Past exam of the physics 1 paper of the University of Oxford / 2017 / 11 / d / Solution

A $90^{\circ }$ gravitational scattering angle has $\cot (\varphi /2)=1$. ThusThe denominator uses the squared speed in $\mathrm{m}{\mathrm{s}}^{-1}$, not its numerical value in $\mathrm{km}{\mathrm{s}}^{-1}$. From b, the corresponding periapsis distance is $r_{\min }=(\sqrt{2}-1)b\simeq 5.53\times 10^8\mathrm{m}$; this differs substantially from the impact parameter because of gravitational focusing.