luminosity of the sun expressed also expressed as a wattage?---the total output of light in all directions.
C)What is the approximate ratio of the two wattages? By what factor is a Ia SN brighter than the sun?
My handbook gives the sun's luminosity in ergs per second.
I'm asking for the answer in watts (one watt = 10^7 ergs per second) because that's more conventional although ergs still seem to be current in a good deal of astronomical writing.
Nobody else feels like answering this so I will. Labguy pointed out a few posts back that the peak Ia absolute magnitude is conventionally taken to be -19.5. And the sun's is 4.8. So the difference is 24.3. Multiplyby 2/5 and you get 9.72 (I will round off later).
C) The wattage ratio, SN versus sun, is 10^9.72 = 5.25 billion.
B) The sun's wattage (standard handbook figure) 3.8E26 watts.
A) Peak supernova luminosity is 5.25 billion times the sun's or
2E36 watts.
2 x 10^36 watts was what I was looking for in the answer.
Since I had to answer my own question the next askership is up for grabs.
The first person who can think of an appropriate-to-the-game type of question to which he/she knows the answer can ask it.
marcus said:
Since I had to answer my own question the next askership is up for grabs.
The first person who can think of an appropriate-to-the-game type of question to which he/she knows the answer can ask it.
How nice, thanks.
If the Sun's orbital velocity around the center of the Milky Way is 220 km s-1 [220 km/sec], roughly how many orbits has it completed since it formed, 4.5 billion years ago? Explain how you arrived at your estimate. You may assume that the Sun's distance from the Galactic center is 8000 pc [parsecs], and that it is following a circular orbit.
Too easy.
It's completed roughly 21 orbits in those 4.5 billion years.
The orbit length l = pi * 2 * 8000 pc.
Divide l (in km) by v = 6937920000 km/year to arrive at an orbit period of 222 million years.
Divide 4.5 billion by 222 million to arrive at roughly 21 orbits.
- Warren
Originally posted by chroot
Too easy.
It's completed roughly 21 orbits in those 4.5 billion years.
The orbit length l = pi * 2 * 8000 pc.
Divide l (in km) by v = 6937920000 km/year to arrive at an orbit period of 222 million years.
Divide 4.5 billion by 222 million to arrive at roughly 21 orbits.
- Warren
Sorry it took so long to get back....
But you are correct (I got between 20 and 21 orbits). Your turn.
Nicool003
May10-03, 08:53 PM
hey guys! I was planning on coming back after vacation but there are tons of exams and tests and quizes and projects..... well you can read all about it in general discussion[:D]
Anyways chroot it is your question.
schwarzchildradius
May14-03, 03:57 PM
Does chroot ask a question?
Nicool003
May14-03, 08:16 PM
He seems to not be coming to this thread anymore so how about you ask schwartz? I don't want to take it because I haven't been able to come to PF as much as you guys so I havent been to this thread in a while either.
Originally posted by Nicool003
He seems to not be coming to this thread anymore so how about you ask schwartz?....
Hear hear! Amen to that I say. Let schwarzschildradius ask a question and god save the empire!
schwarzchildradius
May15-03, 06:21 AM
OK, how do you measure the distance to the moon using lunar eclipse? What other observation of the moon must be made to find the distance to the sun?
schwarzchildradius
May15-03, 11:35 AM
BTW here's a freeware orrery and lunar eclipse predictor that works pretty well..
astropro orrery (http://www.geocities.com/schwartzchildradius999/astropro.zip)
Originally posted by schwarzchildradius
OK, how do you measure the distance to the moon using lunar eclipse? What other observation of the moon must be made to find the distance to the sun?
Hipparchus measure the distance to the moon using eclipse data and got the result that the moon is 30 earth diameters away which is quite close to the right answer.
However Hipparchus method is more complicated than I want to try to explain.
Earlier, Aristarchus (around 250 BC?) estimated the size of the moon compared with the earth by observing an eclipse of the moon.
He judged that the earth's shadow (roughly comparable in size to earth) was twice as big as the moon just by looking at the curve the shadow's edge made on the moon-----actually he should have guessed THREE times but he guessed twice the size.
Aristarchus was very back-of-envelope.
Knowing (in a rough sense) the size of the moon and the angle it made in the sky, Aristarchus could estimate the distance to the moon in earth diameters.
He then observed the lunar dichotomy, which gave him at least a rough lower bound on the distance to the sun as a multiple of the distance to the moon.
Noting that the sun was an order of magnitude farther than the moon and therefore huge compared with the earth, he surmised a heliocentric model.
This guy was from Samos, same birthplace as Pythagoras.
Well Schwarzschild there is at least part of an answer
god save all Samians and the empire of the mind
schwarzchildradius
May16-03, 01:24 PM
There you have it folks, righty-o.
The other observation was of the half moon - the moon is at 90 deg. from the sun when it is half full. You can use geometry to then (since you know the length of one side) find the distance and size of the sun. Your go.
Originally posted by schwarzchildradius
There you have it folks, righty-o.
The other observation was of the half moon - the moon is at 90 deg. from the sun when it is half full. You can use geometry to then (since you know the length of one side) find the distance and size of the sun. Your go.
Schwarzschildradius is a man of cultivation and discernment and has explained to us the method of lunar dichotomy.
In fact the sun is roughly 400 times the distance to the moon and therefore the angle (at halfmoon) is not 90 degrees but
90 degrees MINUS 1/400 RADIANS. Something like 89.8 or 89.9 degrees.
Aristarchus around 250BC measured it and got IIRC something like 89.5 degrees (he was clumsy at measuring angles, this Greek)
and so he judged that the sun was quite far away perhaps IIRC ten times as far as the moon!!!!! All the other greeks were quite astonished by this and could scarcely believe it. He should be awarded the Nobel prize postumously. Lunar dichotomy is a great phrase.
I like responding to questions a great deal better than I like asking them. Am tempted to appoint Schwarzschild my proxy and in invite him to pose a question for me. But perhaps it is better to keep it simple and stick by the rules.
QUESTION:
Describe "co-moving distance"
This is the astronomer's def of distance which works in the Hubble law v=H0 D
So if you ever use that law you need to understand what the comoving distance to an object is.
What is the comoving distance to an object (at this moment in time) which is currently receding from us at 30 thousand kilometers a second?
schwarzchildradius
May16-03, 07:59 PM
The answer to your particular question, is r=c/H, or something like 2e26 meters away. The answer to your general question, is that its the ratio of dx/dt (transverse velocity) to dθ/dt (proper motion), in other words, the distance measured by a "ruler" at specific time t.
Originally posted by schwarzchildradius
The answer to your particular question, is r=c/H, or something like 2e26 meters away. The answer to your general question, is that its the ratio of dx/dt (transverse velocity) to dθ/dt (proper motion), in other words, the distance measured by a "ruler" at specific time t.
Hello, your answer 2E26 meters is off by more than a factor of two so I'm afraid it will not do. But if you recalculate you will almost certainly get it.
Suggest using Hubble parameter of 71 km/s per Megaparsec
since that figure is not considered fairly reliable---the uncertainty has narrowed substantially about H0.
I should have been more explicit about the Hubble Law distance.
There is a particular definition of distance that works in the law
and other types of distance (angular size distance, luminosity distance, light traveltime distance) do not work.
In his javascript distance calculator
http://www.astro.ucla.edu/~wright/CosmoCalc.html
Wright refers to this definition of distance as the
"comoving radial distance, which goes into Hubble's law".
That expression contains a link to where the distance is
defined and described, using diagrams and some detailed
explanation. That "comoving distance which goes into Hubble's law" is what I am looking for.
But I won't insist on the general question---just get the
specific one right and you win the prize.
schwarzchildradius
May16-03, 10:51 PM
Fine
71 km/s/Mpc = 2.3e-18 s-1
R = 1.3e26 m
off by 21/2, much less than the uncertainty in H
or if you like,
3R = 3.9e26 m
Do you (or anyone) know the metric for Homogeneous Isotropic Cosmological Models?
Originally posted by schwarzschildradius
Fine
71 km/s/Mpc = 2.3e-18 s-1
R = 1.3e26 m....
Do you (or anyone) know the metric for Homogeneous Isotropic Cosmological Models?
Still not right. You have to solve the equation v = HD
for D, where H is exactly what you say it is and
v = 30,000 km per second.
You do realize, do you not, that I said 30 thousand km per second----a speed which is not equal to the speed of light?
When I said your answer is off by more than a factor of 2, I meant *more* than a factor of two, also more than a factor of 3.
Your answer is off by more than a factor of 4, hint hint.
Your answer, namely 1.3E26 meters. Is 13.7 billion light years!
Way too big!
schwarzchildradius
May17-03, 04:12 PM
Someone else will have to play with you.
Schwarzschildradius gives up.
Perhaps the problem is too simple for him. [g)]
The problem is, solve v = H0 D, for D
where v is 30 thousand km/sec, i.e. one tenth the speed of light,
and H0, the Hubble parameter, is (1/13.8 billion years).
schwarzchildradius
May17-03, 06:04 PM
Whatever. Its against the rules for me to keep guessing, and the value is in fact irrelevant because of uncertainty in H.
Originally posted by schwarzchildradius
Whatever. Its against the rules for me to keep guessing, and the value is in fact irrelevant because of uncertainty in H.
Both of us are nice people and we just experienced a power-of-ten problem. That's all and it happens quite regularly.
You kept saying 13.7 billion light years when you should have
been saying 1.37 billion light years. And you were off because
you thought that when I said "30 thousand km/sec" I was saying the speed of light.
When actually I was saying c/10.
I blame this on the metric system. If I had simply said "c/10"
then you would have gotten the right result immediately.
c is the natural unit to use in astronomy.
Let's have it be your turn, since you got the right answer except for the dratted order magnitude
General Comment:
Isn't this thread meant to be a "general" astronomy Q&A session instead of math excercises based on someone's assumptions (criteria) regarding some specific set of circumstances, sometimes theoretical and beyond "generally accepted"??
I don't mind the "Here is my theory" stuff, but there is another PF forum named "Theoretical Physics" where that is discussed with attempts to prove / disprove.
Carry on.
Originally posted by Labguy
General Comment:
Isn't this thread meant to be a "general" astronomy Q&A session instead of math excercises based on someone's assumptions (criteria) regarding some specific set of circumstances, sometimes theoretical and beyond "generally accepted"??
I don't mind the "Here is my theory" stuff, but there is another PF forum named "Theoretical Physics" where that is discussed with attempts to prove / disprove.
Carry on.
Mentor hat on...
Overall, this forum is intended for "generally accepted" (dare I say, "mainstream") astronomy. "Here's my theory" goes into the Theory Development forum. Just keep that in mind...no major problems here so far.
As far as the general Q&A vs. math-heavy Q&A. I suppose it's either (1) up to Nicool002 (who started this thread) or (2) free market...if y'all want math, then knock yourselves out...if not, then this topic will evolve or die accordingly. Whatever. As long as it's good clean astronomy, then I'm happy. Certainly, many PF members have mastered general astronomy Q&A and are ready for more. But many PF members are new to astronomy.
Might I recommend 2 topics...one for beginners and one for advanced folks?
Originally posted by Phobos
Mentor hat on...
Might I recommend 2 topics...one for beginners and one for advanced folks?
Hello Phobos, why don't you ask a question at kick off another round?
The last question was about the v=H0 D (Hubble law).
I asked, if v is a tenth of the speed of light (v = 30 thousand km/s)
then solve for D, the distance.
It did not get answered so maybe I can just invite someone to pose a question.
My thought is that the last question was about as basic, mainstream, and beginner-level as you can get in cosmology.
The Hubble parameter has been the main thing to measure and to use in estimating distance for several generations, the law
is fundamental to the field. So I wonder how the issue of "mainstream" arose. Is it too much math to involve a basic
equation like v=H0 D?
Afterthought. I guess the other most basic thing
in cosmology is the density of the universe (rho)
as it compares to the critical density (rhocrit) required for flatness.
For some decades cosmologists have been
talking about how "the fate of the universe" depends
on the ratio
rho/rhocrit-------whether it is less than one
or equal to one (flatness) or greater than one.
Would it by any chance be considered not mainstream to
ask a question about the critical density on this thread?
I don't recall that I have, as yet, but was just now wondering,
since it seems so central to the field that I cant really
picture what conventional cosmology would be like without it.
schwarzchildradius
May18-03, 02:03 PM
Just lets get this thing moving please ;)
Originally posted by schwarzchildradius
Just lets get this thing moving please ;)
OK, since no one else volunteers a question, I will ask:
What famous astronomer's mother was almost burnt as a witch?
(it didnt actually come to lighting the fire but she was tried
and came close to being burned)
god save the Hapsburgs and the holy roman empire!
marcus asked:
What famous astronomer's mother was almost burnt as a witch?
Johannes Kepler
Originally posted by J-Man
Johannes Kepler
Attaboy! Your question
What is a "reflection nebula"?
-Give a definition.
-Give an example.
-Extra points for brown-nosing.
Originally posted by J-Man
What is a "reflection nebula"?
-Give a definition.
-Give an example.
-Extra points for brown-nosing. A reflection nebula is an accumulation of dust and gas where the light, usually blue in color, is reflected off the dust and gas from one or more embedded stars. M20, the Trifid Nebula, is an example where it is seen as blue. It also happens to be an emmission nebula as well, with light being emmitted from ionized Hydrogen, usually seen as red.
I don't know any brown-nose points.
Labguy answered:
A reflection nebula is an accumulation of dust and gas where the light, usually blue in color, is reflected off the dust and gas from one or more embedded stars. M20, the Trifid Nebula, is an example where it is seen as blue. It also happens to be an emmission nebula as well, with light being emmitted from ionized Hydrogen, usually seen as red.
Very good.
I don't know any brown-nose points.
That's okay, it's still your turn.
Originally posted by J-Man
Very good.
That's okay, it's still your turn. Ok, here is an EASY one. Two-parter.
(1) What event, important to observational astronomy, will happen on August 27, 2003? (**Later edit**: If we need very specific, you can use universal time OR EDT, since we are not always on the same day worldwide)
(2) When was this event last seen, to the same or greater degree, by any astronomer using any telescope or optical device to enhance the unaided eye??
Note: I expect this one to be answered instantly by the first PF member to read it....(?)
Ivan Seeking
May19-03, 05:54 PM
Originally posted by Labguy
Ok, here is an EASY one. Two-parter.
(1) What event, important to observational astronomy, will happen on August 27, 2003? (**Later edit**: If we need very specific, you can use universal time OR EDT, since we are not always on the same day worldwide)
(2) When was this event last seen, to the same or greater degree, by any astronomer using any telescope or optical device to enhance the unaided eye??
Note: I expect this one to be answered instantly by the first PF member to read it....(?)
The closest approach of Mars in 73,000 years.
Never.
Originally posted by Ivan Seeking
The closest approach of Mars in 73,000 years.
Never. CORRECT.
Your question.
Ivan Seeking
May19-03, 06:56 PM
Originally posted by Labguy
CORRECT.
Your question.
You tricky devil you. [:D]
Ok. Now if you read my other posts don't worry. I'm not going pseudo or philosophical on you.
What are the hypothetical circumstances, if we can show [mathematically] that one can control his position in time but not in space?
Edit:The language used may be a little optimistic:What are the hypothetical circumstances, if we can show [mathematically] that one can affect his position in time but not in space?
Ivan Seeking
May20-03, 02:24 AM
I quoted since I wasn't sure if edits alert readers of a change.
Originally posted by Ivan Seeking
Edit:The language used may be a little optimistic:
What are the hypothetical circumstances, if we can show [mathematically] that one can affect his position in time but not in space?
schwarzchildradius
May20-03, 02:53 AM
So, are you asking the circumstances under which a particle can move in space but not in time? If v=c, this is the case I think. (T=infinity)
Ivan Seeking
May20-03, 04:12 AM
Originally posted by schwarzchildradius
So, are you asking the circumstances under which a particle can move in space but not in time? If v=c, this is the case I think. (T=infinity)
But that would violate Relativity. [?]
I don't think that qualifies unless you mean some really abstract circumstance that "actually" occurs, or could occur. schwarzchildradius
schwarzchildradius
May20-03, 04:56 AM
only if you consider light abstract, which I guess it kind of is. if you could "ride" light, of course, you wouldn't be moving through time.
Ivan Seeking
May20-03, 05:52 AM
Originally posted by schwarzchildradius
only if you consider light abstract, which I guess it kind of is. if you could "ride" light, of course, you wouldn't be moving through time.
"...one can affect his position in time but not in space"
No voodoo needed. But it wouldn't be pretty! schwarzchildradius
Originally posted by Ivan Seeking
The language used may be a little optimistic:
What are the hypothetical circumstances, if we can show [mathematically] that one can affect his position in time but not in space?
I'm not sure I understand the question but...
f(x) = 0 * x = 0 for all x
f(y) = 0 * y = 0 for all y
f(z) = 0 * z = 0 for all z
f(t) = h(t)
Where h(t) is some function of time independent of the 3 spatial "variables".
Basically, position is invariant, or the coordinate system used moves with the hypothetical person.
Ivan Seeking
May20-03, 06:29 PM
Originally posted by J-Man
I'm not sure I understand the question but...
f(x) = 0 * x = 0 for all x
f(y) = 0 * y = 0 for all y
f(z) = 0 * z = 0 for all z
f(t) = h(t)
Where h(t) is some function of time independent of the 3 spatial "variables".
Basically, position is invariant, or the coordinate system used moves with the hypothetical person.
Clever, but I don't mean to be that tricky. I am trying to ask a hard question not a tricky one; but the thing that makes it hard is tricky.
Hint: Where is our schwarzchildradius?
as soon as one gets inside the event horizon of a black hole
"one can no more avoid the singularity than one can avoid
next Tuesday"
the direction towards the center becomes the time direction.
perhaps this responds appropriately to the question?
Originally posted by marcus
the direction towards the center becomes the time direction.
I can't say I agree with the either this message or the language itself.
It is true that all worldlines inside a black hole at some point intersect the singularity, but it is not true that all worldlines are geodesics. You can still have fuel in your rocket when you cross the horizon, and you can still zoom about inside the horizon in non-inertial motion.
The bottom line is that since two observers can cross the horizon at the same place and follow two different trajectories once inside, there's no way you could even qualitatively think of "the direction towards the center" as being the "time direction."
I think this question is bordering on the philosophical, since it eventually comes down to semantics.
- Warren
schwarzchildradius
May20-03, 06:52 PM
Ah, could it be that Ivan is seeking the conditions on the inside of a black hole event horizon?
Ivan Seeking
May20-03, 06:57 PM
Originally posted by marcus
as soon as one gets inside the event horizon of a black hole
"one can no more avoid the singularity than one can avoid
next Tuesday"
the direction towards the center becomes the time direction.
perhaps this responds appropriately to the question?
I was wondering how many times I could say schwarzchildradius and get away with it. CORRECT!
You're up marcus.
Ivan Seeking
May20-03, 07:09 PM
Originally posted by chroot
You can still have fuel in your rocket when you cross the horizon, and you can still zoom about inside the horizon in non-inertial motion.
The space coordinates exchange roles with time and become imaginary. According to my notes from a senior level GR class - taught by a very good professor - this is the proper interpretation. If this is incorrect or outdated, I am not in a position to argue this point with much proficiency. [:)]
Originally posted by Ivan Seeking
You're up marcus.
The first star to which the distance was measured is_______
(I dont mean the sun I mean a "real" star.)
Who measured the distance? In what year?
How far away is this star?
For extra points, what is the parallax angle by which
the first stellar distance was determined?
Footnote---not part of question---Christian Huygens estimated the distance to the star Sirius in a clever way long before parallax measurement was possible. How he did it is almost funny. But I am not considering that as the first real measurement, so it would
just be distracting to describe his method.
schwarzchildradius
May20-03, 08:14 PM
sun
Originally posted by marcus
The first star to which the distance was measured is_______
(I dont mean the sun I mean a "real" star.)
Who measured the distance? In what year?
How far away is this star?
For extra points, what is the parallax angle by which
the first stellar distance was determined?
Footnote---not part of question---Christian Huygens estimated the distance to the star Sirius in a clever way long before parallax measurement was possible. How he did it is almost funny. But I am not considering that as the first real measurement, so it would
just be distracting to describe his method.
(A) 61 Cygni, a double of red dwarfs.
(B) F.W. Bessel in 1838.
(C) 10.3 LY.
(D) 0.29 arc seconds.
(E) Measured by parallax.
Originally posted by Labguy
(A) 61 Cygni, a double of red dwarfs.
(B) F.W. Bessel in 1838.
(C) 10.3 LY.
(D) 0.29 arc seconds.
(E) Measured by parallax.
Right on! Labguy it is your go.
This is the kind of conversation!
When people know about Bessel, I mean. 1838 a time of giants.
Darwin was just getting Evolution theory written down that year. Faraday visualizing lines of force.
God bless Bessel, put the first yardstick to the stars.
BTW Britannica say he measured 0.31 arc seconds but
so close and dont know who is right you or Britannica,
so call it 0.29.
Bravo. your turn.
schwarzchildradius
May21-03, 04:20 AM
sun was discovered before 1838. Aristarchus, wasn't it? 300 BC?
Originally posted by schwarzchildradius
sun was discovered before 1838. Aristarchus, wasn't it? 300 BC?
the sun was ruled out in the question---we just had a question on the thread relating to Aristarchus----so when I asked the question I said the sun was not the answer I was looking for.
better luck next time [:)]
Labguy's turn
Ok, another easy one: An easy internet look-up.
Re-list, in order of size (Stellar Masses), from biggest to smallest, the following stellar classes with approximate masses and lifetimes. This is based on a total lifetime of our Sun as 10 billion years. Not restricted to just main sequence.
Stellar Class:
A:
F:
B:
MO:
BO:
K0:
G2 (Our Sun)
O3:
M7-8
Just type a list (in order) with class, ~Mass and ~Lifetime in years (Use millions, billions or trillions).
EDIT: I removed "size" typed in by mistake. The first criteria is mass.
I'm not sure this question is well posed.
The order of surface temperatures is OBAFGKM in order of highest to lowest.
However, the spectral type alone is not enough to determine mass. You would also need luminosity. There's no such thing as an "average mass" for a given spectral type.
Lifetime is, of course, tied to mass.
- Warren
Originally posted by chroot
I'm not sure this question is well posed.
The order of surface temperatures is OBAFGKM in order of highest to lowest.
However, the spectral type alone is not enough to determine mass. You would also need luminosity. There's no such thing as an "average mass" for a given spectral type.
Lifetime is, of course, tied to mass.
- Warren The surface temperatures can only be seen from the spectra of the photosphere. While on the main sequence, the larger mass leads to higher temperatures, so there is a direct correlation between temperature and mass until the main sequence is left behind. There is a definite "average" mass for an M type star vs. a star formed as a B type. Also there is a huge difference in "average lifetimes", as you say dependent on mass. Part of a giveaway is that there is no way that an M type star (depending on whether M0 to M9) will have a mass below 0.08 Sm or much above 0.1 Sm. This type of star cannot create a high surface temperature simply because there is not enough mass to generate the fusion processes that would lead to high temperatures.
The original question stands, but nobody has to be very exacting, just good estimates in correct order.
Originally posted by Labguy
Ok, another easy one: An easy internet look-up.
Re-list, in order of size (Stellar Masses), from biggest to smallest, the following stellar classes with approximate masses and lifetimes. This is based on a total lifetime of our Sun as 10 billion years. Not restricted to just main sequence.
Stellar Class:
A:
F:
B:
MO:
BO:
K0:
G2 (Our Sun)
O3:
M7-8
Just type a list (in order) with class, ~Mass and ~Lifetime in years (Use millions, billions or trillions).
EDIT: I removed "size" typed in by mistake. The first criteria is mass.
http://nrumiano.free.fr/Estars/classes.html
You originally wanted the radius, mass, and lifetime of the different spectral classes listed. I found a table on the web that gives figures for these and other characteristics:
spectral type, luminosity, mass, radius, lifetime, surface temperature, and the relative abundance in our galaxy.
No table can be perfect. There is variation which a simple table cannot adequately cover. But we can try to do the best we can by way of giving representative benchmarks of each range.
I dont see any harm in including the radius (that you edited out).
My problem is that I dont know how to copy a table into PF.
It will take time
From the site I gave the link to:
Spectral class
Mass (solar mass)
Radius (solar radius)
Luminosity
Surface temperature (degrees K)
Life time (million of years)
W ---- >40---- 20 ---1.000.000 --- 50.000 ----- <1
O5 ----- 32---- 18 -----600.000 ---- 40.000 ----- 1
B0 ----- 16 -----7.4 ----- 16.000 ---- 28.000 ----- 10
B5 -----6.5---- 3.8 -----600 ------- 15.500 -----100
A0 ----- 3.2----- 2.5----- 60 -------- 9.900 ------500
A5 ----- 2.1 -----1.7 ----20 --------- 8.500 ----- 1.000
F0 ----- 1.75 ---- 1.4 ----- 6 ---------7.400------ 2.000
F5 -----1.25 ------1.2 -----3 -----------6.600 ----- 4.000
G0 -----1.06 ----- 1.1 ----- 1.3 --------6.000 ----- 10.000
G2 Sun -- 1 ----- 1------- 1---------- 5.800 -------- 12.000
G5 ----- 0.92 ---- 0.9 -----0.8 ------- 5.500 --------15.000
K0 ----- 0.80 ---- 0.8 ------ 0.4 ------ 4.900 ------- 20.000
K5 ----- 0.69 ---- 0.7 ------ 0.1 ------ 4.100 ------- 30.000
M0 ----- 0.48 ----0.6 ------ 0.02 ---- 3.500 ------75.000
M5 ----- 0.20 ---- 0.3 ----- 0.001 ---- 2.800 ----- 200.000
Very good summary table. It is not exactly the one I had which showed a few more specific examples, but is close enough to be called correct.
I will "cut and paste" part of where I was looking, but don't want to give the URL now because it has lots of neat stuff I might want to hit you with later....[:)]
Here is part of my question source:
Type O3: 120 Sm 63 thousand years.
Type O: 25 Sm 3.2 million years.
Type B: 10 Sm 32 million years.
Type A: 2.5 Sm 1.0 billion years.
Type F: 1.3 Sm 5.2 billion years.
Type G2:(Sun) 1.0 Sm 10 billion years.
Type K0: 0.7 Sm 24 billion years.
Type M0: 0.5 Sm 57 billion years.
Type M7-8 <0.1 Sm 3.2 trillion years.
Note the mere 63 thousand years for an O3! Also, other readers should note from your site that class and mass are related. The exceptions come when a star leaves the main sequence. For example, a large, massive red giant swells to where the temperature is very low (red spectrum), while a tiny white dwarf has little mass but is hotter than hell; high temperature.
Your question next.
Originally posted by Labguy
Your question next.
In cosmology there is the idea of being at rest with respect to the CMB or the "Hubble flow"----which is the expansion of space.
So there is an absolute rest frame (unlike in special relativity) which corresponds to being at rest with respect to the expansion of space.
The solar system has an absolute velocity with respect to the CMB.
In what direction is it?
What is the solar system's speed in kilometers per second?
Labguy,
LOL.. your question specifically stated: "Not restricted to just main sequence."
As stated, it cannot be answered.
- Warren
Originally posted by marcus
The solar system has an absolute velocity with respect to the CMB.
In what direction is it?
What is the solar system's speed in kilometers per second?
600 km/s roughly in the Leo/Virgo direction.
- Warren
Originally posted by chroot
600 km/s roughly in the Leo/Virgo direction.
- Warren
Virgo and Leo are not in the same direction.
The coordinates of the dipole have been published
but I'd be happy if you would just say in which constellation!
If you cant say in which constellation then pleas give coordinates.
Anybody else have a guess about the km/s speed?
Originally posted by marcus
Virgo and Leo are not in the same direction.
Virgo and Leo share a border.
627 +/- 22 km/s in the direction of (l,b) = (276 +/- 3, 33 +/- 3).
Source: WMAP http://www.arxiv.org/abs/astro-ph/0210165
This seems to put it (just barely) inside the boundaries of the constallation Crater, which shares borders with both Virgo and Leo.
Sure seems I was pretty close to being right the first time, Marcus.
- Warren
Hi chroot, I am very familiar with this "Crater, 600 km/s" velocity vector because I calculated it for myself back in the early 90s as an answer to another question.
I do not doubt that the paper you reference has that vector in it as one of the results. thanks for the link, which will be useful!
However I suspect you may have misread the paper you cited and not have answered the question I asked. So I will have to
double check and get back to you.
Bravo for getting coordinates!
Originally posted by chroot
Virgo and Leo share a border.
627 +/- 22 km/s in the direction of (l,b) = (276 +/- 3, 33 +/- 3).
Source: WMAP http://www.arxiv.org/abs/astro-ph/0210165
This seems to put it (just barely) inside the boundaries of the constallation Crater, which shares borders with both Virgo and Leo.
Sure seems I was pretty close to being right the first time, Marcus.
- Warren
chroot:LOL.. your question specifically stated: "Not restricted to just main sequence." Yes it did, but that followed the statement regarding sun's age of 10 billion years. I should have been more specific as to all the requirements.
Yes I was right.
I printed out the paper you referenced
You made a mistaken interpretation of the paper
Your answer is off by on the order of 100 kilometers/second
and, I estimate, several tens of degrees
The hotspot is not in the constellation Crater
(although the coordinates you gave are I believe in Crater, as you say)
Originally posted by Labguy
chroot: Yes it did, but that followed the statement regarding sun's age of 10 billion years. I should have been more specific as to all the requirements.
Hey Labguy, do you happen to know which direction (which constellation) the microwave hotspot is in? chroot is way off as to the direction and also the solarsystem's speed relative to the CMB.
If you don't, no problem, but I kind of thought you might.
Originally posted by marcus
Hey Labguy, do you happen to know which direction (which constellation) the microwave hotspot is in? chroot is way off as to the direction and also the solarsystem's speed relative to the CMB.
If you don't, no problem, but I kind of thought you might. Maybe..[:)], but it should be someone else's turn.
All I can think of is that the solar system's velocity is different from the local group's. This makes sense, but who the hell cares about the difference?
The paper I cited is from WMAP's data, which is, of course, a satellite with approximately the same orbit as the Earth. The WMAP figure ought to be valid for the solar system. Or did the WMAP team modify this figure specifically to disclude the solar system's motion so as to get the local group's motion?
This all seems rather semantic.
- Warren
I've also seen figures of 390 km/s in the direction of Leo.
It appears, indeed, the 600 km/sec @ Crater figure includes corrections to the observed dipole to include the motions of the Sun and Milky Way.
edit: I cannot find any primary source material to back this conclusion up, however -- only anecdotal second sources.
- Warren
Originally posted by marcus
Hey Labguy, do you happen to know which direction (which constellation) the microwave hotspot is in? chroot is way off as to the direction and also the solarsystem's speed relative to the CMB.
If you don't, no problem, but I kind of thought you might. I just checked two sources, and both said we are moving, with respect to the CMBR, in the direction of Leo. One source said 400 Km/sec. and the other said 600 Km/sec.
It is your call, but it seems that chroot was a correct as most sources indicate. This motion is CMBR motion, and is for our entire "local group" of galaxies.
Originally posted by Labguy
I
It is your call, but it seems that chroot was a correct as most sources indicate. This motion is CMBR motion, and is for our entire "local group" of galaxies.
I did not ask about the motion of the Local Group but about the motion of the solar system relative to CMB.
That is what determines where the hotspot is in the sky.
What I want to know is where is the CMB hotspot, and how hot it is.
(earths orbiting has a small effect because it is a small speed
compared with the solarsystem's)
Somebody please give an unequivocal answer.
So give me some unambiguous definite guess and if you are anywhere close you win.
Chroots coordinates are more than 30 degrees of angle off the mark. His 600 km/sec is more than 200 km/sec off from speed.
I dont think being 50 percent off the mark is good
enough in this league.
Like I said, 370-390 km/s in the direction of Leo.
- Warren
Originally posted by chroot
Like I said, 370-390 km/s in the direction of Leo.
- Warren
roughly correct, the people who actually did the research, publishing
in Astrophysical Journal higher accuracy and confidence
intervals and all that.
http://arxiv.org/PS_cache/astro-ph/pdf/9601/9601151.pdf
369.0 +/- 2.5 km/s
galactic coords (264.31 degrees, +48.05 degrees)
celestial coords (11 h 11' 57", -7.22 degrees)
temperature delta 3.358 millikelvin
Indeed it is in Leo!!!! Not Crater, not Virgo, not Corvus, not Hydra, not Centaurus or any of the other neighboring patches of sky.
the thing is precisely known and continues to be confirmed as more data comes in. What I am quoting are COBE results
written up for 1996 publication by principle members of the COBE team.
Good work.
Or good guesswork [;)].
Your turn!
Here's one that will require a little research, or a little googling.
The Great Ship Argo Navis was a large constellation recognized by the ancient Greeks. It represented the ship sailed by Jason and the Argonauts. (a) In their mythology, what was Jason seeking?
Argo Navis was eventually broken up into six constellations: Puppis, Carina, Vela, Pyxis, Volans, and Columba. (b) Who was responsible for breaking up the great ship Ago Navis, and when was the work first published?
The International Astronomical Union has the duty of being the ultimate authority on the boundaries and names of the constellations. (c) In which meeting did the IAU finally agree up on the set of 88 constellations which included the six pieces of Argo Navis?
However, even after the constellations' names and stars were fixed, their boundaries were not... (d) when was the final set of official boundaries published, and by whom?
- Warren
Cripes chroot!
You have nailed me (us) good on this one. You must have a Voodoo/Astrology book somewhere that sold a total of two copies; one to you and one for the author's mother..[g)] I give up, but someone out there must be able to find it somewhere. I think that you're getting some payback for my question where I was not specific enough. I promise not to do that again if I ever get another question posted..[?]
Originally posted by Labguy
Cripes chroot!
You have nailed me (us) good on this one. You must have a Voodoo/Astrology book somewhere that sold a total of two copies; one to you and one for the author's mother..[g)] I give up, but someone out there must be able to find it somewhere. I think that you're getting some payback for my question where I was not specific enough. I promise not to do that again if I ever get another question posted..[?]
Hold on Labguy, I just looked at the question this morning.
It may not be as hard as you think----just not too interesting and a bit pedantic. Give me a chance to consult sources.
I thought your idea was to give other people a chance--which was why, even tho you knew where the CMB hotspot was and chroot's guess was 20 degrees off (and 200 kilometers a second) you held back from answering. I was going to ignore this one.
But I will give it a try.
We know that de Lacaille broke up and renamed Argo in 1751-2 and that his Planisphere was published in 1756 (the year Mozart was born I believe). It should not be too hard to learn a little something about when the boundaries were drawn---although being interested in such trivia seems silly enough!
Chroot's question has four parts
(a) In their mythology, what was Jason seeking?
Golden Fleece
(b) Who was responsible for breaking up the great ship Ago Navis, and when was the work first published?
Nicolas de Lacaille, published Planisphere in 1756
(c) In which meeting did the IAU finally agree up on the set of 88 constellations which included the six pieces of Argo Navis?
(d) when was the final set of official boundaries published, and by whom?
I will do a bit of googling and see if answers to c and d are easy to come up with.
Official boundaries, voted on by the IAU, were published
in or around 1930
"The following list of constellation names and abbreviations is in accordance with the resolutions of the International Astronomical Union (Trans. IAU, 1, 158; 4, 221; 9, 66 and 77). The boundaries of the constellations are listed by E. Delporte, on behalf of the IAU, in, Delimitation scientifique des constellations (tables et cartes), Cambridge University Press, 1930; they lie along the meridians of right ascension and paralleIs of declination for the mean equator and equinox of 1875"
Don't you like the French title of the 1930 book by Delporte:
"Delimitation scientifique des constellations "
Scientific boundaries of the constellations. Scientific my grandmother---a bunch of arbitrary human conventions. Well
anyway that leaves part (c) of when the IAU or some official body of that ilk decided on ther 88 names
Originally posted by marcus
Chroot's question has four parts
(a) In their mythology, what was Jason seeking?
Golden Fleece
(b) Who was responsible for breaking up the great ship Ago Navis, and when was the work first published?
Nicolas de Lacaille, published Planisphere in 1756
(c) In which meeting did the IAU finally agree up on the set of 88 constellations which included the six pieces of Argo Navis?
(d) when was the final set of official boundaries published, and by whom?
I will do a bit of googling and see if answers to c and d are easy to come up with.
The names of the constels. were decided on in 1922 in Rome
at the First General Assembly of the IAU
" First General Assembly, held at Rome, May 2nd to May 10th, 1922.
Transactions of the International Astronomical Union. Volume 1."
Easy to get on google, just as our good chroot indicated.
Congrats chroot, good question
marcus, you win! Your answers were scattered around, but they are all correct. I've seen various dates on when Lacaille first published his work -- I think 1754 is more of a consensus, but hell, 1756 is close enough.
All of these answers are found in one page: http://users.macunlimited.net/ianrid/startales/startales1c
If you google hard enough, you'll find it! Finding alternate sources to verify this information was, of course, more challenging.
Your turn.
(And yes, I know my question was a bit boring and pedantic, as are all historical questions -- but at the same time, they end up having some allure -- you get to learn how the world came be seen as it is seen today.)
- Warren
Originally posted by chroot
Your turn.
(historical questions ....end up having some allure -- you get to learn how the world came be seen as it is seen today.)
(I agree about historical questions being interesting. For a change here's a non-historical one.)
A black hole was recently observed at the center of the Milky Way galaxy----by tracking a star in tight orbit. The team that made the observations published an estimate of its mass.
(a) What is the estimated mass of the hole?
Assuming this is an ordinary (uncharged, non-rotating) hole...
(b) would the acceleration of gravity at its event horizon be greater than, equal to, or less than normal sealevel gravity on earth? If not equal to a standard gee, what would it be?
For extra credit:
(c) what mass black hole of that type would have surface gravity
equal to standard sealevel earth gravity.
Hey marcus,
It seems you're a fan of the Astronomy Picture of the Day (APOD) site as well. ;)
(a) The black hole at the center of our galaxy is said to have a mass of 2 million solar masses.
(b) The Schwarzschild radius of a 2 million solar mass hole is rs = 2GM/c2, or 5.9 billion meters. The gravitational acceleration is ag = GM/r2, or 7.6 x 106 m/s2, or about 4 orders of magnitude larger than the gravitational acceleration on the surface of the Earth. (Sorry if I offended you by calculation in metric units -- I'm still working on wrapping my brain around the natural units ;))
(c) Simple algebra results in the expression M = c4/(4 G g). Solving for g = 9.8 m/s2, we arrive at M = 1.55 x 1012 solar masses. As a check, this makes sense -- larger black holes have smaller gravitational accelerations.
- Warren
http://arxiv.org/abs/astro-ph/0210426
It has a link to the PDF file of the original journal article.
http://arxiv.org/PS_cache/astro-ph/pdf/0210/0210426.pdf
The journal article says 3.7 million solar masses within the pericenter distance of 124 AU. However it estimates that 2.6 of this is the black hole itself and the remaining 1.1 is a visible cluster of stars and stuff around the black hole.
Originally posted by chroot
(a) The black hole at the center of our galaxy is said to have a mass of 2 million solar masses.
This is close enough to 2.6 million.
Originally posted by chroot
(c) Simple algebra results in the expression M = c4/(4 G g). Solving for g = 9.8 m/s2, we arrive at M = 1.55 x 1012 solar masses. As a check, this makes sense -- larger black holes have smaller gravitational accelerations.
This is entirely right. I got 1.54 trillion solar masses and if I redid my calculation would probably get 1.55 agreeing smack on.
Originally posted by chroot
(b) The Schwarzschild radius of a 2 million solar mass hole is rs = 2GM/c2, or 5.9 billion meters. The gravitational acceleration is ag = GM/r2, or 7.6 x 106 m/s2, or about 4 orders of magnitude larger than the gravitational acceleration on the surface of the Earth.
I don't follow why it is 4 orders of magnitude larger than earth's surface gee. Can you clarify or correct by editing your post? May simply be a typo.
However, mostly right including the last.
Your go!
Yes that picture of the day thing is neat, though Im not
a regular visitor there and always glad to hear of special things
that I may have missed.
Er, 6 orders, not 4 orders. :-x It was a typo.
Let me think, I have to come up with a good question... I like questions involving recent research... *thinks*
How about this:
Can anyone explain (in English is fine) why the vacuum energy density (i.e. cosmological constant, dark energy, quintessence, etc.) is described as being a negative pressure?
- Warren
Originally posted by chroot
Can anyone explain (in English is fine) why the vacuum energy density (i.e. cosmological constant, dark energy, quintessence, etc.) is described as being a negative pressure?
Before I say anything in English think of a cylinder with a piston sliding in it.
The cylinder is a square meter area base, and 2 meters high.
the piston is halfway up
so the volume defined by the piston is one cubic meter.
the cylinder has one cubic meter of *vacuum* in it so it has a certain quantity of dark energy---however much is said to be in a cubic meter.
Imagine we can find a special region where there ISNT any dark energy (we need to be surrounded by really empty space to do this thought experiment). Let's take our cylinder over there.
Now we are going to create some energy just by pulling the piston slowly out one more meter. So now the *vaccuum* inside is TWO cubic meters and has twice as much dark energy----because the special thing about it is the constant presence of so and so much per unit volume----expanding the volume makes there be more energy.
SO WE MUST HAVE DONE WORK in drawing out the piston that extra meter.
And that is what force is-----work per unit distance.
It took a force to draw the piston a meter out.
There is nothing (truly nothing---not even *vacuum*) around the
piston. So the only explanation for the force is that there is negative pressure.
Positive pressure in there would have pushed the piston out by itself, but we had to fight against the negative pressure to expand the volume. It is an application of a local conservation of energy law---if you like.
I've answered the question---why is there negative pressure.
It is because of a constant energy density----every cubic meter of *vacuum* has the same amount of dark energy. that is what makes it work.
But here is something extra. One can say right away how much the negative pressure is.
energy density and pressure are formally the same type of physical quantity.
one says "joule per cubic meter" but joule is newton meter
(just like footpound is foot pound----pound push for foot distance).
So newton meter per cubic meter---energy density--- turns into newton per square meter----pressure. It is just a cancelation.
And if you think about the piston and cylinder a bit you will see that the pressure (except for the minus sign) EQUALS the energy density.
If the energy density were one joule per cubic meter then by pulling the piston out one meter you would create one extra joule of energy and the only way that could be is if the force exerted were one newton ( a joule is a newtonmeter by def).
So if the rho is 1 joule per cubic meter, then p = - ! newton per square meter.
It is the same with realistic numbers---one can say what the dark rho is in joules per volume and the pressure (except for minus sign) is that same number of newtons per area.
schwarzchildradius
May25-03, 04:33 AM
Nice. Thanks for that.
Originally posted by schwarzchildradius
Nice. Thanks for that.
glad you liked it! [:)]
Nicool003
May26-03, 09:04 AM
So your up Marcus-----I think [:)] ----
Originally posted by Nicool003
So your up Marcus-----I think [:)] ----
Hello Nicool,
In fact that explanation with the cylinder has been
used in several papers I've seen----seems to be a favorite
explanation of the negative pressure of dark energy and what
persuaded Einstein to put a Lambda into the equation.
I added metric units to the cylinder to make it seem less abstract
but otherwise just passed it along unchanged.
So it is my turn to ask, and chroot has just calculated that a 1.55 trillion solar mass black hole would have surface gravity of one standard earth gee.
This is a compelling image (a "chroot mass" black hole [;)]).
We ought to have some Astronomy questions about it.
In round numbers it is 1.5 trillion solar mass and the
GM/c^2 is therefore 1.5 trillion miles
and the Schw. radius is therefore 3.0 trillion miles.
An explorer named Sir Edmond McChroot wishes to explore this black hole so he has his Sherpa guides lower him on a 3 trillion mile long cable down from the ship which is dynamically positioned 6 trillion miles from the center of the hole.
The Sherpas lower McChroot to within a few feet of the surface but are careful not to let him touch the surface (the event horizon) as this would cause them the loss of their employer.
Now we have an observer in an essentially one gee field suspended over an apparently flat surface extending as far as the eye can see. What questions can we ask? What are some simple questions concerning McChroot's impressions of his surroundings?
To protect the explorer from light coming from above---blue shifted to quite high energy----we place him a gondola with a protective canopy of special material that keeps out gamma rays and suchlike annoyance. The physical danger to McChroot comes not from the black hole but from light and stuff falling down from above.
I just saw the Nicool post that it was my turn. I want to take a little time to think of a simple question about what an observer would see in this one gee field. The question has to be really simple and straightforward. So I will think a bit over coffee and get back to this.
Originally posted by Nicool003
So your up Marcus-----I think [:)] ----
An uncharged non-rotating black hole is of such a size that it has g (acceleration due to gravity) at its event horizon equal to one earth gee.
Consider a ray of light projected horizontally at an altitude of
4.5 trillion miles from the center of the hole.
What is the radius of curvature of this ray of light?
For extra credit: estimate the radius of curvature of an
initially horizontal ray of light near the event horizon.
Originally posted by marcus
An uncharged non-rotating black hole is of such a size that it has g (acceleration due to gravity) at its event horizon equal to one earth gee.
Consider a ray of light projected horizontally at an altitude of
4.5 trillion miles from the center of the hole.
What is the radius of curvature of this ray of light?
For extra credit: estimate the radius of curvature of an
initially horizontal ray of light near the event horizon.
C'mon! Is this so hard? someone should have replied. Maybe very large black holes boggle the mind? Here is a scaled down version:
An uncharged non-rotating black hole has radius 3 miles.
Consider a ray of light projected horizontally at an altitude of
4.5 miles from the center of the hole.
What is the radius of curvature of this ray of light?
Someone must know this one.
(BTW this would be a black hole of rougly 1 and 1/2 solar mass)
marcus,
It's not hard, I just haven't felt motivated to pull the calculator out today. I'll get to it.
- Warren
Originally posted by chroot
marcus,
It's not hard, I just haven't felt motivated to pull the calculator out today. I'll get to it.
- Warren
darn right it's not hard
for you I would guess it is solve-by-inspection....no calculator needed!
if anybody else is reading this thread, and wants a hint, look over in physics forum, at a thread started by Raavin about gravity and light
Well, the last part is solve-by-inspection -- the radius of curvature of a photon at the event horizon is the radius of the event horizon itself.
Dammit, marcus, when I'm not so busy I'll do the first part.
- Warren
Originally posted by chroot
Well, the last part is solve-by-inspection -- the radius of curvature of a photon at the event horizon is the radius of the event horizon itself.
Dammit, marcus, when I'm not so busy I'll do the first part.
- Warren
Hello chroot, I am happy to hear you are busy and therefore being kept out of trouble.
If everybody had to learn physics and do research there would be a great decline in delinquency and violent crime
Also I am not all in a hurry to see this problem answered.
Howver the FIRST part is a solve-by-inspection if you know a simple BH fact. The second part is harder and not what you say. Maybe I should retract the "extra credit" second part. I was talking about a ray of light projected horizontally near (but not right on) the event horizon. It cant have radius of curvature equal to the hole radius because it is going to arc down and fall into the event horizon. yes, I take that "extra credit" part back. Too hard.
Anybody interested please just consider the ray at 4.5 miles.
But wait, a light ray moving tangentially along the event horizon, will, in fact, orbit the black hole.
- Warren
Originally posted by marcus
Hello chroot, I am happy to hear you are busy and therefore being kept out of trouble.
If everybody had to learn physics and do research there would be a great decline in delinquency and violent crime
Also I am not all in a hurry to see this problem answered.
Howver the FIRST part is a solve-by-inspection if you know a simple BH fact. The second part is harder and not what you say. Maybe I should retract the "extra credit" second part. I was talking about a ray of light projected horizontally near (but not right on) the event horizon. It cant have radius of curvature equal to the hole radius because it is going to arc down and fall into the event horizon. yes, I take that "extra credit" part back. Too hard.
Anybody interested please just consider the ray at 4.5 miles. Without "matherizing", which I hate, I would have to say that the radius of curvature of the photon at 4.5 miles from the "center" would be 4.5 miles, in an attempt to obtain a circular orbit. Your 4.5 miles just happens to be 1.5 R_S.
Originally posted by Labguy
Without "matherizing", which I hate, I would have to say that the radius of curvature of the photon at 4.5 miles from the "center" would be 4.5 miles, in an attempt to obtain a circular orbit. Your 4.5 miles just happens to be 1.5 R_S.
That is right. If a black hole has schwarzschild radius 3 miles then the only distance at which light can go in circular orbits is 4.5 miles.
(this is for the ordinary case of non-rotating uncharged holes.)
The radius of the so-called "photon sphere" around the black hole, where light can orbit, is always 1.5 times the schw. radius.
Your turn Labguy.
(My son once took an introduction to astronomy course where the professor liked to tell students that if you put your head in the photosphere of a BH you would be able to see the back of your own head. for some reason that idea was popular with the students.)
Originally posted by marcus
The radius of the so-called "photon sphere" around the black hole, where light can orbit, is always 1.5 times the schw. radius.
Woops!
- Warren
Originally posted by chroot
But wait, a light ray moving tangentially along the event horizon, will, in fact, orbit the black hole.
- Warren
Not according to the professors and textbooks and stuff. It will dive into the black hole and head for the singularity----that is, if it starts off tangential along the event horizon.
The only altitude where it can orbit is in the "photon sphere" radius 1.5 times schwarzschild.
However it is still interesting to consider projecting a horizontal ray of light a few feet above the event horizon and asking about its initial radius of curvature.
It won't be the schwarzschild radius because then it would orbit, which it doesnt.
but it might initially be a nice fraction of the schwarzsch. radius,
like 2/3 or 1/2 of it----or it might be nearly zero
The first case would mean that for somebody suspended near the surface of a 3 trillion miles radius hole (with its earth-like gee) light seems to go STRAIGHT in his immediate surroundings. It hardly bends at all, or only imperceptibly. I am referring back to maybe 5 or 6 posts ago.
On the other hand the curve might be very tight and the light might take an immediate nosedive into the event horizon, when it is close---in which case the optics would be weird.
Either way it doesnt just orbit along the event horizon. But its interesting to try to figure out what it does.
Originally posted by chroot
Woops!
- Warren
Yeah, I know.
They are tricky objects.
Anyway we must not forget that it is now Labguys turn!
He will undoubtably pose us a question that will be a pleasure
to try answering.
Originally posted by marcus
Yeah, I know.
They are tricky objects.
Anyway we must not forget that it is now Labguys turn!
He will undoubtably pose us a question that will be a pleasure
to try answering. I think you are over-rating my abilities.[:)]... No, the answer to this question will be either very easy, or very difficult to find, rather balck or white.
Question:
Galaxy 3C 66B is a galaxy with huge radio (and other EM) jets shooting out to very large distances. What other recent information, gathered/noticed, about this galaxy shows that it has a property that has been theorized, but not before seen (evidenced) by observations??
a binary pair of supermassive black holes
orbiting around each other
probably the result of the merger of two galaxies
each of which brought a central black hole with it
Originally posted by marcus
a binary pair of supermassive black holes
orbiting around each other
probably the result of the merger of two galaxies
each of which brought a central black hole with it
I forgot to tell the recent observation!
The braided jet.
The smaller of the two is emitting a jet----the way black holes do, from the stuff spiraling in and getting hot on the way and getting sent out along the axis---the obersevers think it is the smaller one making this long bright finger of a jet.
And because of the periodic (about 1 year period) going around there is a kind of CORKSCREW twisted braided look to the jet--nice looking like a unicorn's horn is sometimes shown on medieval tapestry, or a knobbly icicle.
http://www.astronomy.com/Content/Dynamic/Articles/000/000/001/338cudnh.asp
another telling observation
was "elliptical motion"
in the core of the galaxy
it was so short-period (one year)
that what else could it be?
to orbit each other so fast these extremely massive
things would have to be very close together
so most likely two black holes
probably the observers detected a doppler wobble
which plotted not sinusoidally but in a distorted
sinewave that means ellipse
this is cool, let us give it the guy with sunglasses [a)]
Originally posted by Labguy
That's it.
Your go.
Early in the 17th century Kepler discovered that
one thing was "the sesquipotence" of another thing.
That was how he expressed it---i've taken a look at the
original Latin text with a literal translation. We say it
differently now, most likely, but he said "sesquipotence".
What did he mean by sesquipotence?
What were the two quantities he was talking about?
Which one was the sesquipotence of the other?
What year was this?----discovery and publication happend
within a few months of each other.
No one has replied so I will assume its not a good question and ask a different one.
New Question
What z corresponds to a light travel time of 12 billion years?
Assume spatial flatness.
http://www.astro.ucla.edu/~wright/CosmoCalc.html
This requires trying various z until you hit one that gives 12 billion years for the light travel time. When you type in a value for z, press the "flat" button and it will find the light travel time which corresponds to that z.
Unless you explicitly change the default settings, the calculator will assume H0 is 71 in the usual units (km/s per Mpc) and Ωvac = 0.73, and 0.27 for ΩM = 0.27. These are widely accepted values so what I'm looking for is the answer with these default values for H0, Ωvac, and ΩM left unchanged.
The Kepler question seems to have been too obscure [:(] So I withdraw it. For anyone who might be interested, here's a hint:
"sesquicentennial" means the "one-and-a-half century" mark. Sesqui is Latin for 1.5.
Originally posted by marcus
Early in the 17th century Kepler discovered that
one thing was "the sesquipotence" of another thing.
That was how he expressed it---i've taken a look at the
original Latin text with a literal translation. We say it
differently now, most likely, but he said "sesquipotence".
What did he mean by sesquipotence?
What were the two quantities he was talking about?
Which one was the sesquipotence of the other?
What year was this?----discovery and publication happend
within a few months of each other.
Originally posted by Labguy
z = 3.808
Beautiful! right on the button.
with z = 3.808 the light travel time comes out 12.000 billion years
also, as you probably saw, the current distance to an object whose past light is observed with redshift 3.808 is
given by the calculator as 23.362 billion light years.
this is the Hubble law distance D, so the present speed of
recession is 1.7 c.
We will never see that object as it is today (as long as expansion continues) because it is already out of range.
Nevertheless it and the many other redshift 3.8 objects in the sky are considered to belong to the observable universe, obviously, because we observe them. A paradox one must live with.
Your turn, Labguy. Ask a good one!
Ok, but then I will (try) to sit out awhile to get some other players into this game. I think it is fun, but not everyone has the time to find answers or even look up a good question. I say try because sometimes it is so tempting to jump right on an easy one, or one that I happen, by chance, to already know. That's not many..[:(]
Question:
In the final stages of an implosion supernova, with enough energy to make the "heavy" elements to Lead and above (after the BOOM:
(1) Name the supernova type.
(2) Name the progenitor star type.
(3) Name the final two or three fusion-stages, by elements fused, that result in the last element produced before the implosion.
(4) Name the last element formed before the implosion.
This is not a trick question, but "elements" mean simply those as shown and named on any common periodic table, including any isotope of them, if applicable.
(Note: Marcus: Wait at least 4-5 minutes before answering..[t)]..)
Originally posted by Labguy
Ok, but then I will (try) to sit out awhile to get some other players into this game.implosion.
(Note: Marcus: Wait at least 4-5 minutes before answering..[t)]..)
OK Labguy I see your reason for sitting out a few. Maybe I will take a breather too. But we had some good questions going and I enjoyed it.
Originally posted by marcus
OK Labguy I see your reason for sitting out a few. Maybe I will take a breather too. But we had some good questions going and I enjoyed it. No, no. Go ahead on this one (anyone); it has been more than 30 minutes. This question seems simple, and can be, but I bet a lot of posters here would miss one item in particular.
HINT: Some of the "burning shell" graphics on pages discussed earlier might show an answer; but might not....[t)]
Originally posted by Labguy
No, no. Go ahead on this one (anyone); it has been more than 30 minutes. This question seems simple, and can be, but I bet a lot of posters here would miss one item in particular.
HINT: Some of the "burning shell" graphics on pages discussed earlier might show an answer; but might not....[t)]
well I am of two minds about this. I really enjoy this game as a friendly tennis match between the two of us because you know a lot of interesting stuff
on the other hand morally I think it would be good to be more inclusive---so that I should hold back and wait a while before answering each of your questions
I will compromise for now. I will answer ONE PART of your question and then whoever steps in and answers the rest can have a turn.
Let me know if I'm wrong about this----the last element to be made prior to collapse is IRON.
that would, I guess, have to be made from silicon which merges to nickel which emits two betas and decays to iron. I am just
speculating based on the numbers on the atomic table of elements.
Originally posted by marcus
well I am of two minds about this. I really enjoy this game as a friendly tennis match between the two of us because you know a lot of interesting stuff
on the other hand morally I think it would be good to be more inclusive---so that I should hold back and wait a while before answering each of your questions
I will compromise for now. I will answer ONE PART of your question and then whoever steps in and answers the rest can have a turn.
Let me know if I'm wrong about this----the last element to be made prior to collapse is IRON.
that would, I guess, have to be made from silicon which merges to nickel which emits two betas and decays to iron. I am just
speculating based on the numbers on the atomic table of elements. That was the only "hard" part. Most books, and internet sites just say silicon to Iron, when actually it is silicon to 56NI and then to 56Fe.
Anyone else who lists out the other answers will be the next question winner.
Finish the other points on that question, Marcus. (or anyone)!
a) a Supernova Type II
b) a massive red supergiant
- Warren
Originally posted by Labguy
Question:
In the final stages of an implosion supernova, with enough energy to make the "heavy" elements to Lead and above (after the BOOM:
(1) Name the supernova type.
(2) Name the progenitor star type.
(3) Name the final two or three fusion-stages, by elements fused, that result in the last element produced before the implosion.
(4) Name the last element formed before the implosion.
The progenitor star would be a red giant, of some initially very massive class like a spectral type O.
The supernova is type II-----the implosion kind as opposed to Type Ia, the thermonuclear kind.
As we already discussed the last element is iron
which results from decay of nickel and cobalt.
The last fusion stage is silicon to nickel.
I dont remember just how many solar masses the original star has to be in order to be able to make Type II---or even how positive astrophysicists are about that. So I just say very massive, like a spectral type O or thereabouts. Is that right?
We currently don't know the requisite masses for a star to become a Type II SN. We do know that the iron core of the star has to be larger than the Chandrasekhar "limit," but we're not sure how much mass is lost from the envelope in the explosion, and thus are not sure about the star's minimal total mass.
Neener neener boo boo -- I beat you by seconds. :)
- Warren
Originally posted by chroot
We currently don't know the requisite masses for a star to become a Type II SN. We do know that the iron core of the star has to be larger than the Chandrasekhar "limit," but we're not sure how much mass is lost from the envelope in the explosion, and thus are not sure about the star's minimal total mass.
Neener neener boo boo -- I beat you by seconds. :)
- Warren Yes, you "finished the list" within seconds of that "other guy".
chroot ask next question.
Somebody ask another question. I survived the surgery I had yesterday and can sit at my desk again without pain pills...[t)]
Originally posted by Labguy
Somebody ask another question. I survived the surgery I had yesterday and can sit at my desk again without pain pills...[t)]
That's good news. Congratulations! How's the weather in Tampa Bay area----beautiful around SF Bay now and not too hot.
The universe is expanding. It is Chroot's turn.
I know that I have never posted in this thread but it seems to me that it will be a while before chroot comes back from wherever he may be. It has been 7 days since his last post in this thread so perhaps someone else should go or maybe they could simply send a private message to Chroot.
Originally posted by Shadow
I know that I have never posted in this thread but it seems to me that it will be a while before chroot comes back from wherever he may be. It has been 7 days since his last post in this thread so perhaps someone else should go or maybe they could simply send a private message to Chroot. No, I asked the last question, but enough time has gone by for chroot to answer with his question since he got the last one right.
If Shadow is "new" around here, take it and ask away. Otherwise, anyone can jump in now. I think that summer vacations will lose some of the regulars for awhile.
Originally posted by Labguy
No, I asked the last question, but enough time has gone by for chroot to answer with his question since he got the last one right.
If Shadow is "new" around here, take it and ask away. Otherwise, anyone can jump in now. I think that summer vacations will lose some of the regulars for awhile.
I will jump in. My excuse is that I answered almost all of the last question but Chroot beat me to the punch on the final part.
---------------------------------
Use this to answer a simple question:
http://www.astro.ucla.edu/~wright/CosmoCalc.html
How long, before the present, has it taken the universe to expand by a factor of two?
To rephrase that, how long ago were distant galaxies, that are like 1 billion LY away now, just half as far away?
--------------------------
This assumes the current assumptions cosmologists make----namely Hubble parameter 71, spatial flatness, cosmological constant 73 percent. But that is all built in to the CosmoCalculator.
You just have to plug in a value of the redshift z and it will tell you how long.
But be sure to plug in the value of z that goes with the idea of expanding by a factor of two!
I'm sorry I have never used a website like that before. I usually post in the philosophy and political forums and I am an "ameteur" astronomer. I have a telescope and a few books on astronomy that I read in spare time but I have not gotten very far yet so I am hoping to learn by this thread.
-Shadow
Originally posted by Shadow
I'm sorry I have never used a website like that before. I usually post in the philosophy and political forums and I am an "ameteur" astronomer. I have a telescope and a few books on astronomy that I read in spare time but I have not gotten very far yet so I am hoping to learn by this thread.
Shadow would you mind if I talked you thru this? then you would get the answer and the next turn to ask would be yours.
If you would not mind the bother of just mechanically going thru the motions (if you do, say no!) then here it is.
There is only one thing to do.
Go to
*************
http://www.astro.ucla.edu/~wright/CosmoCalc.html
And plug the number 1 into the box that says "z"
It is over on the left side
And then click on the box that says "Flat"
that will do it.
The number of years will show up in the "light travel time" box
on the right of the screen.
Be sure to put the number ONE into the "z" box, and press "Flat".
dont change any of the other settings cause they are usually right. If you have any trouble, post here or write me a PM and I will help.
******the question, quoted from previous post*******
How long, before the present, has it taken the universe to expand by a factor of two?
To rephrase that, how long ago were distant galaxies, that are like 1 billion LY away now, just half as far away?
This assumes the current assumptions cosmologists make----namely Hubble parameter 71, spatial flatness, cosmological constant 73 percent. But that is all built in to the CosmoCalculator.
You just have to plug in a value of the redshift z and it will tell you how long.
But be sure to plug in the value of z that goes with the idea of expanding by a factor of two!
FOOTNOTE: the expansion factor corresponding to redshift z is 1+z
So the expansion factor corresponding to redshift 1 is 1+1 =2
But thats what the question is about! Expansion by a factor of two! So the z to use is z = 1.
If you want to know how long it took for the universe to triple---i.e. expand three-fold---up to the present, then put z = 2 into Ned Wright's calculator. Always put one less than the expansion factor you want to know about. There's nothing wrong it is just how astro conventions work and how the calculator is constructed.
Well I seem to have answered my own question in the course of explaining how to answer it. So here is another
The temp of the CMB has been measured with fine accuracy to be 2.726 kelvin
But Bill Unruh determined that simple acceleration gives space a temperature. A moving box, if it is accelerating, would have inside it some thermal radiation with a temperature proportional to the amount of accelaration.
So I ask you----how fast would you need to accelerate (meters per second per second, feet per second per second, gees, whatever)
in order to produce an Unruh temp of 2.726 kelvin?
Aaargh[g)] ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldnt we know how much acceleration it would take to duplicate the CMB temp?
For sentimental reasons if for no other. The CMB temp is the most prevalent temp in the universe---it IS the temp of the universe.
Aaargh ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldnt we know how much acceleration it would take to duplicate the CMB temp? The word is "Matherize", not arithmeticize, you bumbling human calculator....[t)]
I get 37.14159265 cm/sec./sec. How?; heck, I don't know, I just made it up. But, someone should easily recognize the eight decimal places as something they use a lot.
Sorry for the "No-Answer" post.
Labby
Originally posted by Labguy
The word is "Matherize", not arithmeticize, you bumbling human calculator....[t)]
I get 37.14159265 cm/sec./sec. How?; heck, I don't know, I just made it up. But, someone should easily recognize the eight decimal places as something they use a lot.
Sorry for the "No-Answer" post.
Labby
Some how----a lucky shot in the dark?----you got the right answer! Your turn Labguy.
[:D]
The forumula for the temperatue is T = g/2pi
where g is the acceleration.
It is sort of like the relation of radius and circumference of a circle
the temperature is the radius of the acceleration. whaaaaa???
So I ask you----how fast would you need to accelerate (meters per second per second, feet per second per second, gees, whatever)
in order to produce an Unruh temp of 2.726 kelvin?
Aaargh[g)] ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldnt we know how much acceleration it would take to duplicate the CMB temp?
For sentimental reasons if for no other. The CMB temp is the most prevalent temp in the universe---it IS the temp of the universe.
The CMB has a perfect thermal spectrum that a black body would radiate at a temperature of 1.93 x 10-32 natural.
Only need to multiply that by 2pi to get the desired acceleration.
which is 12 x 10-32 planck.
Evidently this is the acceleration (the answer to the question, cause I said any units you want) which Labguy must have calculated.[;)]
Labguy your only mistake was in trying to convert it to metric, where because of the awkwardness of the system you went way wrong in the fifth decimal place---maybe other places too.
However this will be overlooked in your favor.
Your turn!
I was curious to see how it turned out in metric so I
converted the acceleration
12 x 10-32
into metric and it came out
67 x 1019 meters per second per second.
tho your answer was a few order of magnitudes off I am
delighted to overlook this in the interest of the game
Oh, sorry marcus I have been busy and was unable to reply to your post.
Originally posted by marcus
The forumula for the temperatue is T = g/2pi
where g is the acceleration.
It is sort of like the relation of radius and circumference of a circle
the temperature is the radius of the acceleration. whaaaaa???
The CMB has a perfect thermal spectrum that a black body would radiate at a temperature of 1.93 x 10-32 natural.
Only need to multiply that by 2pi to get the desired acceleration.
which is 12 x 10-32 planck.
Evidently this is the acceleration (the answer to the question, cause I said any units you want) which Labguy must have calculated.[;)]
Labguy your only mistake was in trying to convert it to metric, where because of the awkwardness of the system you went way wrong in the fifth decimal place---maybe other places too.
However this will be overlooked in your favor.
Your turn! So which is it; wrong at the 5th decimal place, or off by "a few orders of magnitude?? To four decimal places seems Ok, but "a few orders of magnitude" sounds way off. Are you giving me "actual" credit as right or wrong?
Originally posted by marcus
I was curious to see how it turned out in metric so I
converted the acceleration
12 x 10-32
into metric and it came out
67 x 1019 meters per second per second.
tho your answer was a few order of magnitudes off I am
delighted to overlook this in the interest of the game Is that a typo, or did you mean 67 x 10-19 m/s/s instead of the +19 ??
Originally posted by Labguy
Is that a typo, or did you mean 67 x 10-19 m/s/s instead of the +19 ??
Hello Labguy and Shadow! Glad to hear you. I want Labguy to have the next turn and ask the next question because he at least came up with some answer.
It was not numerically correct and I was just kidding about being only a "little bit" off. But even if waaaayyy off, some answer is better than none at all! Even one meant jokingly (as I think Labguy's) is better.
So your turn to ask.
(I will tell you about calculating the temperature of acceleration---as in Bill Unruh's 1976 paper----in a later follow-up post. The quantum gravity people are often referring to 3 things from the Seventies: Bekenstein's 1973 BH entropy, Hawking 1975 BH temperature, Unruh 1976 acceleration temperature. It is a famous and surprising result that helped stimulate current directions of research. So you might like to know. But right now the game is more important.)
Unruh discovered that merely accelerating gives space around you a temperature
best to think of an observer in a box. the box is given acceleration so there is thermal radiation inside and some equilibrium temperature
this effect is very very very tiny. Good old mother nature does not confront us with really gross freakishness here. The effect is so tiny that you will never detect it. But it is based on standard physics and is a solid, tho surprising, result.
(essentially just like Hawking's BH temp but in a different context)
No reasonable acceleration is enough to produce a macroscopic temperature, even a small one like 2.726 kelvin!
Anyway the formula for the temperature seen in space by an accelerating observer is T = g/2pi
So to produce a temp of 10-32
requires an acceleration of 2pi x 10-32
You always multiply by 2pi to get the acceleration.
Now Planck temperature is sort of Big Bang temperature and in metric terms it is 1.417 x 1032 kelvin.
So when I say a temp of "10-32 " I mean 1.417 kelvin.
And in metric terms, Planck unit acceleration is 5.56 x 1051 meters per sec per sec. So "10-32 " of acceleration is 5.56 x 1019 meters per sec per sec
You just have to subtract 32 from 51 to get 19.
And 2pi times that is 35 x 1019 meters per sec per sec
That is for making the temperature be 1.4 kelvin or so, earlier I did it for the CMB temperature which is almost twice that.
A more attainable temperature to get by acceleration is a femtokelvin---a quadrillionth of a kelvin
To get 1.4 femtokelvin you need the acceleration to be 35 x 104 meters per second
Well, since I didn't actually get the correct answer on Marcus' last question, I don't feel I should be asking one here. But, I will and it should be a slam-dunk for anyone with eyes.
Easy One:
My last answer was 37.14159265 cm/sec/sec. (That is a statement)
QUESTION:
Based on the number above (forget cm/sec/sec);
(a) Pick one number only to remove: remove the #___ ?
(b) Place the decimal after the number____ ?
(c) Name the resulting number_____ ?
It is a very common number, constant, ratio, measure, etc., used in astronomy every day.
Easy One:
My last answer was 37.14159265 cm/sec/sec. (That is a statement)
QUESTION:
Based on the number above (forget cm/sec/sec);
(a) Pick one number only to remove: remove the #___ ?
(b) Place the decimal after the number____ ?
(c) Name the resulting number_____ ?
It is a very common number, constant, ratio, measure, etc., used in astronomy every day. Come on, guys. This one is too easy to sit for more than a few hours!
Originally posted by Labguy
Come on, guys. This one is too easy to sit for more than a few hours!
OK but you better reciprocate and answer the easy one I ask back!
37.14159265 remove the 7, leave the point after the 3, and
get pi.
My question for you is-------suppose a distant galaxy is observed with cosmological redshift z = 1
then what's the prevailing estimate of the light travel time come out to be?
It depends on values assumed for some parameters and the prevailing assumptions are spatial flatness, Hubble value of around 71 (in the usual units), and a cosmological constant representing 73 percent of the energy density. These will do fine and are the default settings in the calculator provided to calculate just that, among other things, at one of the cosmology websites.
I meant this question especially for Labguy but anyone else is welcome to answer.
I told you (anyone) that the answer to my question was as easy as pi. You did it easily, I see.
For Z=1, the light travel time would be about 7.731 billion years if you keep OmegaM at 0.27, and allow your other assumptions to stand.
Originally posted by Labguy
I told you (anyone) that the answer to my question was as easy as pi. You did it easily, I see.
For Z=1, the light travel time would be about 7.731 billion years if you keep OmegaM at 0.27, and allow your other assumptions to stand.
Bravo Labguy! Your turn to ask.
I guess your answer means that in the past 7.731 billion years the universe has expanded by a factor of two.
Another easy one:
If:
("Blank") = 4.8-[2.5*log(Lstar/Lsol)]
where "L" stands for luminosity;
What is "Blank"? ( Definition, not a number)
Absolute magnitude -- the apparent brightness of a star from a standard distance of 10 parsecs.
- Warren
Originally posted by chroot
Absolute magnitude -- the apparent brightness of a star from a standard distance of 10 parsecs.
- Warren That was too quick, I'll have to start throwing in some tougher ones if I get another go.
Your turn, speed-breath...[6)]
Nearly all models of inflation considered today are "slow-roll" inflationary models. Lots of evidence has been piling up (from the Wilkinson Microwave Anisotropy Probe, for example) that we're at least on the right track.
What does the modifier "slow-roll" mean in the context of the space of inflationary models?
- Warren
Originally posted by chroot
Nearly all models of inflation considered today are "slow-roll" inflationary models. Lots of evidence has been piling up (from the Wilkinson Microwave Anisotropy Probe, for example) that we're at least on the right track.
What does the modifier "slow-roll" mean in the context of the space of inflationary models?
- Warren It can get quite complicated, as there are currently 23 different inflationary models being perused.
But, in simple terms, the "slow roll" term is one model (inflation) that predicts that the universe is flat, and that the spectroscopic variances seen in the CMBR indicate that there was an inflation (singular) where the potential energy slowly "rolled down" with the kinetic energy being damped by the Hubble expansion. Other models predict that this damping (slow rolling) happened more than once, and each had its own "starting and stopping" phase based on the total energy available at the time. In other words, several periods of inflation instead of just the one originally proposed by Guth.
This is a huge simplification.
Sorry Labguy, let me see if I can make my question more specific.
The inflationary model first proposed by Guth involved a particular kind of "exit mode," while slow-roll inflationary models involve a very different one. What is this essential, fundamental difference between the "exit mode" of Guth's model and slow-roll models?
Hint: the fundamental difference can be stated in only one sentence.
- Warren
Originally posted by chroot
Sorry Labguy, let me see if I can make my question more specific.
The inflationary model first proposed by Guth involved a particular kind of "exit mode," while slow-roll inflationary models involve a very different one. What is this essential, fundamental difference between the "exit mode" of Guth's model and slow-roll models?
Hint: the fundamental difference can be stated in only one sentence.
- Warren I would have to say that Guth's original proposal for inflation had no explanation available for the originating and exiting times of the inflationary period. He defined the period, but not why it began or ceased. This is important, because it is why Hawking, and others, doubted inflation to the point that the numerous other models have been developed, even two more by Guth himself.
As I said, there are ~23 models currently being proposed and considered. The only one I know of proposes that a single (scalar) field can be used with no consideration needed for the coordinate system(s). This model assumes that only the Hubble field (size) and the inflationary field (combined wavelength of energies) need to be considered. The "Exit mechanism" (slow roll) occurs when the wavelength of the scalar field becomes larger than the "Hubble Radius", at that time, and that inflationary period stops. Then, the kinetic and potential energy can recombine, only to seperate again and (sometimes) reach the point where kinetic energy "damps" and causes another, short inflationary period, which again will "exit" when the wavelength once more reaches the new Hubble scale.
That wasn't one sentence, and if not what you have in mind, it is all I'll be able to add to this topic.
Hrm, well... :-/
The answer I was looking for is rather simple. I'll go ahead and give it.
Inflationary models that are NOT slow-roll depend upon a quantum-mechanical tunneling mechanism to move from a classically stable minimum of the inflation potential to a classically stable state with zero inflation potential.
Slow-roll models do not depend on quantum-mechanical tunneling. Instead, the inflation potential is classically unstable, and, just like a ball rolling down a hill, the inflation potential slowly rolls to zero.
So the difference between slow-roll and non-slow-roll models is essentially this: slow-roll models do not require a quantum-mechanical tunneling mechanism to end inflation.
- Warren
Okay, I will. Was my last question malformed? Was the answer I gave incorrect or unusual in some respect? I apologize if my question sucked, I thought it was an good one! :)
- Warren
Originally posted by chroot
Okay, I will. Was my last question malformed? Was the answer I gave incorrect or unusual in some respect? I apologize if my question sucked, I thought it was an good one! :)
- Warren No, I thought that it was a good one. But, everybody and his brother have a "theory" of inflation, so it is tough to decide which one is "best" as of this week...[t)]
Okay, here's one that hits close to home. Someone in my introductory adult astronomy classes asks me this question about once every six months or so. The question is "can we see the Lunar rover with your telescope???"
So my question to you is a simple one: how big would my telescope have to be to be able to see the Lunar rover from the surface of the earth? (Neglecting the ever-obnoxious atmosphere, which actually makes it impossible.)
- Warren
The rover is about 3m long and the moon is 3.82x10^8 m away. Using trig you get an angle of 7.85x10^-9 radians.
Using Resolution = wavelength/diameter (I'll use 500nm)
You get d = (500x10^-9/7.85x10^-9)
Which is about 64 meters across. The Keck is only 10 meters!
Excellent work, stuffy. :)
I usually just tell my students it would take a scope about the size of a football field, and that I just don't have room for one in my two-seat roadster. :)
Your turn.
- Warren
Originally posted by chroot
Okay, here's one that hits close to home. Someone in my introductory adult astronomy classes asks me this question about once every six months or so. The question is "can we see the Lunar rover with your telescope???"
So my question to you is a simple one: how big would my telescope have to be to be able to see the Lunar rover from the surface of the earth? (Neglecting the ever-obnoxious atmosphere, which actually makes it impossible.)
- Warren Using the so-called Dawes' Limit at 4.56/diameter of telescope, I get an angular resolution for the 10-foot Rover of 0.00164 arc seconds. 4.56/0.00164 = ~2780 inches; about 70 meters.
But, to see it, as your question asks, we would also need to know its albedo (reflectivity), and I don't think anyone can figure that.
EDIT: Too Late!
I wonder how long it would take that mirror to cool!
If astronauts were to dump a bucket of water on the moon's surface at lunar noon, roughly how long would it take all those water molecules to escape?
Originally posted by stuffy
I wonder how long it would take that mirror to cool!
If astronauts were to dump a bucket of water on the moon's surface at lunar noon, roughly how long would it take all those water molecules to escape? I haven't done any "Matherizing" (see my profile), but with the near-zero atmosphere (yes, the Moon has a small one), and the near-vacuum at the surface, I would say "roughly" less than two seconds if you mean that "escape" is to evaporate to its elemental gasses.
If you mean escape into space beyond the Moon's gravitational hold, I have no idea...[g)]
This does deal with the escape velocity of the moon. [<:)]
As a hint, it also has to do with the mean speed of the particles.
Originally posted by stuffy
This does deal with the escape velocity of the moon. [<:)]
As a hint, it also has to do with the mean speed of the particles.
You said at "noon"
There are craters at the poles of the moon where it is always in shadow and quite cold. I expect ice could exist in a permanently dark crater at the moon's north or south pole.
But you said noon. Let's say at the equator, in vertical sunlight. Then the equilibrium temp is 394 kelvin, I think.
And escape speed from lunar surface is 2370 meters per second.
Let's say they really fling the water wide so it all ends up in intimate contact with hot rock and becomes 394 kelvin.
kT = 5.44 E-21 joules
kT = 0.034 eevee
A water molecule's mass is 3 E-26 kilogram
sqrt (2kT/m) = 600 meters per second (that is just a benchmark: the velocity that would have kT as kinetic energy)
there will be a spread of velocities roughly on that order of magnitude. the bulk will not have escape velocity of 2370 meters per second. so they dont all go off into space instantly
the story gets a mite complex-----UV in sunlight will start to dissociate the water in O and H-----but ignoring that complication....the thermal distribution of velocities will allow some fraction of the water molecules to escape, others will fall back to the surface and get reheated and have another chance.
someone else can have a go at setting up and solving the problem more rigorously-----one could put it in terms of a "half-life" of water at 394 kelvin on lunar surface
You are definitely on the right path. My old Astronomy and Astrophysics book by Zeilik explains that the lifetime of a given gas is related to the value given by v_e/v_rms, probably to avoid the true complexity of the problem. I don't know if I want to give away the relation right now because that will certainly reveal the answer :)
So much for that! v_esc/v_rms approx 3. Which means a lifetime of only a few years.
Next person to reply can ask a question.
*rubs paddles* CLEAR!!!!!! *shock*
Okay, okay, I'll ask one.
It might be depressing to learn that we will never be able to see all of the universe, no matter how patient we are. In fact, as the universe ages, the so-called "particle horizon," a 'boundary' beyond which we cannot see, will approach a specific size as time goes to infinity. (The particle horizon is, of course, not a physical boundary -- it is a mathematical surface that separates those points in space-time from which stars could send light to Earth in infinite time from those from which stars could not.)
The size of our observable universe will "max out" at this particular radius. What is this ultimate radius?
Bonus points: how big is the observable universe today?
- Warren
StephenPrivitera
Jul1-03, 03:10 PM
oooh! I hope I get this one.
I remember reading about this in an old Astronomy magazine.
I believe that the current radius of the observable universe is 10 to 12 billion light years. It will max out at 40 billion light years. I've seen numbers as low as 20 billion LY, but my answer is 40. I trust my Astronomy writers.
Hmmm... sorry Stephen, those are not the correct figures -- at least not as understood from any recent data. Perhaps you could quote from the magazine?
- Warren
StephenPrivitera
Jul1-03, 06:27 PM
See Astronomy, March 2003, page 45. It implicity states that the present radius of the observable universe is 14 billion light-years in the center caption. In a pamphlet produced by Astronomy (Solving Astronomy's 10 Greatest mysteries) that has no date whatsoever, it is explicitly stated that the max size currently is 10 to 12. Looking back at the March issue, page 44, last paragraph, it states, "in an accelerating universe, the speed of galaxies flying away from each other will one day overtake the speed of light - photons won't be able to catch up with the rapidly expanding space through which they travel." On page 45, first paragrah, it states, "there is a horizon . . . beyond which light cannot reach us." Further down the page, it mentions something about 40 billion light-years, but upon closer inspection, it says nothing about this figure being the "maximum" observable universe.
http://www.xs4all.nl/~carlkop/heelal.html
This site gives the radius of the visible universe as 12 billion light-years. See this section: ESA's Hipparcos satellite revises the scale of the cosmos. I have no idea of the date or validity of this source.
http://www.newageuniversity.org/universe.htm
This site gives it at 14.
I'll change my answer to 14 billion light-years.
But that was only the bonus question. I can't find an answer to your other question. Maybe next time!
_____edit
http://www-astronomy.mps.ohio-state.edu/~ryden/ast162_10/notes43.html
a more reliable source, citing the radius of the OU at 14 billion light-years
Well, as usual in cosmology, I'm afraid you're getting confused by some definitions of distance.
The universe is 13.7 billion years old, according to recent experiments. As a result, you might think that the observable universe is exactly 13.7 billion light-years in radius -- but this is not true. Because the universe used to be smaller, we can actually see much further than 13.7 billion light-years.
When I say "how big is the observable universe," what I mean is: what is the comoving distance to the furthest objects which are causally connected to us (i.e. within our past light-cone)?
Perhaps this question is just going to get mired in the confusing definitions that run amuck in cosmology.... argh.
- Warren
No No!!!!
Do not despair or say aaargh!
It is a very clear question. The answer is extremely interesting and depends on recently discovered acceleration in expansion of universe!
Terrific question
Lineweaver's online introduction to cosmology is the greatest.
Wait, I will get it.
Originally posted by chroot
Well, as usual in cosmology, I'm afraid you're getting confused by some definitions of distance.
The universe is 13.7 billion years old, according to recent experiments. As a result, you might think that the observable universe is exactly 13.7 billion light-years in radius -- but this is not true. Because the universe used to be smaller, we can actually see much further than 13.7 billion light-years.
When I say "how big is the observable universe," what I mean is: what is the comoving distance to the furthest objects which are causally connected to us (i.e. within our past light-cone)?
Perhaps this question is just going to get mired in the confusing definitions that run amuck in cosmology.... argh.
- Warren
shucks, I got so interested reading Lineweaver I forgot I was
going to answr the question.
anybody at all interested in cosmology should download
this new (May 2003) tutorial
Inflation and the Cosmic Microwave Background
Charles Lineweaver
it is in arxiv: astro-ph/0305179
and it is also at some Caltech site with a name like
Level 5 knowledge base
the current distance of objects whose light (emitted in the past) we are now seeing can be maximum 47 billion lightyears
Lineweaver's figure 1 on page 6 is the main thing to look at.
because of accelerating expansion the present distance of the farthest objects we will EVER see (even if we live to infinity)
is maximum about 60 billion lightyears.
(this is assuming cosmological constant, causing the acceleration in the expansion rate, is in fact constant. if it is not all bets are off)
fundamental idea is that light travel time, the time it took light to get here, is not a very workable idea of distance because of different rates of expansion at different times in the past.
so many things effect the light travel time
but there is a really clear idea of the PRESENTDAY distance to an object, measured this instant by observers at rest with respect to the expansion process that the universe is undergoing
that is the same as being at rest with respect to the Cosmic Microwave Background, something extremely easy to check by doppler measurement. We know the solar system's speed relative to CMB very accurately, and can allow for it.
So there is a well-defined current distance to objects
as a general rule the light travel time "distance" would not match either the instantaneous distance of the object at present or the instantaneous distance to it at the time the light was emitted.
the instantaneous distance at some definite moment (like a time in the past) is called sometimes "proper distance". With the term "comoving distance" reserved for the proper distance at this moment in year 2003. Terminology seem to me still a bit unsettled, so I just say current or presentday distance because that's what it seems to me that it is, whatever technical term they decide to call it.
You got it, Marcus. Lineweaver's little 30-pager is a great overview of the state of cosmology right now, you're right! It inspired my question. I'm not sure who sent the link to me -- it was probably you. :)
Anyway, it is a shame that Astronomy magazine and others are so haphazard with their details. It's pretty much unacceptable for a magazine like Astronomy to assert in March 2003 that the size of the observable universe is 14 billion light years. Even if you ignore the recent findings about the accelerating expansion, this is patently wrong. Light travel time is just not a good way to describe distances in a universe that has changed in size.
The accelerated expansion, by the way, is the reason why the particle horizon maxes out at 62 billion light-years -- if the expansion were constant, we would eventually be able to see everything in the entire universe.
Your turn, marcus.
- Warren
Originally posted by chroot
The accelerated expansion, by the way, is the reason why the particle horizon maxes out at 62 billion light-years -- if the expansion were constant, we would eventually be able to see everything in the entire universe.
Right! And that is hard to grasp too----without accelerated expansion even objects receding at greater than c could emit light that eventually reaches us----infinity being as it were a long time[;)] The idea of being able to see everything if you wait long enough is staggering. But with accelerated expansion it is only a finite piece that we will ever get to see no matter how long we wait. Cool ideas they get to deal with, these cosmologists.
I had a question ready, which can serve here in the game and also I posted it out in the "lineweaver" thread. Answers there dont count in the game. Here it is:
-----------------
This question is based on Lineweaver's Figure 1, top section'
There are three diagrams in Figure 1, I mean the top one.
If you go to the IPAC-Caltech site then you can enlarge the diagrams to fill the screen.
http://nedwww.ipac.caltech.edu/leve...r_contents.html
It's also convenient to print it out which you can do from the Los Alamos archive site-----tho the diagrams are smaller printed out.
http://arxiv.org/abs/astro-ph/0305179
Either way, what you see in the figure is a tear-drop shaped lightcone
And also you see a curve showing the extent of the "HUBBLE SPHERE" and he explains what he means by that.
Question 1: What are meant by lightcone and hubble sphere?
There is a point in the diagram where the hubble sphere and lightcone intersect. The hubble sphere line crosses the side of the teardrop shaped lightcone.
Question 2: What does the intersection signify? Why does it come at the widest point of the lightcone----where the side of the lightcone is vertical?
I just noticed the rules Nicool posted at the beginning of the thread say I should post a new question since this one wasnt answered in 3 days [:(]
Originally posted by marcus
Question 1: What are meant by lightcone and hubble sphere?
There is a point in the diagram where the hubble sphere and lightcone intersect. The hubble sphere line crosses the side of the teardrop shaped lightcone.
Question 2: What does the intersection signify? Why does it come at the widest point of the lightcone----where the side of the lightcone is vertical?
Just to conclude, the hubble radius is the current distance to where space is receding at c ( Lineweaver, caption to fig. 1).
Within the hubble sphere space is receding at speeds < c. Outside it is receding at > c. the radius of the h. sphere changes with time as shown in the figure.
The past lightcone essentially shows the paths along which light now reaching us has traveled----in a simple 2D spacefime diagram the lightcone line is the path of a ray of light coming towards us for the whole history of the universe and reaching us today.
Where the two curves intersect is where light started out towards us in a region of space receding from us (at that time) at speed c------so for the first few days the light didnt get any closer to us! It's speed towards us was canceled by the hubble flow away from us. So its time-line goes vertically up the diagram parallel to ours. But after a while it begins making progress "swimming upstream" against the current of the hubble flow (another term for the expansion of space).
Below that intersection light starting out towards us is initially swept back (teardrop shape of lightcone, below its widest point)
I'll think of another and post it.
another question
background to question:
the expanding universe model of cosmology was used
in 1948 (Alpher/Herman) to predict the CMB
over 10 years later the CMB was found and it had roughly
the temperature they had predicted-----a classic longshot.
They had predicted 5 kelvin and it turned out to be 2.726.
Another background, with a different temperature, is also
predicted by the model. (Lineweaver p23 section 7.4)
This other cosmic background temp has not as yet been measured.
what is this other background and what (roughly) is its predicted temperature?
subtillioN
Jul4-03, 12:52 PM
the 2.7k CMB is simply the ambient temp of the ubiquitous intersteller Hydrogen. This temp was actually more correctly predicted by the steady state model as I recall.
See this link:
http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
" Abstract.
It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
However, in the standard model of the universe, the simple fundamental Planck's law has been ignored. It is claimed that the observed radiation comes from a combination of complicated hypotheses, involving an elaborate "creation mechanism" called the Big Bang. After this event, the radiation would have been emitted at a single instant when matter became decoupled from radiation. Finally, that radiation would have been shifted increasing its wavelength about 1000 times. We show that the 3 K radiation spectrum observed is simply the Planck's radiation emitted by gaseous matter at 3 K. "
Originally posted by subtillioN
the 2.7k CMB is simply the ambient temp of the ubiquitous intersteller Hydrogen. This temp was actually more correctly predicted by the steady state model as I recall.
See this link:
http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
a stable unexpanding universe in equilibrium at 3 kelvin
is a charming thought but the question was about
something predicted by the prevailing model
there is this other background that mainstream cosmology
predicts is there and that it has a certain black body temperature,
the question is about what temp is predicted by the model
BTW when they do measure the temp it will be a good test
of the standard cosmic model, right?
so what is this other background I am referring to?
and what is the predicted temp?
Originally posted by Nicool002
Hi guys! Most of you know how this works but for the newcomers:
The rules are this: someone will ask a question and if the question is not answered correctly within 3 days then a new question will be posted. If an answer to a question is posted and the person that posted the question does not respond to the answer within 2 to 3 days, then the first person to have answered the question will then be able to post their own question. HAVE FUN AND LEARN!
I will start:
Question: What is the brightest star in the Northern Sky? (excluding the sun)
This game can be a lot of fun and interesting.
Since no one has answered in 3 days I think it is up for grabs.
Anyone can ask a question
or so I think from looking at Nicool's rules
I asked the last question in the Q/A game and it has gone unanswered for more than 3 days so I will answer it myself and maybe pose another.
Originally posted by marcus
another question
background to question:
the expanding universe model of cosmology was used
in 1948 (Alpher/Herman) to predict the CMB
over 10 years later the CMB was found and it had roughly
the temperature they had predicted-----a classic longshot.
They had predicted 5 kelvin and it turned out to be 2.726.
Another background, with a different temperature, is also
predicted by the model. (Lineweaver p23 section 7.4)
This other cosmic background temp has not as yet been measured.
what is this other background and what (roughly) is its predicted temperature?
The other background (besides the microwave background) which the expanding universe model predicts is the NEUTRINO BACKGROUND and this also has a thermal distribution with a definite temperature, just like the microwave background does, but the model predicts a DIFFERENT TEMPERATURE!
The microwave temperature of space has been measured at 2.73 kelvin, corresponding to a redshift z = 1100.
The predicted neutrino background temperature is 1.9 kelvin.
The fact that the two temps differ the way they do provides for a sensitive test of the model at some time in the future when the neutrino background will have been well-enough studied to determine its temperature. I think Lineweaver's discussion of this on page 24 is pretty interesting. It explains why TCMB should be hotter than Tneutrino by a certain amount.
We know that Aristarchus of Samos (who flourished around 270 BC) maintained that the earth travels around the sun and that the cycle of day and night is caused by the earth's rotation.
The main reason we know about Aristarchus and his heliocentric model is because it is described in a book written a few years later by somebody else.
(Unfortunately Aristarchus original writings proposing the heliocentric model have been lost, although his calculation of the relative sizes of the earth, moon, and sun has been preserved)
What is the book which is our primary source about this and who wrote it?
For "extra credit", why did the Greek contemporaries of Aristarchus mostly reject his proposed heliocentric model?
was it De architectura by virtruvis?
Originally posted by Shadow
was it De architectura by virtruvis?
No, the book was by Archimedes.
Archimedes described the heliocentric model (later popularized by Copernicus) around 250 BC in a book called the "Sand-Reckoner."
Copernicus knew about this and made a reference to it in one of his manuscripts, giving Aristarchus the credit for the discovery.
In "Sand-Reckoner" Archimedes describes Aristarchus model of the universe as a sphere with the sun in the middle and he calculates the number of grains of sand that would be needed to fill it.
There is an odd similarity between this and the present excitement among modern cosmologists about how much matter and energy of various sorts the universe contains.
the rules of this "Astronomy" Q/A game suggest that if a question goes unanswered for 3 days or more (which this has) then another should be provided, so I will think of one unless someone else gets to it before I do
What is the most distant object known to astronomers?
I mean a localized object----a certain quasar in fact---rather than diffuse radiation like the microwave background and ancient neutrinos.
What is its observed redshift?
Well there is a baby galaxy that Astronomers have found and it took 122 billion years for it's light to reach Earth but that is, in fact NOT a quasar... when you say local, do you mean in this galaxy?
It's a quasar (I can find neither its name nor its location) at redshift 6.4, discovered by the Sloan DSS.
- Warren
Originally posted by chroot
It's a quasar (I can find neither its name nor its location) at redshift 6.4, discovered by the Sloan DSS.
- Warren
You're right! It is your turn
We both probably saw the announcement in Ned Wright's "news of the universe"-----the quasar was discussed by Bob Becker (of the Sloan sky survey) around November 2002 at a talk that Wright referred to
The direction to it is very close to the star in the Big Dipper called "gamma UMa"
You teach astronomy, I know, but for anybody who doesnt know which star is "gamma" in that constellation, imagine the dipper in standard dipper position and picture where the handle joins on to it: that star is delta and gamma is the one just beneath it.
like, gamma is the lower left corner star of the dipperbowl
the quasar is about a degree below that corner star
and indeed the reshift is a huge 6.4, imagine wavelengths stretched out to more than seven times original length!
Think of a good one Warren.
Ummm...Chroot are you going to post?
Originally posted by Shadow
Ummm...Chroot are you going to post?
Shadow, do you have a question you want to post?
The rules say something like if the thread is inactive for three days then somebody should do something.
I think Nicool wrote something about this at the start.
If you have a question, why not post it?
If you dont have one you want to ask, please reply to that effect and unless chroot shows up I will think of one
Id rather you did though since its time for some new people
Okay then.
1) What is Helioseismology and what do you do? (This may be little unclear so if you dont get it, then just tell me and I will explain what I mean)
2) What two kinds of radiation are Gamma Rays converted into by the time they emerge from the suns photosphere?
Originally posted by Shadow
Okay then.
1) What is Helioseismology and what do you do? (This may be little unclear so if you dont get it, then just tell me and I will explain what I mean)
2) What two kinds of radiation are Gamma Rays converted into by the time they emerge from the suns photosphere?
Two off-the-cuff geusses;
1) Just judging from the composition of the word itself, I would geuss it's the study of turbulance on the Sun's surface (prob. to determine what's happeneing underneath).
2) Heat and light?
Number 1 is close but not exactly right but I will give it to you...it isnt of the suns surface, but the suns interior.
Number two is wrong
1) The study of modes of pressure-wave oscillations in the sun's interior.
2) The gamma-rays emitted by the nuclear reactions at the Sun's core are converted to lower-energy forms of electromagnetic radiation by the non-reacting envelope. There is no 'second' form of radiation attributable to the gamma rays themselves.
- Warren
Wouldn't it be electromagnetic and neutrino? It can't be X-ray, because that is one form of EM. And, neutrinos are produced at the same time as the gamma radiation, but aren't a "product" of the gamma radiation, like Chroot said.
Actually (and I can understand why you dont have these...it was a question with a few possible answers) I was looking for "by the time Gamma Rays emerge ffrom the photosphere, the energy is converted in infrared and optical radiation"
Originally posted by Shadow
Actually (and I can understand why you dont have these...it was a question with a few possible answers) I was looking for "by the time Gamma Rays emerge ffrom the photosphere, the energy is converted in infrared and optical radiation"
Well, this is simply not correct, as it's also converted into ultraviolet, X-ray, radio, etc.
- Warren
Originally posted by Shadow
Actually (and I can understand why you dont have these...it was a question with a few possible answers) I was looking for "by the time Gamma Rays emerge ffrom the photosphere, the energy is converted in infrared and optical radiation"
That's what I said; heat and light!
Actually (and I can understand why you dont have these...it was a question with a few possible answers) I was looking for "by the time Gamma Rays emerge ffrom the photosphere, the energy is converted in infrared and optical radiation"
I clearly said right there that it was a question with a few possible answers but I was also looking for two specific answers and lurch got it right i just missed it because I was looking optical and infrared I think my eyes just skipped over heat and light. Your go LURCH
'K,
When/where on Earth will the next total Solar eclipse be?
23 November 2003, in *gasp* Antarctica.
http://sunearth.gsfc.nasa.gov/eclipse/TSE2003/TSE2003.html
- Warren
[>:)] [;)] (that's bull's eye) Chroot.
Think I might miss that one!
Your go.
Lurch posed a question about a solar eclipse, Chroot answered correctly, Lurch replied and said it was Chroot's turn to ask. But over 3 days have gone by and he has not posted. (People go on vacation or get busy at work etc.)
Nicool's rules cover this kind of situation in other explicit cases.
Originally posted by Nicool002
Hi guys! Most of you know how this works but for the newcomers:
The rules are this: someone will ask a question and if the question is not answered correctly within 3 days then a new question will be posted. If an answer to a question is posted and the person that posted the question does not respond to the answer within 2 to 3 days, then the first person to have answered the question will then be able to post their own question. HAVE FUN AND LEARN!
In the spirit, if not the letter, lets extend the rules of order to this case.
Lurch, you asked the last question. Why dont you give Warren one day to take his turn as asker, and if he doesnt then you take another turn and ask the next question?
Shadow, if Lurch is also on vacation, then its up for grabs. You could pose one
That seems equitable. C'mon, Chroot, let's hear from ya!
Originally posted by LURCH
That seems equitable. C'mon, Chroot, let's hear from ya! He teaches. It is summer time. Teachers do not work in the summer.
Originally posted by Labguy
He teaches. It is summer time. Teachers do not work in the summer.
Well SOMEBODY go!!!!
Originally posted by marcus
Well SOMEBODY go!!!! Ok, I have been away for a long time on "sick-leave". This might have been asked before, but I'm not going to read 20(x) pages to find out.
Question:
When and where (who is a bonus) were the first redshifts measured (taken?) indicating that the Andromeda galaxy was moving with respect to us?
My power's been out the last two days (maybe ya heard about it?).
Thanks for covering for me, Labguy.
schwarzchildradius
Aug17-03, 02:49 AM
Here's my guess:
Humason & Hubble,
Mt. Wilson Palomar observatory
1923
(andromeda is moving towards us)
correct?
Originally posted by schwarzchildradius
Here's my guess:
Humason & Hubble,
Mt. Wilson Palomar observatory
1923
(andromeda is moving towards us)
correct? No, none were correct except for the fact that Andromeda (galaxy) is moving towards us.
Lonewolf
Aug17-03, 08:11 AM
Vesto Melvin Slipher in 1912. Not sure where.
Lonewolf
Aug17-03, 12:45 PM
Lowell Observatory, Flagstaff, Arizona.
Originally posted by Lonewolf
Vesto Melvin Slipher in 1912. Not sure where.
Lowell Observatory, Flagstaff, Arizona. Where is correct and when is very close, but a tiny bit off...[8)]
Lonewolf
Aug17-03, 04:53 PM
He began the observations in 1909, but didn't make the observation until 1912.
Originally posted by Lonewolf
He began the observations in 1909, but didn't make the observation until 1912. That is certainly close enough to be CORRECT! Actually, the first redshift measurement results were determined in 1913.
You ask der next question.
Sorry guys, for derailing the Astronomy Q&A thread again... my weekends are for playing!
- Warren
Lonewolf
Aug18-03, 07:21 AM
Hmm, ok then.
What is stellar interferometry used for, and for a bonus point, what does it involve?
schwarzchildradius
Aug19-03, 04:03 AM
It's used to measure masses and atmospheric content of stars. Interferrometry involves the wave nature of light- start with I believe it was Huygens principle that every wave is the point source of infinitely many subsequent waves. So in a double-slit experiment you can see each slit behaves like a unqiue wave generator, and the waves can interfere with each other (causing alternating bright and dark bands). Not too surprising unless the source is a single photon, in which case the single photon will interfere with itself. But I gladly yield if anyone has a more topical answer.
Lonewolf
Aug19-03, 06:12 AM
It's used for a whole bunch of things, so I'll accept. It's also one of the few methods we have for measuring stellar radii that doesn't rely on a binary system.
Your turn to ask a question, schwarzchild. [:)]
schwarzchildradius
Aug21-03, 12:57 AM
if you've got more details about how that's done, I'd love to hear about it. but anyway,
what's the mass and atmospheric content of the star Sirius?
Lonewolf
Aug21-03, 09:01 AM
It uses a device called the Michelson interferometer. A diagram can be found at http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/michel.html
A.E. Michelson invented and used this interferometer into the 1920s to measure angular diameters of large, nearby stars (always some catch, huh?). In the device mirrors are widely seperated by many meters. The mirrors reflect the starlight to an ordinary focusing telescope, where the wavefronts from different parts of the star form a distinctive interference pattern.
The pattern depends upon the angle between the wavefronts from opposite limbs of the star. The intersection angle increases as the stellar radius increases. We know the angular diameter can be found by the small angle approximation a = D/d where a is the angular diameter in radians, D is the star's physical diameter, and d is the star's distance from Earth.
But the interferometer measured the angular diamter for us, so simple algebra tells us that D = ad (assuming we know the distance, of course).
schwarzchildradius
Aug22-03, 02:48 AM
trying to steal my thunder again I see. Well I say that YOU should ask the next question.
Perhaps this was said and I missed it but:
Interferometry is principally used to achieve the resolution of a very very large instrument while using only two moderate ones.
Two ten-meter telescopes in an interferometer, separated by 100 yards, have the effective resolving ability of an instrument with a 100 yard aperture.
- Warren
Lonewolf
Aug22-03, 04:59 AM
Dude, what's your problem?
if you've got more details about how that's done, I'd love to hear about it.
So I gave more details...
Chroot: Yeah, I forgot to mention that. Thanks for pointing it out.
Originally posted by schwarzchildradius on August 20
what's the mass and atmospheric content of the star Sirius?
that question has been sitting out there for a couple of days
anyone care to answer?
Originally posted by marcus
that question has been sitting out there for a couple of days
anyone care to answer?
Schw. Radius question is "mass and atmosphere makeup of the star Sirius"
I'll take a stab at answering---Sirius is a double star consisting of siriusA and siriusB of mass which IIRC is 2.3 and 1.0 solarmasses.
The smaller one is just a white dwarf and what we think of as the star is actually Sirius A of 2.3 solar masses.
Ionized hydrogen would surround such a massive hot star as Sirius A and provide the main constituent of its atmosphere
Back to you Schw. Radius: is that right?
But, the original question at hand was about an interferometer, and interferometers don't have squat to do with measuring a star's atmospheric content. They do that with spectrometers / spectrographs.
Lonewolf
Aug26-03, 06:15 AM
I think you're confused. The interferometry question was mine, which Schwarzchildradius answered. Schwarzchild's question was:
what's the mass and atmospheric content of the star Sirius?
Originally posted by Lonewolf
I think you're confused. The interferometry question was mine, which Schwarzchildradius answered. Schwarzchild's question was: Yes, I realize that, but Schwartzchild's original answer to your question included: "It's used to measure masses and atmospheric content of stars."
I'm not trying to go for an answer, or to change the order here, but the answer to your first question (interferometer) wasn't but 1/2 or less correct. Sorry to confuse the thread.
Originally posted by Labguy
But, the original question at hand was about an interferometer, and interferometers don't have squat to do with measuring a star's atmospheric content. They do that with spectrometers / spectrographs.
Would this work as a way out.
Lonewolf asked about interferometry and Labguy finally gave the correct answer (it increases resolution so you see greater detail, two telescopes some distance apart are linked so they act in a certain way like a larger telescope with more resolving power)
so it is now Labguy's turn
that seems logical to me, how does it seem to you Lonewolf
meanwhile Schw. Radius answered a different question "what is spectroscopy?" (it is a way to tell atmospheric composition by spectrum of light and can be used to infer the mass under certain assumptions)
And Schw. should be happy because he got a "free turn" and asked about Sirius and I answered his question. But we rescind that little digression and Labguy goes.
Originally posted by marcus
Would this work as a way out.
Lonewolf asked about interferometry and Labguy finally gave the correct answer (it increases resolution so you see greater detail, two telescopes some distance apart are linked so they act in a certain way like a larger telescope with more resolving power)
so it is now Labguy's turn
that seems logical to me, how does it seem to you Lonewolf
meanwhile Schw. Radius answered a different question "what is spectroscopy?" (it is a way to tell atmospheric composition by spectrum of light and can be used to infer the mass under certain assumptions)
And Schw. should be happy because he got a "free turn" and asked about Sirius and I answered his question. But we rescind that little digression and Labguy goes. No, it is not my turn. I never answered the question at all. Your quote given above that "(it increases resolution so you see greater detail, two telescopes some distance apart are linked so they act in a certain way like a larger telescope with more resolving power)" was given by someone else, not me!
I was only trying to get a bit more clear on the "rules of the game".
When someone asks a question, and an answer is incorrect or only slightly correct, is the person asking the question to say "close enough" or wait for a correct answer?? I'm not slighted at all, or trying to "get one" so I can ask a question. I'd rather read yours....[t)]
sometimes there are muddles the rules dont cover
we are in a muddle
I believe I answered both Lonewolf's Q about interferometry AND
S.Radius Q about the mass and atmosph of Sirius
so I will go
two stars differ in absolute magnitude by one step
(say one is magnitude 4 and the other is 5)
if one was comparing WATTAGES of the two stars
roughly by what factor would they differ?
Lonewolf
Aug26-03, 12:15 PM
Ah, ok. Sorry for the mix up, Labguy. I accepted it as an indirect application of interferometry. I was after more what we can use it for than what it does. We can perform spectroscopy on the signal we obtained from the interferometer, and thus obtain atmospheric content. It may seem a dubious decision, but I thought it'd be best to try and get some momentum going on the thread.
Whose turn is it now, then?
Originally posted by Lonewolf
Ah, ok. Sorry for the mix up, Labguy. I accepted it as an indirect application of interferometry. I was after more what we can use it for than what it does. We can perform spectroscopy on the signal we obtained from the interferometer, and thus obtain atmospheric content. It may seem a dubious decision, but I thought it'd be best to try and get some momentum going on the thread.
Whose turn is it now, then? Marcus' turn. He just asked one bottom of previous page.
Lonewolf
Aug27-03, 06:21 PM
So he did. I must have missed that.
Let A be the dimmer star and B be the brighter star.
Using the equation
mB - mA = 2.5 log 10 (IA/IB)
We cab obtain 10(mB - mA)/2.5 = IA/IB
From the question, we know mB - mA = 1. so
101/2.5 = 2.512 (approximate) = IA/IB
or 2.512*IB = IA
Hence, B is approximately 2.512 times as intense as star A. If we multiply both sides of the equation by 1 square meter, we are left with units of power. Hence, their wattages differ by approximately a factor of 2.512.
Thanks for answering, and rightly to boot!
Your go, Lonewolf.
Originally posted by Lonewolf
So he did. I must have missed that.
Let A be the dimmer star and B be the brighter star.
Using the equation
mB - mA = 2.5 log 10 (IA/IB)
We cab obtain 10(mB - mA)/2.5 = IA/IB
From the question, we know mB - mA = 1. so
101/2.5 = 2.512 (approximate) = IA/IB
or 2.512*IB = IA
Hence, B is approximately 2.512 times as intense as star A. If we multiply both sides of the equation by 1 square meter, we are left with units of power. Hence, their wattages differ by approximately a factor of 2.512.
schwarzchildradius
Aug28-03, 03:46 AM
Thanks for the answer, m, lonewolf, I don't think I have a problem, maybe overwork and procrastination (shudder), over.
Lonewolf
Aug28-03, 06:05 AM
It's cool.
Anyway, the question. Which two physical states are in equilibrium in a stable main sequence star?
Do you mean:
(1) Radiative repulsion, and
(2) Gravitational attraction.
??
Lonewolf
Aug28-03, 03:33 PM
That was indeed what I was thinking, Labguy, Your turn.
Originally posted by Lonewolf
That was indeed what I was thinking, Labguy, Your turn. Ok, another easy "Star" question.
In not too many words, why do lower mass stars have radiative cores and convetive outer layers, while massive stars have convective cores and radiative outer layers??
All this is sub-photosphere, of course.
Originally posted by Labguy
Ok, another easy "Star" question.
In not too many words, why do lower mass stars have radiative cores and convetive outer layers, while massive stars have convective cores and radiative outer layers??
All this is sub-photosphere, of course.
Just a geusse;
The more massive stars are denser, rendering normal raditation impossible near the core. So heat from the core is transported via convection to the outer layers, where density is lower (closer to the density of the Sun's inner layers), and radiative heat transfer can occur?
Originally posted by LURCH
Just a geusse;
The more massive stars are denser, rendering normal raditation impossible near the core. So heat from the core is transported via convection to the outer layers, where density is lower (closer to the density of the Sun's inner layers), and radiative heat transfer can occur? Good guess. That's it, 100%.
Your question.
Cool! I love it when I'm right.
Current estimates place the number of potentially deadly Near-Earth Objects (asteroids of 1km or larger who's orbit takes them close to Earth's orbital path) at around 1200. Approximately how many have been officially sighted and cataloged?
schwarzchildradius
Aug31-03, 07:37 AM
Current estimates place the number of potentially deadly Near-Earth Objects (asteroids of 1km or larger who's orbit takes them close to Earth's orbital path) at around 1200. Approximately how many have been officially sighted and cataloged?
Nasa reports 2449 NEOs on 8/28/03 here (http://neo.jpl.nasa.gov/neo/number.html).
The Planetary society reports 2,225 NEOs here (http://planetary.org/html/news/articlearchive/headlines/2003/neoworkshop.html) from 2/19/03.
But objects with absolute magnitude of +18.5 have diameters of 1km or larger, and there are 662 of them by nasa's report, also indicated on this graph:
http://neo.jpl.nasa.gov/gif/web_total.gif
Thor's hammer was supposed to be made of a fallen meteorite.
Originally posted by schwarzchildradius
...objects with absolute magnitude of +18.5 have diameters of 1km or larger, and there are 662 of them by nasa's report, also indicated on this graph:
http://neo.jpl.nasa.gov/gif/web_total.gif
Correct! The "1200" mentioned was originally at the high end of the estimates, but the rate at which new NEO's are being discovered indicates that the original estimates were low, and 1200 may even be an underestimate.
Ask away, Schwarzchild.
schwarzchildradius
Sep1-03, 08:52 PM
ok, Thursday is obviously "thor's day" in Norse mythology. What planet and for ex. crd. Nordic God is represented by "Tuesday"
The saxon Tiu, god of War.
- Warren
Originally posted by chroot
The saxon Tiu, god of War.
- Warren But what is the planet, first point of the question?
God of War -> Mars, the Roman equivalent of the Saxon Tiu.
- Warren
schwarzchildradius
Sep3-03, 01:13 AM
Yep. mars. I've also seen it spelled Tiw. I've got to find out what other days of the week represent planets in said culture.
Go!
What is the largest baseline optical interferometer currently in operation? What is its resolving ability?
- Warren
Originally posted by chroot
What is the largest baseline optical interferometer currently in operation? What is its resolving ability?
- Warren It is (I think) the Navy Prototype Optical Interferometer (NPOI).
Resolving ability is the same as could be obtained with a single 430 to 435 meter mirror, but I don't know what that is in arc seconds, since there are several different ways to figure potential resolving power. It is able to resolve the actual disks of some stars. Latest results I have heard were 9 milliarcseconds.
Hey Labguy,
Unfortunately it is not the NPOI, which has a maximum baseline of 437 meters. The longest baseline used to date at NPOI is only 64 meters.
There are interferometers with both larger maximum baselines, and larger in-use baselines. Would anyone like to give it a shot?
- Warren
Originally posted by chroot
Hey Labguy,
Unfortunately it is not the NPOI, which has a maximum baseline of 437 meters. The longest baseline used to date at NPOI is only 64 meters.
There are interferometers with both larger maximum baselines, and larger in-use baselines. Would anyone like to give it a shot?
- Warren Is the biggest one (correct answer) optical? Now I really want to know...[8)]
Labguy,
It is optical, indeed! :)
- Warren
Originally posted by chroot
Labguy,
It is optical, indeed! :)
- Warren
I believe CHARA on Mt Wilson is an opt. infer. array
with longest possible baseline 330 meters
Originally posted by chroot
Labguy,
It is optical, indeed! :)
- Warren
chara is indeed optical (as well as infrared)
and this Georgia State source
http://www.chara.gsu.edu/CHARA/array.html
gives the resolution as
200 micro-arcseconds
A strong contender would be European Southern's VLT
which is in operation in Chile, the results I have seen
from VLT were in the near infrared I believe, but will check
on the claimed resolution
marcus,
CHARA is correct, with a currently in-use baseline of 330 meters. This is also its maximum possible baseline.
In terms of maximum possible baseline, SUSI in Sydney, Australia is king with a possible 640 meter baseline. Only 140 meters have been used to date, however.
The NPOI is the first-born prince, with a maximum 437 meter baseline, 64 meters used to date.
Reference: Sky & Telescope magazine, May 2003.
marcus, your turn!
- Warren
OK I will ask. Since we are on optical inferometry, here's one:
The plan is to launch an optical interferometry instrument into space in 2009.
You have a lot of latitude in how you answer this question.
What is the name of the mission?
what is one other interesting detail about it? You might want
to pick one of the following suggestions:
1. what outfit is designing it
2. what is the intended orbit
3. how long is the main boom (this sort of determines max baseline)
4. how many interferometers are to be mounted on the boom
5. anything else especially interesting
No need to cover all these, I'm just looking for the name of the mission and some bit of detail about it
marcus,
It is called, quite creatively, the Space Interferometry Mission, or SIM.
It is being built by the JPL.
It will have a 10m baseline separating two 0.3m telescopes, and a maximum resolution of 1 microarcsecond.
The mission should be able to detect proper motion at a level of 2 microarcseconds per year, which corresponds to a movement of 10 m/s at a distance of 1000 pc!
From 10 pc away the Sun would appear to wobble 4.5 milliarcseconds, due mostly to the perturbations of Jupiter and Saturn. The SIM instrument will be able to detect wobbles 5,000 times smaller.
- Warren
Your go, Warren.
BTW what you say is exciting. I hope very much it is launched on schedule (2009) and works as designed.
The doppler exoplanet surveys measure RADIAL wobble speeds and this presumably measure up/down right/left wobble speeds and so one will get a much better grip on the masses of the
100 or so exoplanets found so far. Or so I would imagine.
Besides, one might be able to detect wobble due to smaller planets? I guess that is the idea
Ask one
Originally posted by chroot
marcus,
It is called, quite creatively, the Space Interferometry Mission, or SIM.
It is being built by the JPL.
It will have a 10m baseline separating two 0.3m telescopes, and a maximum resolution of 1 microarcsecond.
The mission should be able to detect proper motion at a level of 2 microarcseconds per year, which corresponds to a movement of 10 m/s at a distance of 1000 pc!
From 10 pc away the Sun would appear to wobble 4.5 milliarcseconds, due mostly to the perturbations of Jupiter and Saturn. The SIM instrument will be able to detect wobbles 5,000 times smaller.
- Warren
Nicool's 3 days have run out
Warren answered (with some nicely turned irony and
utter correctness) on 3 September
and I replied that it was his turn on 4 September
and it is now the 8th so it is incumbent on me to ask another Q
A Greek astronomer measured the distance to the moon (as a multiple of the earth's diameter) with surprising accuracy
his measurement was much more accurate than the earlier one by Aristarchus (the first to propose the heliocentric model) and
indeed it was, I believe, accurate to within 5 percent!
Who was this Greek and roughly when did he live?
For extra credit, indicate how he is believed to have arrived at the ratio of lunar distance to earth diameter.
If you know you are right, dont wait for confirmation, just go ahead and ask a question yourself
It was Hipparchus...(190-120 BC)
He estimated the earth-moon distance at 250,000 miles, less than 5 percent off.
He used an astrolabe...
Astrolabe, instrument used for measuring the positions of heavenly bodies. It consists of a circle or section of a circle, marked off in degrees, with a movable arm pivoted at the center of the circle. When the zero point on the circle has been oriented with the horizon, the altitude or azimuth of any celestial object can be measured by sighting along the arm.
...see ya...
Originally posted by bogdan
It was Hipparchus...(190-120 BC)
He estimated the earth-moon distance at 250,000 miles, less than 5 percent off.
OK bogdan, that is right and it is your turn to ask us a question.
Please dont wait. Just ask, this keeps the game going fast
and makes it more interesting.
BTW I have read somewhere that Hipparchus estimated the earth-moon distance as 30 earth diameters.
I do not know what that is in miles, but you are undoubtably right that it is around what you say and also that it is within a few percent of the true average distance.
I did not ask about his method (you say astrolabe) but some time we could investigate this. People may have different ideas.
Please ask!
Which romanian scientist had his name assigned to a crater from the moon ?
[:))]
Originally posted by bogdan
Which romanian scientist had his name assigned to a crater from the moon ?
[:))]
So the person is a man and a scientist. Hmmmm
I know of a Romanian woman diplomat who has a crater on Venus
named for her-----Elena Vacarescu
Maybe someone else knows
edit: Venus not moon
OK bogdan---the crater is named for the astronomer
Spiru Haret (1851 - 1912)
Now how would you like to ask a question that is more
about astronomical things (planets, stars, galaxies, CMB whatever you like) and less about names?
I would like to hear what question you think of.
I will give you my turn right now, if you ask something soon.
The answer is correct...your turn...I don't know any other questions...[:)]
Originally posted by bogdan
The answer is correct...your turn...I don't know any other questions...[:)]
Thanks bogdan, I'll take a turn.
as all here know, the bulk of the energy released in the sun's core results from fusing 4 protons into a helium nucleus
The question is: for each helium nucleus made, how many
neutrinos or antineutrinos are produced as a byproduct?
For extra credit: write down the series of reactions to show the stage in the process where neutrinos are produced
Lonewolf
Sep14-03, 05:16 PM
OK, I'll have a go.
The standard three stage PP chain is what I'll assume you're looking for.
1H + 1H -> 2H + ν + e+
1H + 1H -> 2H + ν + e+
2H + 1H -> 3He + γ
2H + 1H -> 3He + γ
3He + 3He -> 4He + 1H + 1H
So, there is one neutrino. ν, per helium nucleus. The answer is the same for the PPII and PPIII chains, although they have 2 neutrinos and 2 helium nuclei.
The standard three stage PP chain is what I'll assume you're looking for.
1H + 1H -> 2H + ν + e+
2H + 1H -> 3He + γ
3He + 3He -> 4He + 1H + 1H
So, there is one [correction TWO] neutrinos. ν, per helium-4 nucleus...
full credit if you make the indicated small correction to this otherwise correct answer
in the proton-proton chain the second reaction you list must happen twice (per helium nucleus formed)
so on a per-helium-nucleus-formed basis the first reaction also must occur twice
so TWO neutrinos are emitted for each 4He that is produced (reasonable since two protons changed to neutrons in the process)
You got the proton-proton chain, which is the main thing.
Take it as read that you have made that correction GO. It's your turn.
Dont even wait for confirmation.
Lonewolf
Sep14-03, 06:50 PM
What name is given to the prototype stars thought to be the start of a main sequence star of around one solar mass?
Is it "cheating" if I look the answer up? I know that I know it, I just cant think of it right now.
Lonewolf
Oct3-03, 04:50 PM
I don't think it'd be cheating at all. How would we know if you had anyway? [;)]
Originally posted by Lonewolf
What name is given to the prototype stars thought to be the start of a main sequence star of around one solar mass? T-Tauri (and I didn't look it up!), after the archetype star, T-Tau(rus) [duh]
Lonewolf
Oct3-03, 08:01 PM
It seems the thread has life again. Your turn, Nereid!
Originally posted by Nereid
T-Tauri (and I didn't look it up!), after the archetype star, T-Tau(rus) [duh]
I have to mention that the star's name is T Tauri, not T Taurus.
Name like Tauri, Orionis, Piscium, Andromedae and so on are Latin genitive (possessive) forms of the constellations' names. In other words, T Tauri means "the T star of Taurus."
You'll see a lot of different naming systems in use, like "52 Cygni" (Flamsteed numbers) and "delta Orionis" (Bayer designations) and "RR Lyrae" (variable-star designations) but they all use the Latin genitive name.
- Warren
Is Alpha, Beta, and Maxwell (all three)?
Hint: these three are different from the rest.
Bonus question: why are they different?
Originally posted by Nereid
Is Alpha, Beta, and Maxwell (all three)?
Hint: these three are different from the rest.
Bonus question: why are they different?
No idea, Nereid, nice question. There not all rays? :)[:))] I dunno...
Lonewolf
Oct8-03, 05:58 AM
Would they be on Venus, and the only features on the planet not named after a female? Maxwell Montes being the only feature named after a male.
Correct Lonewolf!
Want to try to bonus question? To re-state a little more clearly: Why were Alpha Regio, Beta Regio, and Maxwell Montes NOT given female names?
Anyway, it's your turn.
Lonewolf
Oct8-03, 04:01 PM
Hmm, Alpha Regio and Beta Regio were already there before. Maxwell Montes was named after James Maxwell. I'm not sure were Alpha and Beta weren't changed, or why Maxwell is the only male on a feminine planet. Perhaps you'd like to explain?
Anyway, question time, sticking to a Venus theme. Earth and Venus are very similar in mass and radius, and other features. Why is it then that their atmospheres are dramatically different?
Maxwell is the most prominent feature on Venus, as seen in radar. Venus was the first target for radar astronomers (after the Moon), starting in 1961. The Alpha and Beta regions were also observed, and confirmed early.
When the IAU got around to formalising the process of assigning names to solar system objects (1973?), they decided to give only female names to features on Venus (and there are other rules too, see the link). However, they made an exception for Maxwell, Alpha, and Beta, because these were already established.
http://astrogeology.usgs.gov/Projects/PlanetaryMapping/VenusMappers/AppendC.html
Originally posted by Lonewolf
Hmm, Alpha Regio and Beta Regio were already there before. Maxwell Montes was named after James Maxwell. I'm not sure were Alpha and Beta weren't changed, or why Maxwell is the only male on a feminine planet. Perhaps you'd like to explain?
Anyway, question time, sticking to a Venus theme. Earth and Venus are very similar in mass and radius, and other features. Why is it then that their atmospheres are dramatically different? The lighter gasses (originally) were lost due to higher temperatures because of proximity to the sun. This left the more dense gasses, leading to higher densities and the start of the (increasing) "greenhouse effect". The pressure of Venus' atmosphere at the surface is `90 atmospheres. It is composed mostly of carbon dioxide. There are several layers of clouds many kilometers thick composed of sulfuric acid. This dense atmosphere produces a run-away greenhouse effect that raises Venus' surface temperature by about 400 degrees to over 740 K (hot enough to melt lead). Venus' surface is actually hotter than Mercury's despite being nearly twice as far from the Sun.
(Answer stolen from internet, just to get specifics right)....[8)]
Lonewolf
Oct9-03, 08:18 AM
Cool, thanks Nereid. Labguy, you got the question right, so ask away!
Originally posted by Lonewolf
Cool, thanks Nereid. Labguy, you got the question right, so ask away! Ok, thanks.
Question(s):
(1)What is the name given to very large wisps of dust which cannot be seen in visible light, and are only ~15-30 K?
(2) Where are they found?
(3) What's the energy source of the (very low) tempreature?
Originally posted by Labguy
Ok, thanks.
Question(s):
(1)What is the name given to very large wisps of dust which cannot be seen in visible light, and are only ~15-30 K?
(2) Where are they found?
(3) What's the energy source of the (very low) tempreature? Ok, guys. You'll find the answer in the same subject/page as the "T-Tauri" answer at the top of this page!
Originally posted by Labguy
Ok, thanks.
Question(s):
(1)What is the name given to very large wisps of dust which cannot be seen in visible light, and are only ~15-30 K?
(2) Where are they found?
(3) What's the energy source of the (very low) tempreature?
(1) dust disks
(2) found around stars (e.g. T Tauri stars)
(3) the starlight heats them to the (very low) temperature
Originally posted by marcus
(1) dust disks
(2) found around stars (e.g. T Tauri stars)
(3) the starlight heats them to the (very low) temperature Close, but only #3 is correct.
Originally posted by Labguy
Close, but only #3 is correct.
well, there are "twisters"
?
1) interstellar cirrus
2) just about everywhere, but they're particularly noticable above the galactic plane
3) the ambient radiation, from stars, keeps them warmer than 2.7oK. As they're more or less transparent, they don't get a chance to cool down below this temperature, and they do absorb starlight, so they are warmer than the CMB.
They were first found (convincingly) by IRAS, and caused astronomers to go back and re-do an awful lot of 'absorption' calculations; previously interstellar absortion was assumed to be more or less uniform; the cirrus showed that is was highly irregular and patchy.
Originally posted by Nereid
1) interstellar cirrus
2) just about everywhere, but they're particularly noticable above the galactic plane
3) the ambient radiation, from stars, keeps them warmer than 2.7oK. As they're more or less transparent, they don't get a chance to cool down below this temperature, and they do absorb starlight, so they are warmer than the CMB.
They were first found (convincingly) by IRAS, and caused astronomers to go back and re-do an awful lot of 'absorption' calculations; previously interstellar absortion was assumed to be more or less uniform; the cirrus showed that is was highly irregular and patchy. That's right on all three points. It was an IRAS site I found this on, several years ago.
That makes it your question!
We've all read about the 'missing mass' or 'dark matter' which is supposed to exist in the galactic halo and bulge.
Red dwarf stars have been observed for a long time (the closest star - apart from the Sun - to us is a red dwarf); more recently even fainter stars - brown dwarfs - have been discovered. There must also be even smaller objects alone out in interstellar space, formed independently or ejected from a nascent planetary system.
How can we be so sure that these isolated faint objects - and others such as cold white dwarfs, neutron stars, even stellar-mass black holes - cannot comprise more than a small fraction of the halo or bulge missing mass?
Originally posted by Nereid
We've all read about the 'missing mass' or 'dark matter' which is supposed to exist in the galactic halo and bulge.
Red dwarf stars have been observed for a long time (the closest star - apart from the Sun - to us is a red dwarf); more recently even fainter stars - brown dwarfs - have been discovered. There must also be even smaller objects alone out in interstellar space, formed independently or ejected from a nascent planetary system.
How can we be so sure that these isolated faint objects - and others such as cold white dwarfs, neutron stars, even stellar-mass black holes - cannot comprise more than a small fraction of the halo or bulge missing mass? I would have to guess that you are looking for "galactic rotation rates", without a website to quote.
If you limit your question as asked, galactic bulge and halo, then any more significant (missing) mass in the bulge would cause stars, and spiral arms such as ours, to rotate faster than observed and faster than the "known mass" formulae would predict. If the halo had significantly more (missing) mass, and the bulge remained as observed, then (1) rotation would be slower than observed and (2) there would be more unstability and disruption in the disk-shaped spiral arms, in the case of a spiral galaxy.
In Elliptical galaxies the same would be true for rotation rates, without a central "bulge". As of now, the only increased rotation rates seen, in most galaxies, is the rapid increase near the cores indicating a massive BH in the center. M-87 is calculated to have a Billion Ms BH in the core, ours (Milky Way Galaxy) is ~1 million Ms.
These are all good points Labguy.
My fault for not asking the question sufficiently clearly.
First, I am only asking about the Milky Way halo and bulge, not other galaxies.
Next, you're right; it is the rotation curves which provide clear indications that the amount of mass is greater than what can be observed as luminous stars, gas, and dust.
What is the missing mass? There are good reasons for thinking it cannot be faint, unobserved objects such as (distant) red dwarfs, brown dwarfs, isolated planets, etc. I'm not counting rocks, pebbles, even smallish asteroid-sized bodies in this; nor 'big dust'.
My question is: why are we confident that these faint objects (brown dwarfs etc) cannot comprise more than a small fraction of the missing mass?
Brown Dwarfs; From a stellar formation graph and formula page:
These so- called "aborted stars" have a mass around .08Ms, and can't convert hydrogen into helium (Their cores never reach the threshold temperature for hydrogen burning.). The only energy radiated is due to gravitational contraction (Kelvin-Helmholtz contraction), which is why they are difficult to detect unless they are located near us. To determine if brown dwarfs contribute substantially to the dark matter mass, we need to estimate their number. We can use the Stellar Mass Function, F(M) (proportional to M-2.33), to extrapolate the number of brown dwarfs from the numbers of more massive stars. If F(M)dM = the number of stars with mass between M and M+dM, then the total mass contribution from brown dwarfs can be written as (M)(F(M))dM. The question is, how does F(M) behave for very low M? F(M) is proportional to M-2.33 only for main sequence stars, not for brown dwarfs. In general, giant clouds of gas and dust collapse, then fragment. The smaller fragments, which become K and M stars, are more abundant then the larger fragments, which become O and B stars. However, the amount of small fragments drops off right near the size needed to make brown dwarfs, limiting the number of brown dwarfs that could exist, and also providing an uncertainty as to that number.
If all the dark matter is composed of brown dwarfs, we would need one brown dwarf every 30 cubic ly of space, many trillions over the entire Milky Way. The number of known brown dwarfs and brown dwarf candidates in our section of the galaxy are exceedingly slim, however, making it unlikely that the density of the galaxy's brown dwarf population accounts for significant dark matter.
(Some original, some stolen text)
Extrapolating the SMF suggests that there would be too few such low mass objects.
What observations have been made that show there are insufficient non-luminous (or low-luminosity) massive objects to account for the missing mass?
Originally posted by Nereid
Extrapolating the SMF suggests that there would be too few such low mass objects.
What observations have been made that show there are insufficient non-luminous (or low-luminosity) massive objects to account for the missing mass? Is this an additional question, or required as an answer to your last question? It wasn't specified.
Here is the original question:
Originally posted by Nereid
...How can we be so sure that these isolated faint objects - and others such as cold white dwarfs, neutron stars, even stellar-mass black holes - cannot comprise more than a small fraction of the halo or bulge missing mass?
"be so sure" suggests that N. is looking for observational evidence, does it not? The model of stellar formation predicts mass distributions, but is a model---it might even have to be modified some day, who knows? Here is the same question rephrased:
Originally posted by Nereid
...What observations have been made that show there are insufficient non-luminous (or low-luminosity) massive objects to account for the missing mass?
This is a really interesting question which I dont think has been addressed yet in this thread. How come we're so sure---on the grounds of what observational evidence? How did we "see" the absence of billions of invisible small-size stars? Someone knows the answer here, I expect (besides Nereid, I mean). Let's hear from some of you others!
Originally posted by marcus
Here is the original question:
"be so sure" suggests that N. is looking for observational evidence, does it not? The model of stellar formation predicts mass distributions, but is a model---it might even have to be modified some day, who knows? Here is the same question rephrased:
This is a really interesting question which I dont think has been addressed yet in this thread. How come we're so sure---on the grounds of what observational evidence? How did we "see" the absence of billions of invisible small-size stars? Someone knows the answer here, I expect (besides Nereid, I mean). Let's hear from some of you others! Ok, then someone else can answer what Nereid "might have thought" instead of answering the question as I did.
I think questions, like my last 3-parter, should all be asked at one time instead of "comebacks" saying that "you're right, but now add more". I can't guess ahead as to what someone "might have meant", just answer the question as asked.
How can we be so sure that these isolated faint objects - and others such as cold white dwarfs, neutron stars, even stellar-mass black holes - cannot comprise more than a small fraction of the [clarification: Milky Way] halo or bulge missing mass? Galaxy rotation curves - including those of the Milky Way - are strong observational evidence that there is missing mass; they don't give any clues as to what it is, or is not.
Extrapolation of the SMF suggests that lower mass compact objects may not make up very much of the missing mass; it's not enough to give much confidence (as marcus correctly noted). It also doesn't address "cold white dwarfs, neutron stars, even stellar-mass black holes"
Hint: evidence for the existence of (non- or sub-)luminous massive, compact, halo objects
Originally posted by Nereid
Galaxy rotation curves - including those of the Milky Way - are strong observational evidence that there is missing mass; they don't give any clues as to what it is, or is not.
Extrapolation of the SMF suggests that lower mass compact objects may not make up very much of the missing mass; it's not enough to give much confidence (as marcus correctly noted). It also doesn't address "cold white dwarfs, neutron stars, even stellar-mass black holes"
Hint: evidence for the existence of (non- or sub-)luminous massive, compact, halo objects Originally posted by nightbat on the stellar astrophysics page; many observations:
Here are a few:
http://spaceflightnow.com/news/n0308/11halo/
http://arxiv.org/abs/astro-ph/0206126
http://aanda.u-strasbg.fr:2002/arti...092/aa1092.html
http://www.strw.leidenuniv.nl/information/normgal.html
http://xxx.lanl.gov/abs/astro-ph/9501068
http://www.star.bris.ac.uk/nam/abs_galaxy_dynamics.html
http://www.blackwell-synergy.com/li...03.06680.x/abs/
You will find many more on the internet. If that's enough, give the credit to nightbat, and notify him that he answered the question here. Emphasis on "You will find many more on the internet."
Originally posted by Nereid
We've all read about the 'missing mass' or 'dark matter' which is supposed to exist in the galactic halo and bulge.
Red dwarf stars have been observed for a long time (the closest star - apart from the Sun - to us is a red dwarf); more recently even fainter stars - brown dwarfs - have been discovered. There must also be even smaller objects alone out in interstellar space, formed independently or ejected from a nascent planetary system.
How can we be so sure that these isolated faint objects - and others such as cold white dwarfs, neutron stars, even stellar-mass black holes - cannot comprise more than a small fraction of the halo or bulge missing mass?
I'll hazard a guess: by "microlensing" observations----which yield an upper bound on the abundance of massive dark objects
the original question is pretty clear---we take for granted that much of the Milkyway mass is dark in the sense that we can't detect radiation from it-----either it is stuff that shines too faintly to detect or that doesnt shine at all. That is not at issue here.
The question is, how do we know (based on what observations do we know) that this invisible mass does not consist of objects in the large-planet-to-small-star range?
The stellar-mass-distribution function is just an empirically arrived at thing with limited predictive power so we cant use it to exclude the possibility that at some time in Milkyway history a lot of un-shining massive objects formed (Nereid mentioned stellar-mass black holes, cold white dwarfs etc)
But people can actually see invisible massive objects by their gravitational-lens effect as they pass in the foreground of visible objects----microlensing. So they have been able to estimate how many dark massive objects there are.
Yep, marcus is right.
Nearly a decade ago now several long term research projects got underway to detect gravitational lensing of background stars. They had several objectives, one of which was to estimate the space density of MAC(H)Os - massive compact (halo) objects. As one of these passes between the line of sight between us and a distant star, it lenses the star's light, and we see a characteristic brightening then fading.
Although it's no longer maintained, the MACHO consortium's homepage (http://wwwmacho.mcmaster.ca/) has good background (including a photo of the Mt Stromlo observatory, now destroyed [:((] ). To read a particular paper - from the publications list - copy and paste the title into Google; to look up the results from other lensing projecs (e.g. OGLE), type the project name into Google.
There have also been dedicated HST searches for red dwarfs in the halo - far fewer were found than would be needed to comprise a significant fraction of the missing mass.
(you can do a simple calculation for yourself - given proxima centuri, distance ~ 4 ly, mag ~11, the nearest red dwarf - how far away would a similar red dwarf be if its observed mag were 25?)
Your turn marcus.
Originally posted by Nereid
Your turn marcus.
Sure. What is the name of Siobahn Morgan's dog?
[prof Morgan was the first person I know of to post an online
cosmology calculator that actually gives the recession speed
of an object of known redshift at the time it emitted the light
we are receiving from it.
http://www.earth.uni.edu/~morgan/ajjar/Cosmology/cosmos.html
be sure to put in 0.27 for matter density and 0.73 for Lambda or cosmological constant, then it will compute recession speeds and distances from whatever redshift you give it
Siobahn's homepage is
http://www.earth.uni.edu/smm.html]
For extra credit, name 3 of Siobahn's favorite superhero comic book characters.
For optional, entirely voluntary [6)] additional extra credit: a quasar has been observed at redshift 6.4---with what speed was the quasar receding from us when it emitted the light?
Originally posted by marcus
Sure. What is the name of Siobahn Morgan's dog?
[prof Morgan was the first person I know of to post an online
cosmology calculator that actually gives the recession speed
of an object of known redshift at the time it emitted the light
we are receiving from it.
http://www.earth.uni.edu/~morgan/ajjar/Cosmology/cosmos.html
be sure to put in 0.27 for matter density and 0.73 for Lambda or cosmological constant, then it will compute recession speeds and distances from whatever redshift you give it
Siobahn's homepage is
http://www.earth.uni.edu/smm.html]
For extra credit, name 3 of Siobahn's favorite superhero comic book characters.
For optional, entirely voluntary [6)] additional extra credit: a quasar has been observed at redshift 6.4---with what speed was the quasar receding from us when it emitted the light? Millie
Batman
Green Lantern
Aquaman
I have no "direct observations" to back this up.
Originally posted by Labguy
Millie
Batman
Green Lantern
Aquaman
I have no "direct observations" to back this up.
[:D] Your go, Labguy, right on all counts
I see you did not opt for the additional extra credit---a redshift 6.4 quasar was receding at 3 times c when it emitted the light we receive from it----or more accurately, by Siobahn's calculator, 2.88 c.
Originally posted by marcus
[:D] Your go, Labguy, right on all counts
I see you did not opt for the additional extra credit---a redshift 6.4 quasar was receding at 3 times c when it emitted the light we receive from it----or more accurately, by Siobahn's calculator, 2.88 c. No, no optional. I was in a hurry to get one right by today (October 14th where I am). Because:
Question: (Easy one)
October 14th is the anniversary of an important event that led to the more rapid development of our ability to make accurate astronomical observations.
(1) What was the event and what year?
(2) How did that event develope into our making better astronomical observations?
Labguy;
aka, The Kid
Originally posted by Labguy
No, no optional. I was in a hurry to get one right by today (October 14th where I am). Because:
Question: (Easy one)
October 14th is the anniversary of an important event that led to the more rapid development of our ability to make accurate astronomical observations.
(1) What was the event and what year?
(2) How did that event develope into our making better astronomical observations?
Labguy;
aka, The Kid No takers? Too easy? Too obscure?
It's about October 14th.
Labguy
Originally posted by Labguy
No takers? Too easy? Too obscure?
It's about October 14th.
Labguy
http://www.wisdomportal.com/History/HistoryDates-1.html
October 14 is the anniversary of the Battle of Hastings.
It was on October 14, 1066 that William the Bastard crossed the
channel and beat the Saxons and became known as "the Conqueror".
Be it known by these tokens that according to Nicool's Rule, if no satisfactory answer is provided within "two or three days" (Nicool left it vague) then another question will be posed.
you mean some boring old comet???
Originally posted by Nereid
you mean some boring old comet???
Nereid, there was a picture of it on the tapestry, was there not?
But I was just being goofy----its up to you to come up with a
serious answer if possible
Originally posted by marcus
Nereid, there was a picture of it on the tapestry, was there not?
But I was just being goofy----its up to you to come up with a
serious answer if possible This particular October 14th had direct implications on the advancement of spaceflight and, therefore, astronomy.
HINT:
Check between WWII and 1950! Anyone can "come up with the answer".
Labguy
Chuck Yeager broke the sound barrier on October 14, 1947. I found this within five minutes of searching history archives when you first posted your question, but I had no idea you were looking for this. I thought it was something directly relating to astronomical phenomena. Yes, I understand your point that breaking the sound barrier was important in the development of space technology and thus science equipment like Hubble, but I really feel it's better categorized as an achievement of engineering, not of astronomy...
- Warren
Originally posted by Labguy
Check between WWII and 1950! Anyone can "come up with the answer".
Labguy
Oh no, I fear you are mistaken! The battle of Hastings (which was when Halley's comet was seen) was in 1066! This is not between WWII and 1950 at all!
The periodic observation of Comet Halley, when Halley put two and two together and computed the orbit, was a great benefit to astronomy and helped to bring about a vast improvement in, as you so rightly say, "the accuracy of our observations".
Your only mistake, I feel, is about when the Battle of Hastings actually occurred!
Originally posted by chroot
Chuck Yeager broke the sound barrier on October 14, 1947. I found this within five minutes of searching history archives when you first posted your question, but I had no idea you were looking for this. I thought it was something directly relating to astronomical phenomena. Yes, I understand your point that breaking the sound barrier was important in the development of space technology and thus science equipment like Hubble, but I really feel it's better categorized as an achievement of engineering, not of astronomy...
- Warren That is the answer. I would think that this event led to the (more rapid) development of aircraft and space technology, let us develope orbital and launch capability, which then "put up" the HST, IRAS, COBE, Probes to the planets, Mars global surveyor, et al and on. Just the HST and IRAS alone "jumped" our astronomical knowledge way above anything we could have known by now from Earth-bases telescopes. Anyway, that was the question and thought behind it.
You ask next question.
Labber
What is the largest supernova remnant visible in the night sky, in terms of solid angle subtended?
- Warren
Originally posted by chroot
What is the largest supernova remnant visible in the night sky, in terms of solid angle subtended?
- Warren That should be the Vela SNR. It is described as a circle of ~8.3°, or 72 parsecs in diameter at its current distance measurement
Labguy
I sure hope I'm not misinterpreting anything I've read. The Vela SNR is not the largest. There's one much, much larger.
- Warren
Originally posted by chroot
I sure hope I'm not misinterpreting anything I've read. The Vela SNR is not the largest. There's one much, much larger.
- Warren I think you are right. It seems that "The Vela Supernova Remnant is just one part of the Gum Nebula (Gum 12), a complex shell of faint nebulous filaments. It is the largest object in the sky, apart from the Milky Way, and measures 36 degrees across! It is a half shell of filaments similar to the Veil Nebula in Cygnus, and was discovered by Australian Astronomer Colin S. Gum in 1952." I hope that GUM NEBULA is right. At 36 degrees, I can't imagine a larger one, in terms of angle subtended.
BUT, the same source says that the Vela SNR is a part of the Gum nebula, but that the Gum nebula itself is not a SNR. (?) It seems that the Vela SNR material collided with, and excited atoms in, an already existing large filament of dust and gas which, in itself, was not a SNR. So I'm confused a bit too. Is it Gum at 36 degrees or Vela at 8.3 degrees? I did find lists of much larger SNR's, but this was by size, and in other galaxies. Their angle subtended is less than 8.3*.
In other words, I have no idea!...[8)]
Labguy
To the best of my knowledge, Labguy, that's it. ;)
Your turn.
- Warren
Hmm Labguy do you have a copy of Burnham's or something authoritative about the Gum? The stuff I'm reading is all online, and it all seems to say that the Gum nebula is indeed an SNR.
- Warren
Originally posted by chroot
Hmm Labguy do you have a copy of Burnham's or something authoritative about the Gum? The stuff I'm reading is all online, and it all seems to say that the Gum nebula is indeed an SNR.
- Warren One source stated what I said in my last post, but others I can find by a search say that the Gum Nebula itself is indeed an "SNR" as you just stated. But, several sources said that the Gum nebula is not well understood, cannot find the date or ~time of "the" supernova. Others say that the Gum nebula is an SNR, but an accumulative SNR from many different supernovae occuring over thousands (many) of years. This might account for its huge size.(?)
In my last post I said: "So I'm confused a bit too. Is it Gum at 36 degrees or Vela at 8.3 degrees? I did find lists of much larger SNR's, but this was by size, and in other galaxies."
Either way, both were mentioned with information from several sources (now), so I'll have to say it is the Gum Nebula at 36-40 degrees subtended. Thanks for the exceedingly HUGE help-hint....[8)]
Labbby
Originally posted by chroot on October 16, 3:15 PM
To the best of my knowledge, Labguy, that's it. ;)
Your turn.
- Warren
Marcus posted:
quote:
--------------------------------------------------------------------------------
Originally posted by chroot on October 16, 3:15 PM
To the best of my knowledge, Labguy, that's it. ;)
Your turn.
- Warren Ok, I'll go since we danced around between both the Vela SNR and/or the Gum nebula.
Question:
I can show at least six (6) clear differences between "properties" of Galactic Clusters and Globular Clusters..
Name at least four (4) differences between these types of clusters. Obviously, any property of one needs to match the same property of the other, showing a difference. Example:
Galctic Cluster:
Property:
1. ------------------(property #1)
2. ------------------
3. ------------------
4. ------------------
Globular Cluster:
Property:
1. ------------------(#1, same as above, but different)
2. ------------------
3. ------------------
4. ------------------
Any other format is fine as long as it lists the same "property", but shows the difference. A "property" might be something like "Chemical composition", but that isn't on the list except as an explanation for a more significant difference.
Labguy
Ok. Two days have come and gone, so here is the answer(s):
Galactic Clusters:
Clusters Located in the disk of our galaxy
Young stars (Population I)
~50 to 10000 stars per cluster
~10 pc in diameter
Star density ~ 0.1 to 10 stars/pc^3
Irregularly shaped
~1000 in the Galaxy
Examples: Pleiades, Hyades, Praesepe
Globular Clusters:
Clusters Located in the halo of our galaxy
Old stars (Population II)
~10^4 to 10^6 stars per cluster
~20-100 pc in diameter
Star density ~ 0.5 to 10^3 stars/pc^3
Roughly spherically shaped
~150 in the Galaxy
Examples: M3, M13 (Hercules), M15
Make it someone else's turn.
Labguy
What mystery about the solar system has the recent HST-based work on the EKB merely served to heighten?
For extra credit, why is Edgeworth's name usually left out when the EKB is mentioned in the popular press?
[edit: Edgeworth as the correct 'E']
Originally posted by Nereid
What mystery about the solar system has the recent HST-based work on the EKB merely served to heighten?
For extra credit, why is Eddington's name usually left out when the EKB is mentioned in the popular press? It took Hubble's ability to resolve as small as 40 milliarcseconds (nothing else can do that) to confirm the size of Quaoar, A KBO (Kuiper Belt Object). Quaoar is about 4 billion miles away from Earth, well over a billion miles farther away than Pluto.
In the mid of last century, the existence of a ring of small planetisemals was first suggested by the astronomers Kenneth Edgeworth (1880-1972) and Gerard P. Kuiper (1905-1973), but the first discovery of an Edgeworth-Kuiper belt object was not until 1992. By now, over 550 KBO's are known.
Hubble's resolution will help confirm the size/mass of more KBO's.
Labguy
Hmm, this is a more difficult question than I'd expected, so here's the answer:
Six astronomers used the new ACS on Hubble to search for faint EKB objects in a small patch of sky, to get an estimate of the size-number distribution of such objects at sizes down to ~15 km. They found far fewer than expected from an extrapolation of the distribution at larger sizes, previously obtained from other studies. While they are careful to point out that their results constrain the size-distribution of the scattered disk (which they call the "excited" objects) only weakly, their finding deepens the mystery of the origin of short period comets*.
Oh, and by the way, Pluto is almost certainly the largest EKB object!
Here's a Hubble site link:
http://hubblesite.org/newscenter/archive/2003/25/
And an arXiv preprint (click on one of the Full-text links to get the full paper):
http://arxiv.org/abs/astro-ph/0308467
*"The KBO populations are far sparser than theoretical estimates of the required precursor population for short period comets." in their own words
Someone else's turn.
OK, an easy one: what are the nearest (known) red dwarf and white dwarf stars?
Bonus question: approx how many red dwarf and white dwarf stars within 100 ly of us remain to be discovered?
Proxima Centauri and Sirius B. Sirius being the white dwarf
OK Shadow, your turn.
As you'll see by looking at earlier posts, the 'rules' are that you post an astronomy question, and the first person to get the answer gets a turn. Of course, as the person who posted the question, you get to decide if the answer is correct or not!
Happy posting!
Originally posted by Nereid
OK Shadow, your turn.
As you'll see by looking at earlier posts, the 'rules' are that you post an astronomy question, and the first person to get the answer gets a turn. Of course, as the person who posted the question, you get to decide if the answer is correct or not!
Happy posting!
there is also (see Nicool's original post) a kind of
vague "three-day rule"
which is that if it is somebody's turn and they dont take it
after a reasonable time like three days
then the last person active gets to go
and if they dont then its up for grabs
so if Shadow doesnt come up with a question in a day or so
then Nereid can pose another one
which, if Nereid doesnt, others can do also
but please folks one at a time
also if you happen to ask a question that is too hard for the
rest of the players and nobody answers
then not to worry because after a while (the "three day" idea again) you can just say what the
answer is and ask an easier one
or, if you dont, someone else can.
this means that its not the end of the world if someone asks something nobody can answer
Shadow yoohoo where are you?
Oh wow i havent been around and completely forgot about this. sorry!
ok heres the question
Give at least two ways (theories) on how to divert an asteroid on a collision course with earth.
Originally posted by Shadow
Give at least two ways (theories) on how to divert an asteroid on a collision course with earth.
Get all the people in China to jump off a chair at the same time to move the Earth out of the way. kidding!
strategies: (thanks to Robert Burnham, Astronomy Mag. Dec 02)
(1) Nuke explosion next to or just below the surface to vaporize a portion. Offgases change the orbit.
(2) Direct impact with fast, massive, inert object to deflect its course.
(3) Use the Yarkovsky effect to generate thrust (one way is to dump lots of reflective material on one side (chalk, ice) and light-absorptive material on the other (charcoal, soot).
(4) If it has a high ice content...use giant mylar mirror to focus sunlight on it. Jetting vapors would alter the orbit.
Lots of lead time is the key to success.
correct! I have to come here more often... seems like i missed quite a bit. Your turn Phobos!
Where has my brain been the past few weeks? No wait...here's the question...
What was the nickname that Carl Sagan (et al.) gave to the solid hydrocarbon snow that is likely to be found on the surface of Saturn's moon Titan? (they recreated it in the lab by zapping a vessel with similar atmospheric gasses)
cragwolf
Mar30-04, 08:55 PM
here is my guess: tholin
(my post has to be longer than ten words for some reason.)
SchrodingersCat
Mar31-04, 12:33 AM
Titan tholins?
schwarzchildradius
Mar31-04, 03:18 AM
Manna from Heaven?
cragwolf got it first....your turn [:)]
(welcome to Physics Forums, SchrodingersCat!)
cragwolf
Mar31-04, 10:29 PM
Here is my question (I hope it hasn't been asked before):
Cepheid variable stars pulsate in a very regular and stable manner. What causes this pulsation? Please include in your answer an explanation of why the Sun doesn't pulsate in this way.
schwarzchildradius
Mar31-04, 11:07 PM
Here is my question (I hope it hasn't been asked before):
Cepheid variable stars pulsate in a very regular and stable manner. What causes this pulsation? Please include in your answer an explanation of why the Sun doesn't pulsate in this way.
Typically there's some Helium in the atmosphere of the star. As radiation ionizes the Helium, the atmosphere expands and the star brightens. The Helium cools and loses its ionization, falling back toward the center of the star, causing it to dim a bit. The correlation between the period of the brightening and the intrinsic brightness of the star allows you to estimate the absolute distance.
cragwolf
Apr1-04, 01:40 AM
Typically there's some Helium in the atmosphere of the star. As radiation ionizes the Helium, the atmosphere expands and the star brightens. The Helium cools and loses its ionization, falling back toward the center of the star, causing it to dim a bit.
Assuming that's true ( :biggrin: ) , why does the atmosphere expand when the Helium ionizes (and contract when it recombines). Also why doesn't this happen to the Sun?
schwarzchildradius
Apr1-04, 05:15 AM
When Helium is ionized it becomes opaque to radiation, the radiation causes the envelope of Helium to expand as it is heated. When the Helium de-ionizes it becomes transparent to radiation & contracts. The sunis not massive enough to be a Cepheid variable
cragwolf
Apr1-04, 11:42 AM
Well done. Your turn to ask a question.
schwarzchildradius
Apr1-04, 08:41 PM
Super! Here it is: What's the name of the Egyptian mathematician who first calculated that the world must be round, and what visible clue did he use to find this out? (If you've seen "Cosmos" by Sagan this one is easy)
SchrodingersCat
Apr2-04, 10:14 AM
Eratosthenes observing the noontime Sun.
SchrodingersCat
Apr2-04, 10:32 AM
Sorry, that was for measuring the circumferance. I belive it was Aristotle using Earth's shadow on the Moon during a lunar eclipse
schwarzchildradius
Apr2-04, 08:26 PM
Naw you were right with the first one. your go/
schwarzchildradius
Apr4-04, 05:29 AM
By the way, I've run across a passage from Cosmos where he names the hydrocarbon sludge
"Such experiments were first performed in the early 1950's by Stanley Miller, then a graduate student of the chemist Harold Urey. Urey had argued compellingly that the early atmosphere of the earth was hydrogen-rich, as is most of the Cosmos; that the hydrogen has since trickled away to space from Earth, but not from massive Jupiter... After Urey suggested that such gasses be sparked, someone asked him wat he expected to make in such and experiment. Urey replied, "Beilstein." Beilstein is the massive German compendium in 28 volumes, listing all the organic molecules known to chemists."
So what is "Tholin?"
So what is "Tholin?"
The hydrocarbon sludge on Saturn's moon Titan (discussed briefly in Sagan's sequel to "Cosmos" called "Pale Blue Dot")
SchrodingersCat
Apr4-04, 03:06 PM
Sorry It took so long to post this. Anyhow, here is the question. What supports a Neutron Star from collapse? (Do not just name the type of support but also explain it)
Sorry It took so long to post this. Anyhow, here is the question. What supports a Neutron Star from collapse? (Do not just name the type of support but also explain it) It is called "neutron degeneracy pressure". Neutron degeneracy is actually at a "core", since a neutron star is not all neutrons from the surface down, they (the stars) actually have a structure. the "mass" is the gravitational mass and is distinct from the baryonic mass, because the gravitational redshift of a neutron star is so great, the gravitational mass is about 20% lower than the baryonic mass.
When the density towards the inner core exceeds the nuclear density of 2.8x10^14 g/cm^3 by a factor of 2 or 3, exotic stuff might be able to form, like pion condensates, lambda hyperons, delta isobars, and quark-gluon plasmas. To me, most likely quark-gluon plasmas. So it is essentially the strong nuclear force in the deep interior preventing further collapse.
The interesting structure can be seen at: http://www.lsw.uni-heidelberg.de/users/mcamenzi/NS_Mass.html
Beyond that (simple explanation) it gets a bit complicated, and several models all work to show one form or another of "condensate" to prevent collapse without additional mass being added.
SchrodingersCat
Apr6-04, 07:52 AM
Well Labguy, it is your turn to ask a question. :wink:
Well Labguy, it is your turn to ask a question. :wink: Hey, S-Cat. I'm going to have to pass and let you ask another because I'll be on a trip and "away from a computer" for several weeks.
Ask another Q or pass to the last guy, etc. I'll be back for more in a few weeks.
SchrodingersCat
Apr7-04, 03:45 AM
In a cluster most stars have the same main sequince turn off point. However, there is another class of bluer stars that seem to be younger. The formation of these stars is uncertain but one of the theories is that they form when one star accretes matter from another star. Another possible formation scenario is the collision of one or more stars. If you look at the HR diagram for some clusters you will see the that most stars lay along the same main sequice and horizontal branch and whatnot but there is also often a seperate main sequince and horizontal branch (containing much less stars) for these strange objects.
What is the name of these objects, who was the first to discovere them and which globular cluster were they first found in?
SchrodingersCat
Apr9-04, 06:18 PM
Need a hint?
The person who discovered these objects was one of the co-discoverers of the quasar.
I guess if no one gets this with in the next couple of days I should post a new question? Or should I just try and think of more hints?
I believe these stars are termed "blue stragglers". They are stars in a globular cluster that are on the main sequence, but to the left of the turn-off point. This means that they should have already burned out into red giants, but haven't yet evolved. In some sense, they appear to be normal stars, but younger than the other stars in the cluster. They were first found in M3 by Sandage in 1953 (had to look his name and date up, couldn't remember them OTOH).
A coalescion of two smaller stars that have not yet evolved would produce a such a blue straggler because, of course, more massive stars tend to be bluer and brighter. Mass transfer from a nearby binary "overflowing" its Roche lobe works in the same way to shift a star left-wards on the main sequence.
Of course, these stars exist everywhere in the universe, but they are only easily identified as BSs in a globular cluster due to the proximity of stars of precisely the same age.
Good question, and mine will follow as soon as I can think of one. :biggrin:
Okay. Is it possible for a pulsar to spin so fast that it breaks up, and are there any known mechanisms for this? How fast (% of c) does the surface need to be going for this to happen?
If it spins so fast... why would it form in the first place?
SchrodingersCat
Apr11-04, 08:17 AM
Congrats on your answer to the last question Cecil.
Well mv^2/r=GMm/r^2 (centripital force= gravity) sets this limit. Thus v=Sqrt[GM/r] where G is the gravitational constant. I will assum M~1.4Msolar and r=10km.
=> ~0.46c.
I belive another factor which actually occurs at lesser velocities involving gravitational radiation. This effect seems to not allow the pulsar to spin so fast and thus saves it from flying apart in such a manor.
Nice question!
FZ++ Well, it spins fast to conserve angular momentum. Neutron Stars have radii of 10e5 cm. Our sun for example has a radius on the order of 1e10 cm. This dramatic loss in radius causes it to have to spin faster in order to acheive conservation of angular momentum.
SchrodingersCat
Apr15-04, 12:37 AM
Sorry I have not been around for a while. I have been really busy. Umm... Cecil, do you want to ask another question? I am afraid I prolly will not be posting for a bit as I have a lot of other stuff to attend to.
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