 Solution

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.