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

[axeeonn]

Hmmm, I think I will pass on that integral. ;)

chroot, you're up.

 

[cragwolf]

[zz)]

Hey chroot!

CHROOT!



Sorry to wake you up, but it's your turn to ask a question.

 

[marcus]

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?

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 interrupt the game, but would appreciate any help.

 

[chroot]

Originally posted by cragwolf
[zz)]

Hey chroot!

CHROOT!



Sorry to wake you up, but it's your turn to ask a question.

Sorry folks. I have been asked to leave this forum by Greg Bernhardt, the owner.

- Warren

 

[Labguy]

Originally posted by chroot
Sorry folks. I have been asked to leave this forum by Greg Bernhardt, the owner.

- Warren

Why?

 

[Greg Bernhardt]

Please use the PM system for personal and off-topic questions.

Thanks

 

[Nicool003]

Hey guys sorry I haven't 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...

 

[marcus]

Axeeonn's turn to ask

Originally posted by Nicool003
Hey guys sorry I haven't 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]

Ok, let's try to get this rolling agian...

How many kilograms of hydrogen is converted to He in our sun per second?

 

[damgo]

6e11 kg/s.

 

[Labguy]

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]

Damgo's turn, go go go!

 

[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)

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?

 

[marcus]

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.

 

[damgo]

^^^ yup... it's pretty amazing, no? Your go!

 

[marcus]

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?

 

[Labguy]

Answer is 2GM/c^2 for radius.

0.4 inches; = 1cm is radius.

Double that for your diameter question; 2cm diameter.

 

[marcus]

Labguy
Right you are! Your go.

 

[Labguy]

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?

 

[marcus]

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.

 

[Labguy]

(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...

 

[marcus]

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]

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

 

[marcus]

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]

Originally posted by marcus
Leo was where, and so they did! Your go.

ok thanks

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)

 

[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.

 

[marcus]

Originally posted by screwball
ok thanks

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.

 

[screwball]

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

 

[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?

 

[marcus]

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.