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

Average the phase probability over the <thermal eccentricity distribution>:
$$
f_{\rm unbound}
=\int_0^1 2e\left(\frac12-\frac e\pi\right)de.
$$
Therefore
$$
\boxed{
f_{\rm unbound}=\frac12-\frac{2}{3\pi}
\simeq0.288}.
$$

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

{title2=$n(e)=2e$}

An isotropic population of bound Kepler orbits has eccentricity probability density $n(e)=2e$ on $0\leq e<1$. It follows from the fixed-energy angular-momentum measure $n(L)dL\propto LdL$ and $L=L_c\sqrt{1-e^2}$.

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 Incoming links: Thermal eccentricity distribution