= Solution
For constant component masses, the <center of mass> frame has $\mathbf r_1=(M_2/M)\mathbf r$ and $\mathbf r_2=-(M_1/M)\mathbf r$. <Newton's law of universal gravitation> gives the relative equation $\ddot{\mathbf r}=-GM\mathbf r/r^3$. Substitution in the component <angular momenta> yields
$$
\boxed{\mathbf J=\mu\mathbf h=\mu\mathbf r\times\dot{\mathbf r},\qquad\dot{\mathbf h}=\mathbf r\times\ddot{\mathbf r}=0.}
$$
Here $\mu=M_1M_2/M$ is the <reduced mass>, while $\mathbf h$ is <specific angular momentum>.
The printed total-energy expression lacks a factor of $\mu$ in its kinetic term. With the stated physical separation coordinate, the dimensionally consistent total <two-body orbital energy> is
$$
\boxed{E=\frac\mu2|\dot{\mathbf r}|^2-\frac{GM\mu}{r}=\mu\varepsilon,\qquad\varepsilon=\frac12|\dot{\mathbf r}|^2-\frac{GM}{r}.}
$$
The second expression is <specific orbital energy>; these two conventions must not be mixed. Its conservation follows directly:
$$
\dot E=\mu\dot{\mathbf r}\cdot\left(\ddot{\mathbf r}+\frac{GM\mathbf r}{r^3}\right)=0.
$$
For the <eccentricity vector>, differentiate $GM\mathbf e=\dot{\mathbf r}\times\mathbf h-GM\widehat{\mathbf r}$. Since $\mathbf h$ is constant,
$$
GM\dot{\mathbf e}=-\frac{GM}{r^3}\mathbf r\times(\mathbf r\times\dot{\mathbf r})-GM\left[\frac{\dot{\mathbf r}}r-\frac{\mathbf r(\mathbf r\cdot\dot{\mathbf r})}{r^3}\right]=0.
$$
The cancellation is the <vector triple product identity>. Thus \b[all three corrected Kepler integrals are conserved]. Dotting the eccentricity relation with $\mathbf r$ also gives the useful orbit identity
$$
\boxed{\mathbf e\cdot\mathbf r=\frac{h^2}{GM}-r.}
$$
For <ballistic streamline focusing by a moving star>, work in the star's rest frame and take the incoming gas to have velocity $v\mathbf e_x$, with downstream behind the star at $x>0$. A streamline with <impact parameter> $d>0$ approaches from $(-\infty,d,0)$. Its upstream <specific angular momentum> and <eccentricity vector> are
$$
\mathbf h=-dv\mathbf e_z,\qquad\mathbf e=\mathbf e_x+\frac{dv^2}{GM}\mathbf e_y.
$$
At its downstream axis crossing, $\mathbf r=b\mathbf e_x$. The orbit identity gives $b=d^2v^2/(GM)-b$, hence \b[the collision distance] is
$$
\boxed{b=\frac{d^2v^2}{2GM}.}
$$
At that point, $h_z=b u_y=-dv$, while the $y$ component of $GM\mathbf e=\mathbf u\times\mathbf h-GM\widehat{\mathbf r}$ gives $dv\,u_x=dv^2$. Therefore
$$
\mathbf u=v\mathbf e_x-\frac{2GM}{dv}\mathbf e_y.
$$
A symmetric streamline with opposite impact parameter has the opposite transverse velocity. Their <shock wave> removes the opposing transverse motion while preserving the common downstream component. Thus \b[the remaining velocity is $v\mathbf e_x$ in the star frame]. In the frame where the undisturbed medium is stationary, adding the star's velocity gives zero velocity immediately after this idealized collision.
After transverse <kinetic energy> is dissipated, the remaining <specific orbital energy> is $\varepsilon_{\rm after}=v^2/2-GM/b$. It is negative if $b<2GM/v^2$, or equivalently if the upstream <impact parameter> satisfies $d<d_{\rm cap}=2GM/v^2$. This is the <post-shock capture criterion in ballistic accretion>; bound axial streams can return to the star. The captured incident <mass flux> through the corresponding disk gives the ballistic <Bondi--Hoyle--Lyttleton accretion rate>:
$$
\boxed{\dot M=\pi d_{\rm cap}^2\rho v=\frac{4\pi(GM)^2\rho}{v^3}.}
$$
This is the pressureless, dissipative capture model, with the shock's transverse energy unavailable to unbind the wake.
For <wind mass transfer in a binary star>, steady isotropic donor mass loss gives the local wind <mass density> at the companion:
$$
\rho_w(a)=\frac{-\dot M_1}{4\pi a^2v_w}.
$$
A fast wind has $v_w\gg2\pi a/P$ and a capture scale small compared with $a$, so its relative incident speed is $v_w$ to leading order. Apply the same <Bondi--Hoyle--Lyttleton accretion rate> with accretor mass $M_2$ to obtain
$$
\boxed{\dot M_{2,\rm acc}=(-\dot M_1)\frac{G^2M_2^2}{a^2v_w^4}.}
$$
Finally, <Kepler's third law> eliminates the unquoted separation, $a=[G(M_1+M_2)P^2/(4\pi^2)]^{1/3}$, giving \b[the rate in terms of the stated <orbital period>]:
$$
\boxed{\dot M_{2,\rm acc}=(-\dot M_1)\frac{(2\pi G/P)^{4/3}M_2^2}{(M_1+M_2)^{2/3}v_w^4}.}
$$
Equivalently, the <fast-wind accretion fraction in a circular binary> is $[M_2/(M_1+M_2)]^2(v_{\rm orb}/v_w)^4$, with $v_{\rm orb}=2\pi a/P$ the relative circular orbital speed. Its smallness is consistent with using an almost undisturbed isotropic <stellar wind>.
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