# derivation of binary mass function for eccentric orbit

When investigating methods to detect exoplanets, I learned about binary mass function(BMF) which could be applied to obtain radial velocity and the mass. I've derived BMF for a circular orbit, but I get wrong answer when it comes to an eccentric orbit.

$$\frac{M_2^3}{M_{tot}^2}=\frac{4\pi^2}{G P^2}a_1^3 \\$$ from: $$-\frac{GM_1M_2}{a(1-e)}+ \frac{1}{2}M_1v_1^2 + \frac{1}{2}M_2v_2^2 =-\frac{GM_1M_2}{a(1+e)}+ \frac{1}{2}M_1v_1^{{'2}} + \frac{1}{2}M_2v_2^{'2} \\ M_1v_1a_1(1-e)=M_1v_1^{'}a_1(1+e)\\ M_2v_2a_2(1-e)=M_2v_2^{'}a_2(1+e)\\ M_1v_1=M_2v_2\\ a = a_1\frac{M_{tot}}{M_2}$$ I got: $$a_1=\frac{1+e}{1-e}\frac{GM_2^3}{M_{tot}^2 v_1^2}$$ so: $$\frac{M_2^3}{M_{tot}^2}=\frac{Pv_1^3}{2\pi G}(\frac{1-e}{1+e})^{3/2}=\frac{PK^3}{2\pi G \sin^3 i}(\frac{1-e}{1+e})^{3/2}$$ However, result from Wikipedia is: $$f = \frac{M_2^3 \sin^3 i}{M_{tot}^2} = \frac{P K^3}{2\pi G}(1-e^2)^{3/2}$$

I'm wondering what's wrong with my calculation...

• Hello and welcome to Astronomy SE. What exactly is your question here? Jul 17 at 18:47
• @PierrePaquette the question has been nicely updated; I think it was just accidentally posted before it was finished.
– uhoh
Jul 18 at 16:23

let z = star position alone the line of sight = $$r(t) \sin(\omega + \theta)\sin i$$
then radial velocity $$v_r = \dot{z} = \sin{i}[\dot{r}\sin(\omega + \theta)+r \dot{\theta}\cos(\omega+\theta)]$$.
on the other hand, we have: $$r = \frac{a_1(1-e^2)}{1+e\cos{\theta}} \\ \cos{\theta} = \frac{\cos{E}-e}{1-e\cos{E}} \\ E - e\sin{E} = \frac{2\pi}{P}t$$ where $$E=E(t)$$ is the so called eccentric anomaly.
After lengthy calculation, we get: $$v_r(t) = \frac{2\pi a_1}{P}\frac{\sin{i}}{\sqrt{1-e^2}}[e\cos{\omega} + \cos(\omega + \theta(t))]$$ and here appears K (amplitude):
$$K = \frac{2\pi a_1}{P}\frac{\sin{i}}{\sqrt{1-e^2}}$$
use this to substitute $$a_1$$ in Kepler's Third Law: $$\frac{M_2^3}{M^2_{tot}}=\frac{4π^2}{GP^2}a_1^3$$ at last there is: $$\frac{M_2^3 \sin^3 i}{M_{tot}^2}=\frac{PK^3}{2\pi G}(1-e^2)^{3/2}$$