Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 10

About a chosen origin $O$, a particle has angular momentum $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. A body has $\mathbf{L}={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$, or the corresponding integral for a continuous mass distribution. For a fixed origin in an inertial frame,Internal central pair forces cancel in the total torque. About a fixed origin, a central force conserves a particle's angular momentum; zero net force alone is insufficient if an external couple remains.

Put $\mu =GM$. At the turning point of a radial launch the speed vanishes, so conservation of energy givesThis result is exact within the exterior Newtonian gravitational potential. The condition $r_0\gg R$ means $v_0$ is just below the escape velocity: $1-v_0^2/v_{\mathrm{esc}}^2\ll 1$. Replacing gravity by a constant would be inappropriate in this regime.

For a tangential launch, conserved specific angular momentum is $j=Rv$. At the outer turning point the radial velocity vanishes but the tangential speed is $j/r_0$. ThusUsing a eliminates the gravitational potential difference and givesThe expression printed in the source is the leading correction for $r_0\gg R$, rather than an exact identity. Equivalently, $v^2=2\mu r_0/[R(r_0+R)]$, so the launch speed exceeds the circular value and the surface is the periapsis of the nonintersecting ideal orbit.

At a fixed launch radius the escape threshold is zero total mechanical energy. It depends on the initial speed alone:A direction aimed into the planet can of course cause impact before escape. If the planet rotates, the ground-relative launch velocity must be added to $\mathbf{\Omega }\times \mathbf{R}$ to obtain the inertial velocity. The condition becomes $|\mathbf{u}+\mathbf{\Omega }\times \mathbf{R}{|}^2\ge 2GM/R$: prograde launch needs less ground-relative speed than retrograde launch. For a spherical planet the threshold expressed as inertial speed at the same radius remains independent of direction.