<p>I found the answer under the assumption of circular orbits of a binary system around their center of mass:</p>
<p>Let <span class="math-container">$a_1$</span> and <span class="math-container">$a_2$</span> be the radius of the orbit around the center of mass of <span class="math-container">$M$</span> and <span class="math-container">$M^*$</span> respectively. It follows that: <span class="math-container">$$a = a_1 + a_2$$</span>
The center of mass satisfies the equation: <span class="math-container">$$ a_1M  = a_2M^*  $$</span>
From the above two equations we get:
<span class="math-container">$$ a_2 = a\frac{M}{M + M^*} $$</span></p>
<p>Newton's second law for body <span class="math-container">$M^*$</span> is:
<span class="math-container">$$ M^* \omega² a_2 = \frac{G M M^*}{a²} $$</span>
Where <span class="math-container">$\omega$</span> is the angular velocity of the circular orbit.
Solving for <span class="math-container">$\omega$</span> using the expression for <span class="math-container">$a_2$</span> we get:</p>
<p><span class="math-container">$$ \omega = \sqrt{ \frac{G (M+M^*)}{a^3}}$$</span></p>
<p>For the orbital mean motion <span class="math-container">$n$</span> there should be a factor of <span class="math-container">$2 \pi$</span>, but I guess it was just omitted</p>

CC BY-SA 4.0
2021-11-29T22:15:57.420Z
The standard units of mean motion are radians per unit time,  so no, there isn't a factor of $2\pi$ missing.  Mean motion is just $2\pi$  divided by the orbital period. (Had wrong value, corrected now)
notovny
CC BY-SA 4.0
2021-11-30T10:06:40.413Z
A useful quantity in two-body calculations is the [reduced mass](https://en.wikipedia.org/wiki/Reduced_mass).
PM 2Ring
CC BY-SA 4.0
2021-11-29T22:10:33.447Z
2021-11-29T22:10:33.447Z
ValientProcess
<p>I'm reading an <a href="https://academic.oup.com/mnras/article/458/3/2890/2589321" rel="nofollow noreferrer">article</a> about tidal forces, and the expression for the mean motion is given by:</p>
<p><span class="math-container">$$ n = \sqrt{ \frac{G(M + M^*)}{a^3}}$$</span></p>
<p>Where <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$M$</span> is the mass of the primary body, <span class="math-container">$M^*$</span> is the mass of the perturbing body, and <span class="math-container">$a$</span> is the semi-major axis. I know that Kepler laws were used to derive this equation, but I'm not sure exactly how and what are the underlying assumptions</p>

<orbital-mechanics><kepler>
Orbital mean motion expression derivation
ValientProcess
# Orbital mean motion expression derivation

(Asked by: username_0 on 2021-11-29)

I'm reading an [article](https://academic.oup.com/mnras/article/458/3/2890/2589321) about tidal forces, and the expression for the mean motion is given by:

$$ n = \sqrt{ \frac{G(M + M^\*)}{a^3}}$$

Where $G$ is the gravitational constant, $M$ is the mass of the primary body, $M^\*$ is the mass of the perturbing body, and $a$ is the semi-major axis. I know that Kepler laws were used to derive this equation, but I'm not sure exactly how and what are the underlying assumptions

**Tags:** orbital-mechanics, kepler

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## 1 answers

### Answer by username_0 on 2021-11-29. Score: 1 | Accepted answer

I found the answer under the assumption of circular orbits of a binary system around their center of mass:

Let $a\_1$ and $a\_2$ be the radius of the orbit around the center of mass of $M$ and $M^\*$ respectively. It follows that: $$a = a\_1 + a\_2$$
The center of mass satisfies the equation: $$ a\_1M = a\_2M^\* $$
From the above two equations we get:
$$ a\_2 = a\frac{M}{M + M^\*} $$

Newton's second law for body $M^\*$ is:
$$ M^\* \omega² a\_2 = \frac{G M M^\*}{a²} $$
Where $\omega$ is the angular velocity of the circular orbit.
Solving for $\omega$ using the expression for $a\_2$ we get:

$$ \omega = \sqrt{ \frac{G (M+M^\*)}{a^3}}$$

For the orbital mean motion $n$ there should be a factor of $2 \pi$, but I guess it was just omitted

> **Comments:**
> **username_1**: The standard units of mean motion are radians per unit time,  so no, there isn't a factor of $2\pi$ missing.  Mean motion is just $2\pi$  divided by the orbital period. (Had wrong value, corrected now)  
> **username_2**: A useful quantity in two-body calculations is the [reduced mass](https://en.wikipedia.org/wiki/Reduced_mass).  

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