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

[marcus]

Originally posted by marcus
A)What is the peak luminosity of a Type Ia supernova expressed as a wattage?

Don't count other ways the supernova might be releasing energy, like neutrinos, I guess that's implicit when one asks about the luminosity.

B)What is the luminosity of the sun expressed also expressed as a wattage?---the total output of light in all directions.

C)What is the approximate ratio of the two wattages? By what factor is a Ia SN brighter than the sun?

My handbook gives the sun's luminosity in ergs per second.
I'm asking for the answer in watts (one watt = 10^7 ergs per second) because that's more conventional although ergs still seem to be current in a good deal of astronomical writing.

Nobody else feels like answering this so I will. Labguy pointed out a few posts back that the peak Ia absolute magnitude is conventionally taken to be -19.5. And the sun's is 4.8. So the difference is 24.3. Multiplyby 2/5 and you get 9.72 (I will round off later).

C) The wattage ratio, SN versus sun, is 10^9.72 = 5.25 billion.

B) The sun's wattage (standard handbook figure) 3.8E26 watts.

A) Peak supernova luminosity is 5.25 billion times the sun's or
2E36 watts.

2 x 10^36 watts was what I was looking for in the answer.

Since I had to answer my own question the next askership is up for grabs.

The first person who can think of an appropriate-to-the-game type of question to which he/she knows the answer can ask it.

 

[J-Man]

marcus said:
Since I had to answer my own question the next askership is up for grabs.

The first person who can think of an appropriate-to-the-game type of question to which he/she knows the answer can ask it.

How nice, thanks.

If the Sun's orbital velocity around the center of the Milky Way is 220 km s-1 [220 km/sec], roughly how many orbits has it completed since it formed, 4.5 billion years ago? Explain how you arrived at your estimate. You may assume that the Sun's distance from the Galactic center is 8000 pc [parsecs], and that it is following a circular orbit.

 

[chroot]

Too easy.

It's completed roughly 21 orbits in those 4.5 billion years.

The orbit length l = pi * 2 * 8000 pc.

Divide l (in km) by v = 6937920000 km/year to arrive at an orbit period of 222 million years.

Divide 4.5 billion by 222 million to arrive at roughly 21 orbits.

- Warren

 

[J-Man]

Originally posted by chroot
Too easy.

It's completed roughly 21 orbits in those 4.5 billion years.

The orbit length l = pi * 2 * 8000 pc.

Divide l (in km) by v = 6937920000 km/year to arrive at an orbit period of 222 million years.

Divide 4.5 billion by 222 million to arrive at roughly 21 orbits.

- Warren

Sorry it took so long to get back...
But you are correct (I got between 20 and 21 orbits). Your turn.

 

[Nicool003]

hey guys! I was planning on coming back after vacation but there are tons of exams and tests and quizes and projects... well you can read all about it in general discussion


Anyways chroot it is your question.

 

[schwarzchildradius]

Does chroot ask a question?

 

[Nicool003]

He seems to not be coming to this thread anymore so how about you ask schwartz? I don't want to take it because I haven't been able to come to PF as much as you guys so I haven't been to this thread in a while either.

 

[marcus]

Originally posted by Nicool003
He seems to not be coming to this thread anymore so how about you ask schwartz?...

Hear hear! Amen to that I say. Let schwarzschildradius ask a question and god save the empire!

 

[schwarzchildradius]

OK, how do you measure the distance to the moon using lunar eclipse? What other observation of the moon must be made to find the distance to the sun?

 

[schwarzchildradius]

BTW here's a freeware orrery and lunar eclipse predictor that works pretty well..
astropro orrery

 

[marcus]

Originally posted by schwarzchildradius
OK, how do you measure the distance to the moon using lunar eclipse? What other observation of the moon must be made to find the distance to the sun?

Hipparchus measure the distance to the moon using eclipse data and got the result that the moon is 30 Earth diameters away which is quite close to the right answer.

However Hipparchus method is more complicated than I want to try to explain.

Earlier, Aristarchus (around 250 BC?) estimated the size of the moon compared with the Earth by observing an eclipse of the moon.

He judged that the Earth's shadow (roughly comparable in size to earth) was twice as big as the moon just by looking at the curve the shadow's edge made on the moon-----actually he should have guessed THREE times but he guessed twice the size.
Aristarchus was very back-of-envelope.

Knowing (in a rough sense) the size of the moon and the angle it made in the sky, Aristarchus could estimate the distance to the moon in Earth diameters.

He then observed the lunar dichotomy, which gave him at least a rough lower bound on the distance to the sun as a multiple of the distance to the moon.

Noting that the sun was an order of magnitude farther than the moon and therefore huge compared with the earth, he surmised a heliocentric model.

This guy was from Samos, same birthplace as Pythagoras.
Well Schwarzschild there is at least part of an answer
god save all Samians and the empire of the mind

 

[schwarzchildradius]

There you have it folks, righty-o.

The other observation was of the half moon - the moon is at 90 deg. from the sun when it is half full. You can use geometry to then (since you know the length of one side) find the distance and size of the sun. Your go.

 

[marcus]

Originally posted by schwarzchildradius
There you have it folks, righty-o.

The other observation was of the half moon - the moon is at 90 deg. from the sun when it is half full. You can use geometry to then (since you know the length of one side) find the distance and size of the sun. Your go.

Schwarzschildradius is a man of cultivation and discernment and has explained to us the method of lunar dichotomy.

In fact the sun is roughly 400 times the distance to the moon and therefore the angle (at halfmoon) is not 90 degrees but
90 degrees MINUS 1/400 RADIANS. Something like 89.8 or 89.9 degrees.

Aristarchus around 250BC measured it and got IIRC something like 89.5 degrees (he was clumsy at measuring angles, this Greek)
and so he judged that the sun was quite far away perhaps IIRC ten times as far as the moon! All the other greeks were quite astonished by this and could scarcely believe it. He should be awarded the Nobel prize postumously. Lunar dichotomy is a great phrase.

I like responding to questions a great deal better than I like asking them. Am tempted to appoint Schwarzschild my proxy and in invite him to pose a question for me. But perhaps it is better to keep it simple and stick by the rules.

QUESTION:
Describe "co-moving distance"
This is the astronomer's def of distance which works in the Hubble law v=H₀ D

So if you ever use that law you need to understand what the comoving distance to an object is.

What is the comoving distance to an object (at this moment in time) which is currently receding from us at 30 thousand kilometers a second?

 

[schwarzchildradius]

The answer to your particular question, is r=c/H, or something like 2e26 meters away. The answer to your general question, is that its the ratio of dx/dt (transverse velocity) to dθ/dt (proper motion), in other words, the distance measured by a "ruler" at specific time t.

 

[marcus]

Originally posted by schwarzchildradius
The answer to your particular question, is r=c/H, or something like 2e26 meters away. The answer to your general question, is that its the ratio of dx/dt (transverse velocity) to dθ/dt (proper motion), in other words, the distance measured by a "ruler" at specific time t.

Hello, your answer 2E26 meters is off by more than a factor of two so I'm afraid it will not do. But if you recalculate you will almost certainly get it.

Suggest using Hubble parameter of 71 km/s per Megaparsec
since that figure is not considered fairly reliable---the uncertainty has narrowed substantially about H₀.

I should have been more explicit about the Hubble Law distance.

There is a particular definition of distance that works in the law
and other types of distance (angular size distance, luminosity distance, light traveltime distance) do not work.
In his javascript distance calculator
http://www.astro.ucla.edu/~wright/CosmoCalc.html
Wright refers to this definition of distance as the
"comoving radial distance, which goes into Hubble's law".

That expression contains a link to where the distance is
defined and described, using diagrams and some detailed
explanation. That "comoving distance which goes into Hubble's law" is what I am looking for.

But I won't insist on the general question---just get the
specific one right and you win the prize.

 

[schwarzchildradius]

Fine
71 km/s/Mpc = 2.3e-18 s^-1
R = 1.3e26 m
off by 2^1/2, much less than the uncertainty in H
or if you like,
3R = 3.9e26 m

Do you (or anyone) know the metric for Homogeneous Isotropic Cosmological Models?

 

[marcus]

Originally posted by schwarzschildradius
Fine
71 km/s/Mpc = 2.3e-18 s^-1
R = 1.3e26 m...


Do you (or anyone) know the metric for Homogeneous Isotropic Cosmological Models?

Still not right. You have to solve the equation v = HD

for D, where H is exactly what you say it is and
v = 30,000 km per second.

You do realize, do you not, that I said 30 thousand km per second----a speed which is not equal to the speed of light?
When I said your answer is off by more than a factor of 2, I meant *more* than a factor of two, also more than a factor of 3.
Your answer is off by more than a factor of 4, hint hint.

Your answer, namely 1.3E26 meters. Is 13.7 billion light years!
Way too big!

 

[schwarzchildradius]

Someone else will have to play with you.

 

[marcus]

an algebra problem

Schwarzschildradius gives up.

Perhaps the problem is too simple for him.

The problem is, solve v = H₀ D, for D

where v is 30 thousand km/sec, i.e. one tenth the speed of light,

and H₀, the Hubble parameter, is (1/13.8 billion years).

 

[schwarzchildradius]

Whatever. Its against the rules for me to keep guessing, and the value is in fact irrelevant because of uncertainty in H.

 

[marcus]

Originally posted by schwarzchildradius
Whatever. Its against the rules for me to keep guessing, and the value is in fact irrelevant because of uncertainty in H.

Both of us are nice people and we just experienced a power-of-ten problem. That's all and it happens quite regularly.

You kept saying 13.7 billion light years when you should have
been saying 1.37 billion light years. And you were off because
you thought that when I said "30 thousand km/sec" I was saying the speed of light.

When actually I was saying c/10.

I blame this on the metric system. If I had simply said "c/10"
then you would have gotten the right result immediately.
c is the natural unit to use in astronomy.

Let's have it be your turn, since you got the right answer except for the dratted order magnitude

 

[Labguy]

General Comment:

Isn't this thread meant to be a "general" astronomy Q&A session instead of math excercises based on someone's assumptions (criteria) regarding some specific set of circumstances, sometimes theoretical and beyond "generally accepted"??

I don't mind the "Here is my theory" stuff, but there is another PF forum named "Theoretical Physics" where that is discussed with attempts to prove / disprove.

Carry on.

 

[Phobos]

Originally posted by Labguy
General Comment:

Isn't this thread meant to be a "general" astronomy Q&A session instead of math excercises based on someone's assumptions (criteria) regarding some specific set of circumstances, sometimes theoretical and beyond "generally accepted"??

I don't mind the "Here is my theory" stuff, but there is another PF forum named "Theoretical Physics" where that is discussed with attempts to prove / disprove.

Carry on.

Mentor hat on...

Overall, this forum is intended for "generally accepted" (dare I say, "mainstream") astronomy. "Here's my theory" goes into the Theory Development forum. Just keep that in mind...no major problems here so far.

As far as the general Q&A vs. math-heavy Q&A. I suppose it's either (1) up to Nicool002 (who started this thread) or (2) free market...if y'all want math, then knock yourselves out...if not, then this topic will evolve or die accordingly. Whatever. As long as it's good clean astronomy, then I'm happy. Certainly, many PF members have mastered general astronomy Q&A and are ready for more. But many PF members are new to astronomy.

Might I recommend 2 topics...one for beginners and one for advanced folks?

 

[marcus]

Originally posted by Phobos
Mentor hat on...


Might I recommend 2 topics...one for beginners and one for advanced folks?

Hello Phobos, why don't you ask a question at kick off another round?


The last question was about the v=H₀ D (Hubble law).

I asked, if v is a tenth of the speed of light (v = 30 thousand km/s)
then solve for D, the distance.

It did not get answered so maybe I can just invite someone to pose a question.

My thought is that the last question was about as basic, mainstream, and beginner-level as you can get in cosmology.
The Hubble parameter has been the main thing to measure and to use in estimating distance for several generations, the law
is fundamental to the field. So I wonder how the issue of "mainstream" arose. Is it too much math to involve a basic
equation like v=H₀ D?

 

[marcus]

Afterthought. I guess the other most basic thing
in cosmology is the density of the universe (rho)
as it compares to the critical density (rho_crit) required for flatness.

For some decades cosmologists have been
talking about how "the fate of the universe" depends
on the ratio

rho/rho_crit-------whether it is less than one
or equal to one (flatness) or greater than one.

Would it by any chance be considered not mainstream to
ask a question about the critical density on this thread?
I don't recall that I have, as yet, but was just now wondering,
since it seems so central to the field that I can't really
picture what conventional cosmology would be like without it.

 

[schwarzchildradius]

Just let's get this thing moving please ;)

 

[marcus]

Originally posted by schwarzchildradius
Just let's get this thing moving please ;)

OK, since no one else volunteers a question, I will ask:

What famous astronomer's mother was almost burnt as a witch?


(it didnt actually come to lighting the fire but she was tried
and came close to being burned)

god save the Hapsburgs and the holy roman empire!

 

[J-Man]

marcus asked:
What famous astronomer's mother was almost burnt as a witch?

Johannes Kepler

 

[marcus]

Originally posted by J-Man
Johannes Kepler

Attaboy! Your question

 

[J-Man]

What is a "reflection nebula"?
-Give a definition.
-Give an example.
-Extra points for brown-nosing.

 

[Labguy]

Originally posted by J-Man
What is a "reflection nebula"?
-Give a definition.
-Give an example.
-Extra points for brown-nosing.

A reflection nebula is an accumulation of dust and gas where the light, usually blue in color, is reflected off the dust and gas from one or more embedded stars. M20, the Trifid Nebula, is an example where it is seen as blue. It also happens to be an emmission nebula as well, with light being emmitted from ionized Hydrogen, usually seen as red.

I don't know any brown-nose points.

 

[J-Man]

Labguy answered:
A reflection nebula is an accumulation of dust and gas where the light, usually blue in color, is reflected off the dust and gas from one or more embedded stars. M20, the Trifid Nebula, is an example where it is seen as blue. It also happens to be an emmission nebula as well, with light being emmitted from ionized Hydrogen, usually seen as red.

Very good.

I don't know any brown-nose points.

That's okay, it's still your turn.

 

[Labguy]

Originally posted by J-Man
Very good.


That's okay, it's still your turn.

Ok, here is an EASY one. Two-parter.

(1) What event, important to observational astronomy, will happen on August 27, 2003? (**Later edit**: If we need very specific, you can use universal time OR EDT, since we are not always on the same day worldwide)

(2) When was this event last seen, to the same or greater degree, by any astronomer using any telescope or optical device to enhance the unaided eye??

Note: I expect this one to be answered instantly by the first PF member to read it...(?)

 

[Ivan Seeking]

Originally posted by Labguy
Ok, here is an EASY one. Two-parter.

(1) What event, important to observational astronomy, will happen on August 27, 2003? (**Later edit**: If we need very specific, you can use universal time OR EDT, since we are not always on the same day worldwide)

(2) When was this event last seen, to the same or greater degree, by any astronomer using any telescope or optical device to enhance the unaided eye??

Note: I expect this one to be answered instantly by the first PF member to read it...(?)

The closest approach of Mars in 73,000 years.
Never.

 

[Labguy]

Originally posted by Ivan Seeking
The closest approach of Mars in 73,000 years.
Never.

CORRECT.
Your question.