Measure the distance to a star using magnitude and extinction

In summary, A star is observed close to the center of the Milky Way and from its spectrum we find that it is a type A3 star. Its observed magnitude is m_v = 25. There is only a diffusive gas between us and the star, so we can assume an extinction, of 1 magnitude per 1 kpc. Using the distance formula, we find that the star is 8 kpc away.
  • #1
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Homework Statement
Measure distance using magnitude and extinction
Relevant Equations
m - M = 5log_{10}(d/10 pc) + 10^{-3}
Hello, I'm struggling with this question
  • A star is observed close to the center of the Milky Way and from its spectrum we find that it is a type A3 star. Its observed magnitude is m_v = 25. There is only a diffusive gas between us and the star, so we can assume an extinction, of 1 magnitude per 1 kpc.
  • How far away is the star and what is its (un-redded) magnitude?
  • In the H-band, which is 1.63 , the reddening law is . How big is the reddening in the H-band?

I think I got the generel idea right for this problem, but I must be messing up my SI units somewhere. I use the distance formula when working with magnitudes and extinction. I look up the absolute magnitude for a type A3 star, which is roughly 2. I solve for d and end up with a distance of 398kpc, which doesn’t make sense, as the star is at the center of the Milky Way.
I use this formula and I divide A by 10^3 to get units of pc.
Udklip.JPG

Udklip.JPG


I hope someone can spot my mistake o_O
 
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  • #2
The last term should be d/1000, not 1/1000.
NSolve[25-2.0 == 5*Log10[d/10]+d/1000, d] gives d -> 8383. About 8 kpc.
 
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Likes Mikkel
  • #3
Thanks for your fast reply!
Why should I multiply by d when considering the extinction term?
 
  • #4
Mikkel said:
Thanks for your fast reply!
Why should I multiply by d when considering the extinction term?

The extinction is 1 magnitude per kpc. So the farther away the star is, the more it's brightness is diminished. This is because you are looking through dust that absorbs the light from the star. The farther away the star is, the more dust is between you and the star, and the more light is absorbed. The way you had it, the extinction would be the same regardless of how far away the star is.
 
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  • #5
phyzguy said:
The extinction is 1 magnitude per kpc. So the farther away the star is, the more it's brightness is diminished. The way you had it, the extinction would be the same regardless of how far away the star is.
Ahh yea that makes sense, thank you very much!
Do you perhaps have a clue for the other question as well?
 
  • #6
Mikkel said:
Ahh yea that makes sense, thank you very much!
Do you perhaps have a clue for the other question as well?
Y ou should be able to find the unreddened magnitude. The last question isn't complete.
 
  • #7
whoops.. sorry

  • In the H-band, which lies in 1.63 μm , the reddening law is A_H = 0.591 E_{B-V}. How big is the reddening in the H-band?
I'm not sure that the "reddening" is in terms of symbol? Not sure how to list it.
 
  • #8
Well, you should be able to look up (B-V) for an A3 star. I'm not sure what E is in this context. Also, what are the units here. Is it also magnitudes/kpc?
 
  • #9
phyzguy said:
Well, you should be able to look up (B-V) for an A3 star. I'm not sure what E is in this context. Also, what are the units here. Is it also magnitudes/kpc?
YEs, that's what I'm in doubt of. I don't know which units it is and don't know what symbol it is represented by, but I will try and look up (B-V) for an A3 star. I think it is written as E(B-V) = A, but I'm not able to find a correct formula in my book.
 

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