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Consider a binary system on a circular orbit, consisting of stars of masses  M_0, M_1 . We shall assume that the star  M_0 explodes suddenly and symmetrically (thus, receiving no "kick" due to the mass ejection) and its remaining mass is  M_f . The following dynamics of the system are determined by the conservation of energy and angular momentum after the explosion. From energy conservation we get:

 \mu_f \frac{v_{i}^2}{2}-\frac{GM_1 M_f}{a_i}=-\frac{G M_1 M_f}{2 a_f}

where  \mu_f=M_1 M_f/(M_1+M_f) is the reduced mass,  v_{i} is the stars' relative velocity and  a_i,a_f are the semi-major axis' before and after the explosion correspondingly. The L.H.S. is the gravitational + kinetic energy right after the explosion, and the R.H.S. is the Keplerian energy of the resulting orbit. From this equation we get that: \frac{v_{i}^2}{2(M_1+M_f)}-\frac{G}{a_i}=-\frac{G}{2a_f}

Plugging in  v_{i}^2=G(M_0+M_1)/a_i and defining  \chi=(M_0+M_1)/(M_f+M_1)=1+\Delta M/M_{tot,f} (where  \Delta M is the ejected mass and  M_{tot,f} is the total final mass of the two stars), we have:

 \frac{a_i}{a_f}=1-\frac{\Delta M}{M_{tot,f}} .

From angular conservation, we have:

 \mu_f a_i v_{i}=\mu_f \sqrt{G(M_1+M_f)a_f(1-e^2)}

where, again, the L.H.S. is the angular momentum right after the explosion and the R.H.S. is the Keplerian angular momentum of the resulting orbit. Squaring this equation and plugging in  \chi we get:

 1-e^2=\frac{a_i}{a_f} (1+\frac{\Delta M}{M_{tot,f}}) .

Using the ratio for the semi major axis in the energy equation, we get:

 1-e^2=1-\bigg( \frac{\Delta M}{M_{tot,f}} \bigg)^2 .

It follows that that for symmetrical explosions, the resulting eccentricity follows the simple relation:  e=\frac{\Delta M}{M_{tot,f}} . In particular there is no bound orbit when  \Delta M > M_{tot,f} , i.e when the system losses more than half its original mass (as shown in Maximum mass loss in a supernova that leaves a bound binary).