Newtonian Gravity – Are Orbital Eccentricities in Binary System Always the Same?

Question

Some excercises on Kepler laws and binary system use this relation $$\frac{r_1}{r_2} = \frac{a_1}{a_2},$$
where $r$ is the distance from the center of mass to each object and $a$ is the semi-major axis of its orbit. Given this equation
$$ r = \frac{a(1 – e^2)}{1 + e\cos\theta} $$
and the fact that both bodies are always opposite to each other, the only way the first relation holds is if $e_1 = e_2$, but I don't know how to prove that.

Answer 1

Place the center of your coordinates at the center of mass; this says that $M_{1}\vec{r}_{1}+M_{2}\vec{r}_{2}=0$. Since there are no external forces, the center of mass does not move, and this equation holds for all time. Therefore, we can express the position of one body in terms of the other and the mass ratio: $$\vec{r}_{2}(t)=-\frac{M_{1}}{M_{2}}\vec{r}_{1}(t).$$ Therefore, the second mass traces out an orbit that is exactly the same shape as the orbit of the first mass. The sizes of the orbits differ by the factor $M_{1}/M_{2}$, but their eccentricities are identical.

Note that the statement that the two trajectories have the same shape (just rescaled) is true in any two-body problem, not just the Kepler problem. However, only in the Kepler problem are the orbits described by closed ellipses (and thus describable with an eccentricity and major axis).