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

Let $M_0=M_1+M_2$, $M^{\prime }=M_{\mathrm{NS}}+M_2=M_0-\Delta M$, and let the initial circular separation be $A$. The initial relative speed is $v_0=v_1+v_2$, with $v_0^2=G{M}_0/A$. Instantaneous symmetric mass loss changes neither the separation nor either surviving star's velocity at that instant. In the new centre-of-mass frame, use the reduced mass ${\mu }^{\prime }=M_{\mathrm{NS}}{M}_2/M^{\prime }$; the orbital energy isConsequentlyLess than half the total initial binary mass lost gives a bound orbit, not necessarily a circular one. Exactly half gives zero orbital energy and the marginal parabolic escape case. The threshold concerns the total mass of both stars, not just the exploding star. Using the old centre-of-mass kinetic energy without subtracting the new systemic motion would give the wrong binding criterion; the relative-velocity form performs that subtraction automatically.