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Homework Help: Apparent magnitude of two stars

  1. Feb 8, 2007 #1
    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.
  2. jcsd
  3. Feb 8, 2007 #2


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    Magnitudes don't add. Luminosities add. What's the relation between magnitude and luminosity?
  4. Feb 8, 2007 #3
    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?
    Last edited: Feb 8, 2007
  5. Feb 8, 2007 #4


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    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.
  6. Feb 8, 2007 #5
    Thanks for the help :)
  7. Jan 17, 2008 #6
    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.
    Last edited: Jan 17, 2008
  8. Jan 17, 2008 #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.
  9. Jan 18, 2008 #8
    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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