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

[Labguy]

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.

 

[Labguy]

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!

 

[marcus]

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.

 

[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 Omega_M at 0.27, and allow your other assumptions to stand.

 

[marcus]

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 Omega_M 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.

 

[Labguy]

Another easy one:

If:
("Blank") = 4.8-[2.5*log(Lstar/Lsol)]
where "L" stands for luminosity;

What is "Blank"? ( Definition, not a number)

 

[chroot]

Absolute magnitude -- the apparent brightness of a star from a standard distance of 10 parsecs.

- Warren

 

[Labguy]

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...

 

[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

 

[Labguy]

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.

 

[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

 

[Labguy]

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 separate 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.

 

[chroot]

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

 

[Labguy]

Ask a new one!

 

[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

 

[Labguy]

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...

 

[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

 

[stuffy]

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!

 

[chroot]

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

 

[Labguy]

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!

 

[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?

 

[Labguy]

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...

 

[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.

 

[marcus]

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 don't 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

 

[stuffy]

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 :)

 

[stuffy]

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.

 

[stuffy]

*rubs paddles* CLEAR! *shock*

 

[chroot]

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]

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.

 

[chroot]

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]

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

 

[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

 

[marcus]

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

 

[marcus]

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)

 

[marcus]

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.