= Solution
{c}
Far from the planet, the magnitude of <angular momentum> per unit <mass> is the magnitude of the <cross product> of <position vector> and <velocity>. The perpendicular distance from the incoming <asymptote> is the <impact parameter> $d$, so
$$
\boxed{h=dv.}
$$
Choose $\theta=0$ at closest approach. The general solution of the <differential equation> in <5/a>, with $u'(0)=0$, is
$$
u=\frac\mu{h^2}(1+e\cos\theta).
$$
The <specific orbital energy> is
$$
\varepsilon=\frac12(\dot r^2+r^2\dot\theta^2)-\frac\mu r
=\frac{h^2}2(u'^2+u^2)-\mu u
=\frac{\mu^2}{2h^2}(e^2-1).
$$
At infinity $\varepsilon=v^2/2$, giving $e=\sqrt{1+h^2v^2/\mu^2}>1$. Thus the <hyperbolic orbit> has
$$
\boxed{u(\theta)=\frac{GM}{d^2v^2}\left[1+\sqrt{1+\frac{d^2v^4}{G^2M^2}}\cos\theta\right].}
$$
A different zero of <polar angle> simply replaces $\theta$ by $\theta-\theta_0$.
The incoming and outgoing <asymptotes> have radial directions $-\theta_\infty$ and $\theta_\infty$, where $\cos\theta_\infty=-1/e$ and $\pi/2<\theta_\infty<\pi$. The incoming <velocity> points opposite its radial direction, so the <deflection angle> between incoming and outgoing <velocities> is $\delta=2\theta_\infty-\pi$. Hence
$$
\sin\frac\delta2=\frac1e,\qquad
\tan\frac\delta2=\frac{GM}{dv^2}.
$$
Putting $\delta=\pi/2$ gives
$$
\boxed{d=\frac{GM}{v^2}.}
$$
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