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Consider Kepler's third law

The square of the orbital period $\tau$ of a planet is directly proportional to the cube of the semi-major axis of its orbit $a$.

$\frac{\tau^2}{a^3}=k$

The constant depends on both the mass of the star $M$ and the mass of the planet $m$.

$k=\frac{4\pi^2}{(M+m) \gamma}$

Where $\gamma$ is the universal gravitation constant.

Nevertheless (here is my doubt) considering the simple case of a circular orbit the following must hold.

$\frac{-\gamma M m}{r^2}=m \omega^2 r \implies \frac{\tau^2}{r^3}=k=\frac{4 \pi^2}{M \gamma}$

In this case $m$ is not involved in the expression of $k$. Is this just the consequence of an approximation in considering that the star $M$ is not moving and it is an inertial frame of reference? Or is there something different here? I'm a bit confused about it.

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    $\begingroup$ The problem is that you assumed the star doesn't move. This is a pretty good approximation for our solar system, but fails in general; look up the term 'effective mass'. $\endgroup$ – knzhou Mar 23 '16 at 21:29
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    $\begingroup$ Reduced mass. $\endgroup$ – Qmechanic Mar 23 '16 at 21:35
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The assumption you made in your derivation is that the planet mass is far less than the star mass, i.e. $m\ll M$. In that limit, the Kepler's constant

$$k=\frac{4\pi^2}{(M+m)\gamma}\to \frac{4\pi^2}{M\gamma},$$

is indeed independent of the planet mass $m$. Away from that limit, the star is not an inertial frame, and you need to replace the planet (kinetic) mass by its reduced mass $m\to Mm/(M+m)$, and still for the simple circular orbit, you can already get the correct Kepler constant

$$\frac{-\gamma Mm}{r^2}=\frac{Mm}{M+m}\omega^2 r\implies k=\frac{4\pi^2}{(M+m)\gamma}.$$

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