 Solution

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.