Past exam of the physics 1 paper of the University of Oxford / 2016 resit / 10

Use the test-particle approximation, neglecting the planet's recoil. The conserved specific angular momentum is $j=bv$ and the specific energy at infinity is $v^2/2$. At closest approach the radial velocity vanishes, so the effective potential givesThe positive root and the tangential speed areThe second form for $r_0$ avoids subtracting two nearly equal numbers for a small impact parameter. These formulas describe a nonradial orbit; $b=0$ gives a collision rather than a finite nonzero pericentre in the point-planet model.

Put $V=v_{\max }$ and let $u_c=\sqrt{GM/r_0}$ be the circular orbit speed. Take the captured half to move in the original tangential direction, the minimum-energy choice. Conservation of momentum during the short explosion gives the other half speed $2V-u_c$. The gravitational potential energy is unchanged at the instant of separation, so the necessary generated energy is the increase in kinetic energy:The captured half has zero radial velocity and exactly the required tangential speed, so it can enter the stable Newtonian circular orbit at this radius. Internal self-binding and any mass lost to radiation are neglected.

For the prograde captured orbit, the departing half has instantaneous velocity tangent to the original orbit, in the original direction, with speedIt exceeds the local escape speed: since $V\ge \sqrt{2}{u}_c$, $u_{\mathrm{depart}}\ge (2\sqrt{2}-1)u_c>\sqrt{2}{u}_c$. This is its speed immediately after the explosion, not its later speed at infinity.

If a counterrotating circular captured orbit is allowed, its signed velocity is $-u_c$, the outgoing speed is instead $2V+u_c$, and the required generated energy is $m(V+u_c{)}^2/2$. The usual answer takes the prograde orbit; specifying that choice removes the directional ambiguity.