= 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>:
$$
\boxed{E_{\rm explosion}=\frac m4[u_c^2+(2V-u_c)^2]-\frac m2V^2=\frac m2(V-u_c)^2.}
$$
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.
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