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I want to find a star's magnitude in a triple star. I already knew the the magnitude of the triple star and the other two star, but I don't quit know how to solve it. Is there a way to find it?

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Basically you need to convert between luminosities (which you can add) and magnitudes using $$M-M_\odot=-2.5\log_{10}(L/L_\odot)$$ Let's call the total luminosity $L_0$ and magnitude $M_0$ and the individual luminosities and magnitudes $L_1$, $L_2$ and $L_3$ and $M_1$, $M_2$ and $M_3$. Then, you have the total luminosity of the system, directly $$L_0/L_\odot=10^{-0.4(M_0-M_\odot)}$$ and as the sum of the components $$L_0/L_\odot=(L_1+L_2+L_3)/L_\odot=10^{-0.4(M_1-M_\odot)}+10^{-0.4(M_2-M_\odot)}+10^{-0.4(M_3-M_\odot)}$$ Solving these equations for $M_3$ gives $$M_3-M_\odot=-2.5\log_{10}\left(10^{-0.4(M_0-M_\odot)}-10^{-0.4(M_1-M_\odot)}-10^{-0.4(M_2-M_\odot)}\right)$$ I'm assuming you have absolute magnitudes, but you can rewrite the formulae in terms of apparent magnitudes using $$M=m+5(1-\log_{10}d)$$ but I think the result then also depends on the distance.

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    $\begingroup$ Surely the absolute magnitude of the Sun needs to be subtracted from the LHS of eqn. 4? $\endgroup$ – Rob Jeffries Apr 20 '15 at 15:36
  • $\begingroup$ @RobJeffries: You're right! Fixed. $\endgroup$ – Warrick Apr 21 '15 at 5:00

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