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

[chroot]

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

 

[Labguy]

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.

 

[marcus]

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.

(Earth's 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 don't think being 50 percent off the mark is good
enough in this league.

 

[chroot]

Like I said, 370-390 km/s in the direction of Leo.

- Warren

 

[marcus]

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!

 

[chroot]

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

 

[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.. 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..[?]

 

[marcus]

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.. 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!

 

[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.

 

[marcus]

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.

 

[marcus]

first meeting of the IAU, in 1922, named them

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

 

[chroot]

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

 

[marcus]

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.

 

[chroot]

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 r_s = 2GM/c², or 5.9 billion meters. The gravitational acceleration is a_g = 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

 

[marcus]

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 r_s = 2GM/c², or 5.9 billion meters. The gravitational acceleration is a_g = 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 I am not
a regular visitor there and always glad to hear of special things
that I may have missed.

 

[chroot]

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

 

[marcus]

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.

 

[marcus]

p = -rho

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]

Nice. Thanks for that.

 

[marcus]

Originally posted by schwarzchildradius
Nice. Thanks for that.

glad you liked it!

 

[Nicool003]

So your up Marcus-----I think ----

 

[marcus]

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.

 

[marcus]

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.

 

[marcus]

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)

 

[chroot]

marcus,

It's not hard, I just haven't felt motivated to pull the calculator out today. I'll get to it.

- Warren

 

[marcus]

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

 

[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

 

[marcus]

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 can't 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.

 

[chroot]

But wait, a light ray moving tangentially along the event horizon, will, in fact, orbit the black hole.

- Warren

 

[Labguy]

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 can't 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.

 

[marcus]

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.)

 

[chroot]

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

 

[marcus]

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 doesn't just orbit along the event horizon. But its interesting to try to figure out what it does.

 

[marcus]

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.

 

[Labguy]

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??