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).