Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2 / iv / Solution

At release,At large radius the potential vanishes, soThe asymptote satisfies $1+e\cos f_{\infty }=0$, where $f=\theta -\varpi$. If the angular displacement from the release direction is $\Delta =\pi -f_{\infty }$, thenFor $\beta \gg 1$, $-\cos \Delta \simeq -1+{\Delta }^2/2$, and therefore