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Apparent magnitude of two stars |
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| Feb8-07, 02:25 PM | #1 |
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Apparent magnitude of two stars
1. The problem statement, all variables and given/known data
Two stars have apparent magnitudes of V = 5.1 and V = 4.6 but are too close together to be resolved with the naked eye and appear to be a single object. What is its apparent magnitude? 2. Relevant equations I don't know what the relevant equations are if there are any. 3. The attempt at a solution Apparent magnitude = the sum of the two magnitudes = 9.7 but i know that's probably wrong, any help would be appreciated thanks! Sorry if this is in the wrong place by the way. |
| Feb8-07, 02:35 PM | #2 |
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Magnitudes don't add. Luminosities add. What's the relation between magnitude and luminosity?
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| Feb8-07, 03:01 PM | #3 |
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m = -2.5*log(L/4*pi*d^2) + c
So L = 10^((m-c)/-2.5) * (4*pi*d^2) So errm i think i was given the c for the visual in another question i think it's -21.58... So L1+L2 = 10^((5.1+21.58)/-2.5)*(4*pi*d^2)+10^((4.6+21.58)/-2.5)*(4*pi*d^2) = 2.13*10^-11*(4*pi*d^2) + 3.37*10^-11*(4*pi*d^2) = 5.50*10^-11*(4*pi*d^2) m = -2.5*log(L/4*pi*d^2)+ c m = -2.5*log(5.50*10^-11*(4*pi*d^2)/4*pi*d^2) -21.58 m = -2.5*log(5.50*10^-11) -21.58 m = 4.07 Is that right? I just thought woohoo i have a number which is different from the others but i don't know if it's right? |
| Feb8-07, 03:15 PM | #4 |
Recognitions:
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Apparent magnitude of two stars
It's right. You've got a lot of extra factors you don't really need in this particular problem, but I wouldn't worry about it.
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| Feb8-07, 03:17 PM | #5 |
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Thanks for the help :)
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| Jan17-08, 06:26 AM | #6 |
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I have two questions that are very very similar to the one above but I'm pretty sure I have no value for the constant.
Question 1, parts a & b A binary star has a total apparent magnitude of 15.00. One component star is twice as bright as the other. (a) Show that the apparent magnitude of the brighter star is 15.44. [2] This one I can do, but don't understand why it works. Ie - m + 2.5log(F) = mbright --> F= 1.5 15 + 2.5log(1.5) = 15.44 So why does that work? (b) The fainter star has an absoloute magnetude of 4.50. How far away is the binary system in Kpc? Then this next question I'm sure I could do if it wasn't for the confusion with part a but I have a mental block because of that. I know the distance equation D = 10 ^ (m-M)/5 x 10, but I need the apparent magnitude of the fainter star to work that out and I can't do it. I don't think you need to find the total apparent magnitude to find the answer but I could be wrong. EDIT - Going on a complete guess is the apparent mag of the faint star 16.19 (using 15 + 2.5log(3) ) giving a distance of 2.18Kpc? Even if this is right, it's revision for an exam so I'd like to know how to do it rather than fluking my way there. |
| Jan17-08, 10:20 AM | #7 |
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You are getting the right answers but you don't know why? For problems like this you can simplify the luminosity magnitude relation to just M=(-2.5)*log(L). The real formula has an additive constant and the L is divided by some stuff, but that all cancels out. You have a bright star of luminosity 2L and a dimmer one at L. So the total magnitude of the system is 15=(-2.5)log(3L). The magnitude of the brighter is M2=(-2.5)log(2L). Take the difference of the two. M2-15=(-2.5)*(log(2)-log(3))=0.44 (use stuff like log(3L)=log(3)+log(L)). Similarly the magnitude of the dimmer is M1=(-2.5)log(L). Take the difference again, M1-15=(-2.5)*(-log(3))=1.19.
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| Jan18-08, 10:30 AM | #8 |
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Thanks, that was really helpful. Knowing my luck it probably won't come up on the exam now I understand it :D.
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