 Solution

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.