Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 3 / vii / Solution

For a phase-mixed orbit, probability in a short arc is $dt/T=ds/(Tv)$, so the line density on a Kepler orbit is inversely proportional to speed. At true anomaly $\theta$,Using the vis-viva equation, $v^2=(G{M}_{\star }/a)[1+2e\cos \theta ]+O(e^2)$ and hence $v=na[1+e\cos \theta ]+O(e^2)$. ThereforeFor one particle, the normalized line probability is $(2\pi a{)}^{-1}(1-e\cos \theta )+O(e^2)$. This is density per arc length; density per unit polar angle has a further factor $ds/d\theta \simeq a(1-e\cos \theta )$ and is proportional to $1-2e\cos \theta$.