Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 5

Write $\mu =GM$. In polar coordinates, Newton's second law per unit mass gives the radial and transverse equations of motionThe transverse equation impliesThis is conservation of angular momentum per unit mass. For the nonradial trajectory in the next part, choose orientation so $h>0$. With $u=1/r$ and primes denoting derivatives with respect to $\theta$, the chain rule givesSubstituting into the radial equation yields $-h^2{u}^2(u^{\prime \prime }+u)=-\mu u^2$, henceThe change of independent variable requires $h\ne 0$; a purely radial trajectory instead has constant $\theta$.

Far from the planet, the magnitude of angular momentum per unit mass is the magnitude of the cross product of position vector and velocity. The perpendicular distance from the incoming asymptote is the impact parameter $d$, soChoose $\theta =0$ at closest approach. The general solution of the differential equation in a, with $u^{\prime }(0)=0$, isThe specific orbital energy isAt infinity $\epsilon =v^2/2$, giving $e=\sqrt{1+h^2{v}^2/{\mu }^2}>1$. Thus the hyperbolic orbit hasA different zero of polar angle simply replaces $\theta$ by $\theta -{\theta }_0$.

The incoming and outgoing asymptotes have radial directions $-{\theta }_{\infty }$ and ${\theta }_{\infty }$, where $\cos {\theta }_{\infty }=-1/e$ and $\pi /2<{\theta }_{\infty }<\pi$. The incoming velocity points opposite its radial direction, so the deflection angle between incoming and outgoing velocities is $\delta =2{\theta }_{\infty }-\pi$. HencePutting $\delta =\pi /2$ gives