= 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> $\theta$,
$$
r=\frac{a(1-e^2)}{1+e\cos\theta}=a[1-e\cos\theta]+O(e^2).
$$
Using the <vis-viva equation>, $v^2=(GM_\star/a)[1+2e\cos\theta]+O(e^2)$ and hence $v=na[1+e\cos\theta]+O(e^2)$. Therefore
$$
\boxed{\lambda_\ell(\theta)\propto1-e\cos\theta+O(e^2).}
$$
For 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 $1-2e\cos\theta$.
Back to article page
