1
$\begingroup$

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...

$\endgroup$
2
  • 4
    $\begingroup$ Hello and welcome to Astronomy SE. What exactly is your question here? $\endgroup$ Jul 17 at 18:47
  • 1
    $\begingroup$ @PierrePaquette the question has been nicely updated; I think it was just accidentally posted before it was finished. $\endgroup$
    – uhoh
    Jul 18 at 16:23
1
$\begingroup$

I tried to find detailed derivation online after my fail. Finally I found it in The Exoplanet Handbook by Michael Perryman:

Actrually K, or the so called radial velocity semi-amplitude, is not the peak radial velocity (which I understand before) but some kind of amplitude of the fluctuation of the radial velocity. And it must be considered in 3D space.
let z = star position alone the line of sight = $r(t) \sin(\omega + \theta)\sin i$ sketch

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: $$ \frac{M_2^3 \sin^3 i}{M_{tot}^2}=\frac{PK^3}{2\pi G}(1-e^2)^{3/2} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.