Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook