At fixed energy, the circular orbit has $L_c^2=GMa$, while
$$
L=L_c\sqrt{1-e^2}.
$$
An isotropic <galactic distribution function> has conditional angular-momentum density $n(L)\,dL\propto L\,dL$. Therefore
$$
n(e)\,de\propto L\left|\frac{dL}{de}\right|de
=L_c^2e\,de.
$$
Normalization on $0\leq e<1$ gives the <thermal eccentricity distribution>
$$
\boxed{n(e)=2e}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 320 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-320/1/b/solution