= Solution
At a fixed launch radius the escape threshold is zero total <mechanical energy>. It depends on the initial <speed> alone:
$$
\boxed{v_{\rm esc}=\sqrt{2GM/R}\quad\text{independent of outward launch direction in the nonrotating model}.}
$$
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 $\boldsymbol\Omega\times\mathbf R$ to obtain the inertial <velocity>. The condition becomes $|\mathbf u+\boldsymbol\Omega\times\mathbf R|^2\ge2GM/R$: 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.
Back to article page
