# Kepler's law for comparable masses

I've recently been studying about Kepler's 3rd law... I went through the modification of it when the orbiting masses are comparable in terms of their mass.It was almost symmetric with the one I learnt first, just had the sum of two masses downstairs. $$T^2= \frac{4π^2a^3}{G(M+m)}$$ But, maybe I have some misconceptions about the semi major axis in it. Isn't it half of the distance between the masses?

• en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes – G. Smith Sep 4 '19 at 14:32
• In an elliptical orbit, the distance between the masses isn’t constant. – G. Smith Sep 4 '19 at 14:34
• In the derivation, I only used the fact that the area of the eclipse is πab. And when i use highest distance/2 as a, it gives the wrong answer. – Golam Ishtiak Sep 4 '19 at 17:19
• $a$ is half of the greatest separation, so your derivation must have an error. See Wikipedia for confirmation of the formula. – G. Smith Sep 4 '19 at 19:26