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

[schwarzchildradius]

Here's my guess:
Humason & Hubble,
Mt. Wilson Palomar observatory
1923
(andromeda is moving towards us)
correct?

 

[Labguy]

Originally posted by schwarzchildradius
Here's my guess:
Humason & Hubble,
Mt. Wilson Palomar observatory
1923
(andromeda is moving towards us)
correct?

No, none were correct except for the fact that Andromeda (galaxy) is moving towards us.

 

[Lonewolf]

Vesto Melvin Slipher in 1912. Not sure where.

 

[Lonewolf]

Lowell Observatory, Flagstaff, Arizona.

 

[Labguy]

Originally posted by Lonewolf
Vesto Melvin Slipher in 1912. Not sure where.

Lowell Observatory, Flagstaff, Arizona.

Where is correct and when is very close, but a tiny bit off...

 

[Lonewolf]

He began the observations in 1909, but didn't make the observation until 1912.

 

[Labguy]

Originally posted by Lonewolf
He began the observations in 1909, but didn't make the observation until 1912.

That is certainly close enough to be CORRECT! Actually, the first redshift measurement results were determined in 1913.

You ask der next question.

 

[chroot]

Sorry guys, for derailing the Astronomy Q&A thread again... my weekends are for playing!

- Warren

 

[Lonewolf]

Hmm, ok then.

What is stellar interferometry used for, and for a bonus point, what does it involve?

 

[schwarzchildradius]

It's used to measure masses and atmospheric content of stars. Interferrometry involves the wave nature of light- start with I believe it was Huygens principle that every wave is the point source of infinitely many subsequent waves. So in a double-slit experiment you can see each slit behaves like a unqiue wave generator, and the waves can interfere with each other (causing alternating bright and dark bands). Not too surprising unless the source is a single photon, in which case the single photon will interfere with itself. But I gladly yield if anyone has a more topical answer.

 

[Lonewolf]

It's used for a whole bunch of things, so I'll accept. It's also one of the few methods we have for measuring stellar radii that doesn't rely on a binary system.

Your turn to ask a question, schwarzchild.

 

[schwarzchildradius]

if you've got more details about how that's done, I'd love to hear about it. but anyway,
what's the mass and atmospheric content of the star Sirius?

 

[Lonewolf]

It uses a device called the Michelson interferometer. A diagram can be found at http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/michel.html

A.E. Michelson invented and used this interferometer into the 1920s to measure angular diameters of large, nearby stars (always some catch, huh?). In the device mirrors are widely separated by many meters. The mirrors reflect the starlight to an ordinary focusing telescope, where the wavefronts from different parts of the star form a distinctive interference pattern.

The pattern depends upon the angle between the wavefronts from opposite limbs of the star. The intersection angle increases as the stellar radius increases. We know the angular diameter can be found by the small angle approximation a = D/d where a is the angular diameter in radians, D is the star's physical diameter, and d is the star's distance from Earth.

But the interferometer measured the angular diamter for us, so simple algebra tells us that D = ad (assuming we know the distance, of course).

 

[schwarzchildradius]

trying to steal my thunder again I see. Well I say that YOU should ask the next question.

 

[chroot]

Perhaps this was said and I missed it but:

Interferometry is principally used to achieve the resolution of a very very large instrument while using only two moderate ones.

Two ten-meter telescopes in an interferometer, separated by 100 yards, have the effective resolving ability of an instrument with a 100 yard aperture.

- Warren

 

[Lonewolf]

Dude, what's your problem?

if you've got more details about how that's done, I'd love to hear about it.

So I gave more details...

Chroot: Yeah, I forgot to mention that. Thanks for pointing it out.

 

[marcus]

Originally posted by schwarzchildradius on August 20

what's the mass and atmospheric content of the star Sirius?

that question has been sitting out there for a couple of days
anyone care to answer?

 

[marcus]

Originally posted by marcus
that question has been sitting out there for a couple of days
anyone care to answer?

Schw. Radius question is "mass and atmosphere makeup of the star Sirius"

I'll take a stab at answering---Sirius is a double star consisting of siriusA and siriusB of mass which IIRC is 2.3 and 1.0 solarmasses.

The smaller one is just a white dwarf and what we think of as the star is actually Sirius A of 2.3 solar masses.

Ionized hydrogen would surround such a massive hot star as Sirius A and provide the main constituent of its atmosphere

Back to you Schw. Radius: is that right?

 

[Labguy]

But, the original question at hand was about an interferometer, and interferometers don't have squat to do with measuring a star's atmospheric content. They do that with spectrometers / spectrographs.

 

[Lonewolf]

I think you're confused. The interferometry question was mine, which Schwarzschildradius answered. Schwarzschild's question was:

what's the mass and atmospheric content of the star Sirius?

 

[Labguy]

Originally posted by Lonewolf
I think you're confused. The interferometry question was mine, which Schwarzschildradius answered. Schwarzschild's question was:

Yes, I realize that, but Schwartzchild's original answer to your question included: "It's used to measure masses and atmospheric content of stars."

I'm not trying to go for an answer, or to change the order here, but the answer to your first question (interferometer) wasn't but 1/2 or less correct. Sorry to confuse the thread.

 

[marcus]

Originally posted by Labguy
But, the original question at hand was about an interferometer, and interferometers don't have squat to do with measuring a star's atmospheric content. They do that with spectrometers / spectrographs.

Would this work as a way out.

Lonewolf asked about interferometry and Labguy finally gave the correct answer (it increases resolution so you see greater detail, two telescopes some distance apart are linked so they act in a certain way like a larger telescope with more resolving power)

so it is now Labguy's turn


that seems logical to me, how does it seem to you Lonewolf



meanwhile Schw. Radius answered a different question "what is spectroscopy?" (it is a way to tell atmospheric composition by spectrum of light and can be used to infer the mass under certain assumptions)
And Schw. should be happy because he got a "free turn" and asked about Sirius and I answered his question. But we rescind that little digression and Labguy goes.

 

[Labguy]

Originally posted by marcus
Would this work as a way out.

Lonewolf asked about interferometry and Labguy finally gave the correct answer (it increases resolution so you see greater detail, two telescopes some distance apart are linked so they act in a certain way like a larger telescope with more resolving power)

so it is now Labguy's turn


that seems logical to me, how does it seem to you Lonewolf



meanwhile Schw. Radius answered a different question "what is spectroscopy?" (it is a way to tell atmospheric composition by spectrum of light and can be used to infer the mass under certain assumptions)
And Schw. should be happy because he got a "free turn" and asked about Sirius and I answered his question. But we rescind that little digression and Labguy goes.

No, it is not my turn. I never answered the question at all. Your quote given above that "(it increases resolution so you see greater detail, two telescopes some distance apart are linked so they act in a certain way like a larger telescope with more resolving power)" was given by someone else, not me!

I was only trying to get a bit more clear on the "rules of the game".

When someone asks a question, and an answer is incorrect or only slightly correct, is the person asking the question to say "close enough" or wait for a correct answer?? I'm not slighted at all, or trying to "get one" so I can ask a question. I'd rather read yours...

 

[marcus]

sometimes there are muddles the rules don't cover
we are in a muddle
I believe I answered both Lonewolf's Q about interferometry AND
S.Radius Q about the mass and atmosph of Sirius
so I will go

two stars differ in absolute magnitude by one step

(say one is magnitude 4 and the other is 5)

if one was comparing WATTAGES of the two stars
roughly by what factor would they differ?

 

[Lonewolf]

Ah, ok. Sorry for the mix up, Labguy. I accepted it as an indirect application of interferometry. I was after more what we can use it for than what it does. We can perform spectroscopy on the signal we obtained from the interferometer, and thus obtain atmospheric content. It may seem a dubious decision, but I thought it'd be best to try and get some momentum going on the thread.

Whose turn is it now, then?

 

[Labguy]

Originally posted by Lonewolf
Ah, ok. Sorry for the mix up, Labguy. I accepted it as an indirect application of interferometry. I was after more what we can use it for than what it does. We can perform spectroscopy on the signal we obtained from the interferometer, and thus obtain atmospheric content. It may seem a dubious decision, but I thought it'd be best to try and get some momentum going on the thread.

Whose turn is it now, then?

Marcus' turn. He just asked one bottom of previous page.

 

[Lonewolf]

So he did. I must have missed that.

Let A be the dimmer star and B be the brighter star.

Using the equation

m_B - m_A = 2.5 log 10 (I_A/I_B)

We cab obtain 10^{(m_B - m_A)/2.5} = I_A/I_B

From the question, we know m_B - m_A = 1. so

10^1/2.5 = 2.512 (approximate) = I_A/I_B

or 2.512*I_B = I_A

Hence, B is approximately 2.512 times as intense as star A. If we multiply both sides of the equation by 1 square meter, we are left with units of power. Hence, their wattages differ by approximately a factor of 2.512.

 

[marcus]

Thanks for answering, and rightly to boot!

Your go, Lonewolf.

Originally posted by Lonewolf
So he did. I must have missed that.

Let A be the dimmer star and B be the brighter star.

Using the equation

m_B - m_A = 2.5 log 10 (I_A/I_B)

We cab obtain 10^{(m_B - m_A)/2.5} = I_A/I_B

From the question, we know m_B - m_A = 1. so

10^1/2.5 = 2.512 (approximate) = I_A/I_B

or 2.512*I_B = I_A

Hence, B is approximately 2.512 times as intense as star A. If we multiply both sides of the equation by 1 square meter, we are left with units of power. Hence, their wattages differ by approximately a factor of 2.512.

 

[schwarzchildradius]

Thanks for the answer, m, lonewolf, I don't think I have a problem, maybe overwork and procrastination (shudder), over.

 

[Lonewolf]

It's cool.

Anyway, the question. Which two physical states are in equilibrium in a stable main sequence star?

 

[Labguy]

Do you mean:
(1) Radiative repulsion, and
(2) Gravitational attraction.

??

 

[Lonewolf]

That was indeed what I was thinking, Labguy, Your turn.

 

[Labguy]

Originally posted by Lonewolf
That was indeed what I was thinking, Labguy, Your turn.

Ok, another easy "Star" question.

In not too many words, why do lower mass stars have radiative cores and convetive outer layers, while massive stars have convective cores and radiative outer layers??

All this is sub-photosphere, of course.

 

[LURCH]

Originally posted by Labguy
Ok, another easy "Star" question.

In not too many words, why do lower mass stars have radiative cores and convetive outer layers, while massive stars have convective cores and radiative outer layers??

All this is sub-photosphere, of course.

Just a geusse;

The more massive stars are denser, rendering normal raditation impossible near the core. So heat from the core is transported via convection to the outer layers, where density is lower (closer to the density of the Sun's inner layers), and radiative heat transfer can occur?

 

[Labguy]

Originally posted by LURCH
Just a geusse;

The more massive stars are denser, rendering normal raditation impossible near the core. So heat from the core is transported via convection to the outer layers, where density is lower (closer to the density of the Sun's inner layers), and radiative heat transfer can occur?

Good guess. That's it, 100%.

Your question.