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

[marcus]

A different question, instead of the Kepler one

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 Ω_vac = 0.73, and 0.27 for Ω_M = 0.27. These are widely accepted values so what I'm looking for is the answer with these default values for H₀, Ω_vac, and Ω_M 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.

 

[Labguy]

z = 3.808

 

[marcus]

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!

 

[Labguy]

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.. ..)

 

[marcus]

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.. ..)

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.

 

[Labguy]

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...

 

[marcus]

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...

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.

 

[Labguy]

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.

 

[Labguy]

Finish the other points on that question, Marcus. (or anyone)!

 

[chroot]

a) a Supernova Type II
b) a massive red supergiant

- Warren

 

[marcus]

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 don't 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?

 

[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

 

[Labguy]

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.

 

[Labguy]

Somebody ask another question. I survived the surgery I had yesterday and can sit at my desk again without pain pills...

 

[marcus]

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...

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.

 

[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.

 

[Labguy]

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.

 

[marcus]

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 into 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!

 

[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

 

[marcus]

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 into 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 that's 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.

 

[marcus]

another riddle

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 ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldn't 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.

 

[Labguy]

Aaargh ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldn't we know how much acceleration it would take to duplicate the CMB temp?

The word is "Matherize", not arithmeticize, you bumbling human calculator...

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

 

[marcus]

Originally posted by Labguy
The word is "Matherize", not arithmeticize, you bumbling human calculator...

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.

 

[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?

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 ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldn't 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!

 

[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

 

[Shadow]

Oh, sorry marcus I have been busy and was unable to reply to your post.

 

[Labguy]

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?

 

[Labguy]

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 ??

 

[marcus]

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.)

 

[marcus]

Footnote about T = g/2pi

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

 

[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!