Kepler's law

The Two Body Problem: Kepler's Harmonic Law

For simplicity we assume circular orbits.
In the center of gravity frame we have m₁ at distance r₁ and m₂ at distance r₂ from the center of gravity and
m₁ r₁ = m₂ r₂.
Note that a=r₁+r₂ is the semimajor axis.
Both bodies have the same sidereal period w=2p/p (and of course a line through the bodies always includes the center of gravity).

The magnitude of the gravitational force on both bodies is Gm₁m₂/a² (action=reaction).

The centripital force on body 1 is m₁w²r₁ so setting gravitational = centripetal we have

Gm₁m₂/a² = m₁w²r₁     for m₁
Gm₁m₂/a² = m₂w²r₂     for m₂
or
Gm₂/a² = w²r₁
Gm₁/a² = w²r₂ Adding the two gives     G(m₁+m₂)/a² = w²a     (a=r₁+r₂)
or (w=2p/p)
(m₁+m₂)p² = (4p²/G)a³

If we measure m in solar masses, p in years and a in AU the (4p²/G) factor is unity and we get
(m₁+m₂)p² = a³
a form used in getting stellar masses from binary stars. In fact this is the only way we can get stellar masses!

If m₂ is much less than m₁ we can simplify the equation to

mp² = a³
a form used to estimate the distances to planets in other stellar systems if p is known (from spectroscopic line shifts like in spectroscopic binary stars or from periodic light dimming like in eclipsing binary stars). It is interesting to note that while the mass m can be reliably estimated from the spectrum of the star, the distance obtained is only sensitive to the cube root of the mass so the estimate can be quite accurate.

Finally, if we can take m₁+m₂ to be one we get Kepler's harmonic law, quite accurate for solar system objects, p² = a³.