Past exam of the astrophysics course of the University of Oxford / 2018 / 6 / d / Solution

Write $f=M^{\prime }/M_0$ for the retained fraction and $q=v^{-}/v_0$ for an opposite kick. With more than half the mass lost, $0<f<1/2$. A same-direction kick cannot bind the pair, because $1+v^{+}/v_0\ge 1>\sqrt{2f}$. An opposite kick can reduce the relative speed enough to leave negative energy:Both endpoints are marginal zero-energy cases. Kicks somewhat larger than $v_0$ can still satisfy this interval, but then the direction of the relative orbital motion is reversed.

At $v^{-}=v_0=v_1+v_2$, the two surviving stars have identical instantaneous velocities, so their relative velocity and orbital angular momentum vanish. Their energy is nevertheless negative,They fall radially toward each other from rest in their new centre-of-mass frame. In the point-mass idealization the collision orbit has semimajor axis $A/2$ and zero pericentre; real finite-size stars collide or interact before reaching zero separation. It is not a stable circular binary. Zero relative angular momentum, rather than the negative energy alone, determines this plunge outcome.