= Solution
Work in the <planet>'s rest frame, neglecting its recoil and <tidal forces> during the close encounter. A <hyperbolic Kepler orbit> has positive <specific orbital energy> $\varepsilon=v_\infty^2/2$ and <specific angular momentum> $h=bv_\infty$, where $b$ is the <impact parameter>. Its <orbital eccentricity> is
$$
e_h=\sqrt{1+\frac{2\varepsilon h^2}{\mu^2}}
=\sqrt{1+\frac{b^2v_\infty^4}{\mu^2}}>1.
$$
At the <pericentre distance>, $r_{\min}=h^2/[\mu(1+e_h)]$. Substituting $h^2=\mu^2(e_h^2-1)/v_\infty^2$ yields
$$
e_h=1+\frac{r_{\min}v_\infty^2}{\mu}.
$$
The two asymptotes have <true anomalies> $\pm f_\infty$, where $\cos f_\infty=-1/e_h$. Their directions of motion give the <gravitational scattering angle> $\theta=2f_\infty-\pi$, so $\sin(\theta/2)=-\cos f_\infty=1/e_h$. Therefore
$$
\boxed{\sin\frac{\theta}{2}=\left(1+\frac{r_{\min}v_\infty^2}{\mu}\right)^{-1}}.
$$
As a useful independent form, $b^2=r_{\min}^2+2\mu r_{\min}/v_\infty^2$, displaying <gravitational focusing>. Here $\mu=GM_p$ to the stated small-comet accuracy; the exact relative-coordinate formula replaces it by $G(M_p+M)$.
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