= Solution
Let $M_0=M_1+M_2$, $M'=M_{\mathrm{NS}}+M_2=M_0-\Delta M$, and let the initial circular separation be $A$. The initial relative speed is $v_0=v_1+v_2$, with $v_0^2=GM_0/A$. 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> $\mu'=M_{\mathrm{NS}}M_2/M'$; the <orbital energy> is
$$
E'=\mu'\left(\frac12v_0^2-\frac{GM'}A\right)
=\frac{G\mu'}{2A}(2\Delta M-M_0).
$$
Consequently
$$
\boxed{\Delta M>\frac{M_0}{2}\ \Longrightarrow\ E'>0\quad\text{and disruption}.}
$$
Less 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.
Back to article page
