<p>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 <em>think</em> the result then also depends on the distance.</p>

CC BY-SA 3.0
2015-04-20T15:36:30.870Z
Surely the absolute magnitude of the Sun needs to be subtracted from the LHS of eqn. 4?
ProfRob
CC BY-SA 3.0
2015-04-21T05:00:30.760Z
@RobJeffries: You're right! Fixed.
Warrick
CC BY-SA 3.0
2015-04-17T07:47:13.010Z
2015-04-21T05:00:26.220Z
2015-04-21T05:00:26.220Z
Warrick
<p>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?</p>

<star><apparent-magnitude>
How to calculate magnitude of a star in a triple star system?
dogtag
# How to calculate magnitude of a star in a triple star system?

(Asked by: username_0 on 2015-04-16)

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?

**Tags:** star, apparent-magnitude


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## 1 answers

### Answer by username_1 on 2015-04-17. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_2**: Surely the absolute magnitude of the Sun needs to be subtracted from the LHS of eqn. 4?  
> **username_1**: @RobJeffries: You're right! Fixed.  

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