<p>I tried to find detailed derivation online after my fail. Finally I found it in <strong>The Exoplanet Handbook</strong> by <em>Michael Perryman</em>:</p>
<p>Actrually K, or the so called <strong>radial velocity semi-amplitude</strong>, is not the peak radial velocity (which I understand before) but <strong>some kind of amplitude of the fluctuation of the radial velocity</strong>. And it must be considered in 3D space.<br />
let z = star position alone the line of sight = <span class="math-container">$r(t) \sin(\omega + \theta)\sin i$</span>
<a href="https://i.sstatic.net/mUvSn.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mUvSn.jpg" alt="sketch" /></a></p>
<p>then radial velocity <span class="math-container">$v_r = \dot{z} = \sin{i}[\dot{r}\sin(\omega + \theta)+r \dot{\theta}\cos(\omega+\theta)]$</span>.<br />
on the other hand, we have:
<span class="math-container">$$
r = \frac{a_1(1-e^2)}{1+e\cos{\theta}} \\
\cos{\theta} = \frac{\cos{E}-e}{1-e\cos{E}} \\
E - e\sin{E} = \frac{2\pi}{P}t
$$</span>
where <span class="math-container">$E=E(t)$</span> is the so called <strong>eccentric anomaly</strong>.<br />
After lengthy calculation, we get:
<span class="math-container">$$
v_r(t) = \frac{2\pi a_1}{P}\frac{\sin{i}}{\sqrt{1-e^2}}[e\cos{\omega} + \cos(\omega + \theta(t))]
$$</span>
and here appears K (amplitude):<br />
<span class="math-container">$$ K = \frac{2\pi a_1}{P}\frac{\sin{i}}{\sqrt{1-e^2}}$$</span><br />
use this to substitute <span class="math-container">$a_1$</span> in Kepler's Third Law:
<span class="math-container">$$
\frac{M_2^3}{M^2_{tot}}=\frac{4π^2}{GP^2}a_1^3
$$</span>
at last there is:
<span class="math-container">$$
\frac{M_2^3 \sin^3 i}{M_{tot}^2}=\frac{PK^3}{2\pi G}(1-e^2)^{3/2}
$$</span></p>

CC BY-SA 4.0
2021-07-18T17:46:38.113Z
2021-07-18T18:19:14.190Z
2021-07-18T18:19:14.190Z
Kyle
<p>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.</p>
<p><span class="math-container">$$
\frac{M_2^3}{M_{tot}^2}=\frac{4\pi^2}{G P^2}a_1^3 \\
$$</span>
from:
<span class="math-container">$$
-\frac{GM_1M_2}{a(1-e)}+ \frac{1}{2}M_1v_1^2 + \frac{1}{2}M_2v_2^2
=-\frac{GM_1M_2}{a(1+e)}+ \frac{1}{2}M_1v_1^{{'2}} + \frac{1}{2}M_2v_2^{'2} \\
M_1v_1a_1(1-e)=M_1v_1^{'}a_1(1+e)\\
M_2v_2a_2(1-e)=M_2v_2^{'}a_2(1+e)\\
M_1v_1=M_2v_2\\
a = a_1\frac{M_{tot}}{M_2}
$$</span>
I got:
<span class="math-container">$$
a_1=\frac{1+e}{1-e}\frac{GM_2^3}{M_{tot}^2 v_1^2}
$$</span>
so:
<span class="math-container">$$
\frac{M_2^3}{M_{tot}^2}=\frac{Pv_1^3}{2\pi G}(\frac{1-e}{1+e})^{3/2}=\frac{PK^3}{2\pi G \sin^3 i}(\frac{1-e}{1+e})^{3/2}
$$</span>
However, result from <a href="https://en.wikipedia.org/wiki/Binary_mass_function" rel="nofollow noreferrer">Wikipedia</a> is:
<span class="math-container">$$
f = \frac{M_2^3 \sin^3 i}{M_{tot}^2} = \frac{P  K^3}{2\pi  G}(1-e^2)^{3/2}
$$</span></p>
<p>I'm wondering what's wrong with my calculation...</p>

CC BY-SA 4.0
2021-07-17T18:47:01.053Z
Hello and welcome to Astronomy SE. What exactly is your question here?
Pierre Paquette
CC BY-SA 4.0
2021-07-18T16:23:42.660Z
@PierrePaquette the question has been nicely updated; I think it was just accidentally posted before it was finished.
uhoh
<eccentric-orbit><radial-velocity>
derivation of binary mass function for eccentric orbit
Kyle
# derivation of binary mass function for eccentric orbit

(Asked by: username_0 on 2021-07-17)

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](https://en.wikipedia.org/wiki/Binary_mass_function) 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...

**Tags:** eccentric-orbit, radial-velocity

> **Comments:**
> **username_1**: Hello and welcome to Astronomy SE. What exactly is your question here?  
> **username_2**: @username_1 the question has been nicely updated; I think it was just accidentally posted before it was finished.  

---

## 1 answers

### Answer by username_0 on 2021-07-18. Score: 1

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](https://i.sstatic.net/mUvSn.jpg)](https://i.sstatic.net/mUvSn.jpg)

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:
$$
\frac{M\_2^3 \sin^3 i}{M\_{tot}^2}=\frac{PK^3}{2\pi G}(1-e^2)^{3/2}
$$


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