Astronomy Trivia Challenge: Can You Answer These Questions About the Night Sky?

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The Astronomy Trivia Challenge engages participants in a question-and-answer format about various astronomical topics. Participants take turns asking questions, with rules stipulating time limits for responses to keep the game moving. Discussions cover a range of topics, including the brightest stars, celestial bodies, and cosmic phenomena like supermassive black holes and cold dark matter. Players share knowledge and insights, often referencing their studies or experiences in astronomy. The thread fosters a collaborative learning environment while maintaining a fun and competitive spirit.
  • #101
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.
 
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  • #102
Damgo's turn, go go go!
 
  • #103
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?
 
  • #104
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[/URL]

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.
 
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  • #105
^^^ yup... it's pretty amazing, no? Your go!
 
  • #106
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?
 
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  • #107
Answer is 2GM/c^2 for radius.

0.4 inches; = 1cm is radius.

Double that for your diameter question; 2cm diameter.
 
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  • #108
Labguy
Right you are! Your go.
 
  • #109
Originally posted by marcus
Labguy
Right you are! Your go.
Cool. Ok, more Black Hole stuff.

The "original" Schwarzschild 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 Schwarzschild 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?
 
  • #110
Originally posted by Labguy
Cool. Ok, more Black Hole stuff.

The "original" Schwarzschild 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 Schwarzschild 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.
 
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  • #111
(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...:smile:
 
  • #112

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...:smile: [/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?
 
  • #113
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
 
  • #114
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.
 
  • #115
Originally posted by marcus
Leo was where, and so they did! Your go.

ok thanks:smile:

i don't 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)
 
  • #116
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.
 
  • #117
Originally posted by screwball
ok thanks:smile:

i don't 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.
 
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  • #118
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

but cragwolf buzzed in first so its your turn buddy
 
  • #119
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?
 
  • #120
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.
 
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  • #121
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
 
  • #122
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.
 
  • #123
Oops, I didn't see this last page, Marcus. OK, Labguy is right, and can ask the next question.
 
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  • #124
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".
 
  • #125
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.
 
  • #126
Correct.

Your go.
 
  • #127
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?
 
  • #128
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"
 
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  • #129
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
 
  • #130
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?
 
  • #131
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...
 
  • #132
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...:smile:
 
  • #133
Originally posted by Labguy
(1) No.
(2) No.
(3) No.

Keep trying folks...:smile:

Hubble was wrong? the date too? ok
 
  • #134
Originally posted by screwball
Hubble was wrong? the date too? ok
Yes, no Hubble, no 1929. No comment on the location...:smile:
 
  • #135
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.
 
  • #136
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.
 
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  • #137
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.
 
  • #138
lol u beat me to it
 
  • #139
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.
 
  • #140
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.
 
  • #141
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.
 
  • #142
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.
 
  • #143
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
 
  • #144
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.
 
  • #145
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


Anyways chroot it is your question.
 
  • #146
Does chroot ask a question?
 
  • #147
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 haven't been to this thread in a while either.
 
  • #148
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!
 
  • #149
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?
 
  • #150
BTW here's a freeware orrery and lunar eclipse predictor that works pretty well..
astropro orrery
 

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