Orbital Period Equation According to Kepler’s Third Law





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Orbital Period Equation

In general, two masses, $m$ and $M$ will orbit around the center of mass of the system and the system can be replaced by the motion of the reduced mass, $\mu \equiv mM/(m+M)$. It is important to understand that the following equations are valid for elliptical orbits (i.e., not just circular), and for arbitrary masses (i.e., not just for the case for one mass much less than the other). The only resitriction is that the motion is purely due to gravity and that the motion in nonrelativistic (i.e., orbital speeds negligible compared to the speed of light).

Let’s define:
$T = \ $ orbital period,
$a = \ $ semimajor axis of the elliptical orbit (i.e., half of the largest symmetry axis of the ellipse),
$m, \ M \ = \ $ masses of objects participating in the orbit (e.g., a planet and a star, two stars, etc.)

For arbitary units:

\[
T \ = \ 2\pi \left[\frac{a^3}{G(M+m)}\right]^{\frac{1}{2}}
\]

For convenient units of years and AU:

Since we know that for the Sun-Earth system $T$ is 1 year for $a=1 \ {\rm AU}$ (AU $=$ astronomical units), and for $M+m$ in units of solar mass,

\[
T ({\rm years}) \ = \ \left[\frac{a_{\rm AU}^3}
{(M+m) ({\rm in \ solar \ masses})}\right]^{\frac{1}{2}}
\]

where $a_{\rm AU}$ is the semimajor axis in units of AU.