Quick question on derivation of mass of star

Question

How do I show that for a binary star system, if one star has mass $M_s$, speed $V_s$, period $P$, the mass of the other star is given by:

$$M_P^3 \approx \frac{V_s^3}{2\pi G} PM_s^2$$

I showed it using Kepler's second law and using centripetal force motion. But I made the huge assumption of $(M_p+M_s) \approx M_s$. Starting with centripetal force equation:

$$\frac{GM_pM_s}{(a_p+a_s)^2} = \frac{M_s v_s^2}{a_s} $$

Substituting in $PV_s = 2\pi a_s$:

$$ m_p = \frac{2\pi(a_p+a_s)^2V_s}{PG} $$:

Using kepler's second law: $P^2 = \frac{(a_p+a_s)^3(2\pi)^2}{G(M_p+M_s)} $ :

$$M_p^3 = \frac{V_s^3}{2\pi G} P (M_p+M_s)^2 $$

I feel like I'm missing something here..

Answer 1

I think your derivation is fine. In general we cannot measure the velocity of the secondary directly, but only the "projected" velocity $V_s \sin(i) = K_s$, where $i$ is the inclination of the orbital axis to the line of sight ($i=90^{\circ}$ is an orbit seen edge-on).

In those circumstances your derivation becomes $$ \frac{M_{p}^{3} \sin^3 i}{(M_p + M_s)^2} = \frac{P K_{s}^{3}}{2\pi G} $$

The right hand side contains observable quantities, the left hand side contains the masses and the inclination angle. If only the secondary (projected) velocity can be measured this is as far as you can go without making assumptions about the relative masses of the components and the inclination. To obtain the formula you quote then indeed you have to say that $M_p \ll M_s$ and that $V_s$ is the measured projected velocity (i.e. that $i=90^{\circ}$).