 Solution

In centre-of-mass and relative coordinates, the kinetic energy separates into centre-of-mass motion plus $\mu v^2/2$, while the mutual potential is $-G{M}_1{M}_2/r=-GM\mu /r$. In the centre-of-mass frame the orbital energy is therefore

The force is central, so the specific relative angular momentum $h=r^2\dot{\theta }$ is conserved. With $u=1/r$, the radial equation becomes the Binet equationwhose solution isHere $\theta$ is the true anomaly measured from periapsis and $e$ is the orbital eccentricity. Substitution into the energy, or evaluation at an apsis, givesand hence

Immediately before the supernova the circular relative speed obeys $v^2=GM/a$. The impulsive supernova kick in a binary star changes it toApplying the energy formula just after the explosion at the unchanged separation $r=a$ givesUsing the original circular-speed relation,The post-explosion system is bound exactly when $1+2\alpha \cos \psi +{\alpha }^2<2M^{\prime }/M$. The left side ranges from $(1-\alpha {)}^2$ to $(1+\alpha {)}^2$, so every kick direction remains bound ifwhereas every direction unbinds the stars if

For an unbound orbit, conservation of relative specific energy at infinity gives $V^2/2=v^{\prime 2}/2-G{M}^{\prime }/a$. Therefore