Astronomy C

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Re: Astronomy C

Post by astronomybuff » September 15th, 2020, 5:04 am

a. 1. Brightness is proportional to distance (its inversely proportional). 2. It's inversely proportional to the square of the distance. 2. Compare brightness to the Sun (in terms of its brightness), and use inverse square law.

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Re: Astronomy C

Post by RiverWalker88 » September 15th, 2020, 6:39 am

astronomybuff wrote:
September 15th, 2020, 5:04 am
a. 1. Brightness is proportional to distance (its inversely proportional). 2. It's inversely proportional to the square of the distance. 2. Compare brightness to the Sun (in terms of its brightness), and use inverse square law.
Yep, all correct (I actually made a mistake in the question, Newton originally assumed that brightness was inversely proportional, but didn't account for luminosity. Just for historical accuracy, I figured I'd point out my mistake).
Your turn!
Last edited by RiverWalker88 on September 15th, 2020, 6:42 am, edited 1 time in total.
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Re: Astronomy C

Post by astronomybuff » September 18th, 2020, 4:40 am

A supermassive blackhole within a quasar consumes 2 solar masses a year with a mass-to-energy conversion efficiency of 12%.

a. What is its energy output per seconds, in Joules?

b. What is the lower bound to its mass, in solar masses?

c. How do you know the answer to b?

d. If it has an apparent magnitude of 11.3, how far is it from Earth?

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Re: Astronomy C

Post by RiverWalker88 » September 19th, 2020, 6:15 pm

astronomybuff wrote:
September 18th, 2020, 4:40 am
A supermassive blackhole within a quasar consumes 2 solar masses a year with a mass-to-energy conversion efficiency of 12%.

a. What is its energy output per seconds, in Joules?

b. What is the lower bound to its mass, in solar masses?

c. How do you know the answer to b?

d. If it has an apparent magnitude of 11.3, how far is it from Earth?

a. 2 solar masses * .12 = 0.24 solar masses. 0.24 solar masses = ~1.51*10^22 kg. 1.51*10^22 kg*c^2 = 1.36J
b. I converted the luminosity calculated in part a to solar luminosities and plugged that into (L/Lsun)=3.2*10^4(M/Msun) (an approximate equation for Eddington luminosity) and got... 1.113*10^8 solar masses. That doesn't quite seem right... 
c. The Eddington Luminosity is the maximum luminosity a body can radiate at, so it can be used to determine a limit for mass, as well.
d. Black holes really can't be seen (they emit no light), but theoretically, we can still make these calculations work. Using the luminosity of the black hole, I calculated the absolute magnitude to be -26.549. Using the Distance modulus, I get 3.714*10^6 parsecs. 
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Re: Astronomy C

Post by RiverWalker88 » Yesterday, 4:10 pm

Oops... I forgot about posting a question (sorry).
Just a basic one, I can't properly think right now.

A MACHO passes in front of a star.
  1. What is a MACHO?
  2. What is the effect on the brightness of this star of the MACHO passing in front of it?
    1. The star appears brighter.
    2. The star appears dimmer.
    3. The star appears unchanged.
    4. The star actually blinks out of existence.
  3. What is the effect that caused the change in brightness known as?
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Re: Astronomy C

Post by astronomybuff » Today, 4:41 am

RiverWalker88 wrote:
September 19th, 2020, 6:15 pm
astronomybuff wrote:
September 18th, 2020, 4:40 am
A supermassive blackhole within a quasar consumes 2 solar masses a year with a mass-to-energy conversion efficiency of 12%.

a. What is its energy output per seconds, in Joules?

b. What is the lower bound to its mass, in solar masses?

c. How do you know the answer to b?

d. If it has an apparent magnitude of 11.3, how far is it from Earth?

a. 2 solar masses * .12 = 0.24 solar masses. 0.24 solar masses = ~1.51*10^22 kg. 1.51*10^22 kg*c^2 = 1.36J
b. I converted the luminosity calculated in part a to solar luminosities and plugged that into (L/Lsun)=3.2*10^4(M/Msun) (an approximate equation for Eddington luminosity) and got... 1.113*10^8 solar masses. That doesn't quite seem right... 
c. The Eddington Luminosity is the maximum luminosity a body can radiate at, so it can be used to determine a limit for mass, as well.
d. Black holes really can't be seen (they emit no light), but theoretically, we can still make these calculations work. Using the luminosity of the black hole, I calculated the absolute magnitude to be -26.549. Using the Distance modulus, I get 3.714*10^6 parsecs. 
Uh for A where is your 10 to some power? Or am I just blind :P. B and C are spot on! For D, I believe it should be 10^8 instead of 10^6

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Re: Astronomy C

Post by astronomybuff » Today, 4:45 am

RiverWalker88 wrote:
Yesterday, 4:10 pm
Oops... I forgot about posting a question (sorry).
Just a basic one, I can't properly think right now.

A MACHO passes in front of a star.
  1. What is a MACHO?
  2. What is the effect on the brightness of this star of the MACHO passing in front of it?
    1. The star appears brighter.
    2. The star appears dimmer.
    3. The star appears unchanged.
    4. The star actually blinks out of existence.
  3. What is the effect that caused the change in brightness known as?
No, no it's my fault for not replying quickly!
An object that has an almost 0 luminosity. So they appear nearly invisible, and as they pass in front of a massive body, they bend it's light rays. 2. Make it brighter. 3. Gravitational lensing.

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Re: Astronomy C

Post by RiverWalker88 » Today, 6:42 am

astronomybuff wrote:
Today, 4:41 am
RiverWalker88 wrote:
September 19th, 2020, 6:15 pm
astronomybuff wrote:
September 18th, 2020, 4:40 am
A supermassive blackhole within a quasar consumes 2 solar masses a year with a mass-to-energy conversion efficiency of 12%.

a. What is its energy output per seconds, in Joules?

b. What is the lower bound to its mass, in solar masses?

c. How do you know the answer to b?

d. If it has an apparent magnitude of 11.3, how far is it from Earth?

a. 2 solar masses * .12 = 0.24 solar masses. 0.24 solar masses = ~1.51*10^22 kg. 1.51*10^22 kg*c^2 = 1.36J
b. I converted the luminosity calculated in part a to solar luminosities and plugged that into (L/Lsun)=3.2*10^4(M/Msun) (an approximate equation for Eddington luminosity) and got... 1.113*10^8 solar masses. That doesn't quite seem right... 
c. The Eddington Luminosity is the maximum luminosity a body can radiate at, so it can be used to determine a limit for mass, as well.
d. Black holes really can't be seen (they emit no light), but theoretically, we can still make these calculations work. Using the luminosity of the black hole, I calculated the absolute magnitude to be -26.549. Using the Distance modulus, I get 3.714*10^6 parsecs. 
Uh for A where is your 10 to some power? Or am I just blind :P. B and C are spot on! For D, I believe it should be 10^8 instead of 10^6
Ahh... I dropped those orders of magnitude very badly. I can't find my work on them, so I'm not sure what they were supposed to be.
astronomybuff wrote:
Today, 4:45 am
RiverWalker88 wrote:
Yesterday, 4:10 pm
Oops... I forgot about posting a question (sorry).
Just a basic one, I can't properly think right now.

A MACHO passes in front of a star.
  1. What is a MACHO?
  2. What is the effect on the brightness of this star of the MACHO passing in front of it?
    1. The star appears brighter.
    2. The star appears dimmer.
    3. The star appears unchanged.
    4. The star actually blinks out of existence.
  3. What is the effect
    that caused the change in brightness known as?
No, no it's my fault for not replying quickly!
An object that has an almost 0 luminosity. So they appear nearly invisible, and as they pass in front of a massive body, they bend it's light rays. 2. Make it brighter. 3. Gravitational lensing.
a. Yep (I was looking for MAssive Compact Halo Object, but your definition is better)
b. Yep
c. Yep
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