= Solution
In the <center of mass> frame, the velocities are $\mathbf v_1=(M_2/M)\mathbf v$ and $\mathbf v_2=-(M_1/M)\mathbf v$. Hence the <kinetic energy> is $T=\frac12\mu v^2$, while the <Newtonian gravitational potential energy> is $U=-GM_1M_2/r=-GM\mu/r$. The <two-body orbital energy> is therefore
$$
\boxed{E=\mu\varepsilon=\mu\left(\frac12v^2-\frac{GM}{r}\right),}\qquad\mu=\frac{M_1M_2}{M}.
$$
The relative equation of <Newtonian gravity> is $\ddot{\mathbf r}=-GM\mathbf r/r^3$. It is a <central force>, so <conservation of angular momentum> confines the motion to a plane and preserves the <specific angular momentum> $h=|\mathbf r\times\mathbf v|=r^2\dot\theta$.
For completeness, derive the <polar equation of a Kepler orbit>. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot r=-hw'$ and $\ddot r-r\dot\theta^2=-h^2w^2(w''+w)$. The radial equation gives the <Binet equation>
$$
w''+w=\frac{GM}{h^2}.
$$
Choose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos\theta)$, so
$$
\boxed{r=\frac{l}{1+e\cos\theta},\qquad l=\frac{h^2}{GM}.}
$$
For the <ellipse>, the <orbital eccentricity> satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The <semi-major axis> is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The <true anomaly> $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a <circular Kepler orbit> the angular origin is arbitrary.
The radial and transverse relative speeds are $\dot r=(GM/h)e\sin\theta$ and $r\dot\theta=h/r$. Substituting these into the <specific orbital energy> gives
$$
\varepsilon=\frac{GM}{2l}\left[e^2\sin^2\theta+(1+e\cos\theta)^2-2(1+e\cos\theta)\right]
=\frac{GM}{2l}(e^2-1).
$$
Therefore \b[the conserved orbital energy is]
$$
\boxed{E=-\frac{GM\mu}{2a}.}
$$
The negative <specific orbital energy> and fixed <angular momentum> characterize the bound <Kepler orbit>; neither should be assumed unchanged through a mass-ejecting explosion.
Initially the <circular Kepler orbit> has $r=a$ and $v^2=GM/a$. Treat the <supernova kick in a binary star> as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new <neutron star> receives $\mathbf u$ with $|\mathbf u|=\alpha v$. Thus $\mathbf v'=\mathbf v+\mathbf u$, and with $\psi$ the angle between these vectors before the kick,
$$
|\mathbf v'|^2=v^2S,\qquad S=1+2\alpha\cos\psi+\alpha^2.
$$
Write $M'=M_1'+M_2$, $\beta=M'/M$ and $\mu'=M_1'M_2/M'$. The post-explosion <specific orbital energy> is
$$
\varepsilon'=\frac12v^2S-\frac{GM'}a=\frac{GM}{2a}(S-2\beta).
$$
Using $\varepsilon'=-GM'/(2a')$ gives
$$
\boxed{\frac{M'}{a'}=\frac{2M'}a-\frac Ma(1+2\alpha\cos\psi+\alpha^2).}
$$
Here $a'>0$ for a bound <Kepler orbit>. For a <hyperbolic Kepler orbit>, this formula uses the signed energy parameter $a'<0$, rather than the positive geometric magnitude of the hyperbola's <semi-major axis>. At the <parabolic Kepler orbit> boundary, $1/a'=0$.
The <kick-direction binary survival criterion> is \b[bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$]. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos\psi=0$: then $S=1+\alpha^2$, so $M'<\frac12(1+\alpha^2)M$ produces positive <specific orbital energy>. This is sufficient, not the sharp existence threshold. Since
$$
(1-\alpha)^2\le S\le(1+\alpha)^2,
$$
the complete <kick-direction binary survival criterion> is
$$
\boxed{\begin{aligned}
\text{some direction gives }\varepsilon'>0&\iff\beta<\tfrac12(1+\alpha)^2,\\
\text{some direction remains bound}&\iff\beta>\tfrac12(1-\alpha)^2.
\end{aligned}}
$$
The second inequality follows by taking a kick directly opposite to $\mathbf v$. All directions remain bound if $2\beta>(1+\alpha)^2$, while every direction has positive escape energy if $2\beta<(1-\alpha)^2$. For $\alpha>0$, the marginal direction obeys $\cos\psi=(2\beta-1-\alpha^2)/(2\alpha)$ when the right-hand side lies in $[-1,1]$. Equalities give marginal <parabolic Kepler orbits> for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the <circular Kepler orbit>.
For an escaping pair, the <Newtonian gravitational potential energy> approaches zero at infinite separation. Conservation of the post-explosion <two-body orbital energy> then gives $\frac12\mu'V^2=\mu'\varepsilon'$, or the <asymptotic relative speed of a disrupted binary>
$$
\boxed{V=v\sqrt{1+2\alpha\cos\psi+\alpha^2-\frac{2M'}M}.}
$$
This is the relative recession speed, not the velocity of the new <center of mass> or either star's individual velocity in the original <inertial frame>. A marginal <parabolic Kepler orbit> separates with $V=0$ at infinity.
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