## FANDOM

138 Pages

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).