= Solution
For an attractive <inverse-square potential>, let the <reduced mass> be $\mu$, potential $-k/r$, and nonzero orbital <angular momentum> $\ell$. Conservation of <angular momentum> makes the orbit planar. The orbit equation is
$$
\boxed{r=\frac{\ell^2/(\mu k)}{1+e\cos\varphi},\qquad
e^2=1+\frac{2E\ell^2}{\mu k^2}.}
$$
Positive relative <energy> gives $e>1$ and an unbound <hyperbolic Kepler orbit>, arriving from and returning to infinity. Its asymptotes satisfy $\cos\varphi_\infty=-1/e$. At infinity $E=\mu v_\infty^2/2$ and $\ell=\mu b v_\infty$ for <impact parameter> $b$, so the deflection angle is
$$
\boxed{\chi=2\arcsin(1/e)
=2\arctan\frac{k}{\mu b v_\infty^2}.}
$$
This describes gravitational encounters in the Newtonian two-body approximation. The body speeds up near <periapsis> while total <energy> remains positive. It is distinct from an elliptic bound orbit and from the zero-energy parabolic limit; repulsive Coulomb scattering also gives hyperbolae but uses a different orientation of the force and orbit.
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