PX158 - G1 - two masses in a mutual circular orbit

generally difficult to measure the mass of a distant object
binary stars give a way to measure mass
assume a binary star system in the plane of the sky and that the orbits are circular (ie: $e c c e n t r i c i t y (e) = 0$ )
separation between the stars, $R = r_{1} + r_{2}$
by definition of the centre of mass:

M_{1} r_{1} = M_{2} r_{2}

\begin{aligned} F_{G} & = \frac{G M_{1} M_{2}}{R^{2}}; & F & = m a \\ F_{C} & = m Ω^{2} r; & Ω & = \frac{2 π}{P} \\ m_{1} Ω^{2} r_{1} & = \frac{G M_{1} M_{2}}{R^{2}}; & m_{2} Ω^{2} r_{2} & = \frac{G M_{1} M_{2}}{R^{2}} \end{aligned}

\begin{aligned} Ω^{2} (r_{1} + r_{2}) & = \frac{G}{R^{2}} (M_{1} + M_{2}) \\ Ω^{2} & = \frac{G}{R^{3}} (M_{1} + M_{2}) \\ ∴ \frac{4 π^{2}}{P^{2}} & = \frac{G}{R^{3}} (M_{1} + M_{2}) \end{aligned}

generalized kepler's law
eg: what is the mass of the sun based on earth's orbit
$\frac{4 π^{2}}{P^{2}} = \frac{G}{R_{S E}^{3}} (M_{S} + M_{E})$
$M_{S} >>> M_{E} ⟹ M_{S} + M_{E} \approx M_{S}$
$P = 365.25 d a y s$
$R_{S E} = 1 A U = 1.5 \times 10^{11} m$

∴ M_{S} \approx 2 \times 10^{30} k g

\frac{4 π^{2}}{P^{2}} = \frac{G}{R^{3}} (M_{1} + M_{2}) ⟹ M_{1} + M_{2} = 1.17 M_{⊙}

M_{1} R_{1} = M_{2} R_{2} ⟹ \frac{M_{1}}{M_{2}} = \frac{R_{2}}{R_{1}} = \frac{2}{1}

∴ M_{1} = \frac{2}{3} 1.17 M_{⊙}; M_{2} = \frac{1}{3} 1.17 M_{⊙}