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Given a pair of binary stars rotating in a circular orbit where one of the stars undergoes a supernova explosion, there is a maximal amount of mass that may be lost without unbinding the system.

Assume the masses of the stars are $ M_1^0 $ and $ M_2 $ and the radius of their orbit is $ a_0 $. The first star explodes in a supernova, and loses an amount of mass $ \Delta M=M_1^0-M_1 $. We further assume that the explosion is instantaneous and spherically symmetric and so due to conservation of momentum, no "kick" is given to the exploding star, so its velocity does not change.

The energy before the explosion is:

$ E_0=-\frac{G M_1^0 M_2}{2a_0}=-\frac{1}{2}\mu_0 v_0^2 \rightarrow \frac{GM}{a_0}=v_0^2 $

where $ \mu_0 $ is the reduced mass, $ M $ the total mass and $ v_0 $ is the relative velocity before the explosion. In the second equality we use the fact that in a circular orbit $ E_p=-\frac{1}{2}E_k $.

After the explosion, the energy is

$ E=-\frac{G M_1 M_2}{2a}=-\frac{G M_1 M_2}{a_0} +\frac{1}{2}\mu v_0^2 $

where $ a $ is the semi major axis of the system after the explosion and we have used the fact that right after the explosion the separations and relative velocities have not yet changed.

The limiting case for a post-explosion bound orbit ($ E<0 $) is given when $ E \rightarrow 0 $ (or equivalently $ a \rightarrow \infty $). At this limit

$ -\frac{G M_1 M_2}{a_0} +\frac{1}{2}\mu v_0^2=0 \longrightarrow \frac{G (M-\Delta M)}{a_0}=\frac{1}{2} v_0^2 $

Plugging in $ v_0^2 $ we obtain $ \frac{G (M-\Delta M)}{a_0}=\frac{G M}{2a_0} $, leading to $ \Delta M=M/2 $. Therefore, if the exploding star loses more than half of the mass of the system in the supernova, it will become unbound.