= Solution
Interpret projection from very far away as the asymptotic incoming state at $r\to\infty$. For $v>0$ and <impact parameter> $b>0$, <conservation of energy> and <conservation of angular momentum> give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial <velocity> vanishes. With $k=-\kappa<0$, the <central-force radial turning point> equation is
$$
\frac12v^2=\frac{b^2v^2}{2p^2}-\frac\kappa p,\qquad v^2p^2+2\kappa p-b^2v^2=0.
$$
Only the positive root is physical. Put $q=\kappa/v^2$. The <pericentre distance> and the purely tangential <speed> there are
$$
\boxed{p=\sqrt{b^2+q^2}-q,\qquad u=\frac{bv}{p}=\frac v b\left(\sqrt{b^2+q^2}+q\right).}
$$
Thus $p<b$ and $u>v$: attraction bends the path inward and increases the <speed>. With positive <energy>, the <Kepler orbit> is a <hyperbola>. Measured from the pericentre direction, its <polar coordinates> satisfy $r=h^2/[\kappa(1+e\cos\phi)]$ with $e=\sqrt{1+b^2v^4/\kappa^2}>1$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-4-scattering-trajectories.png]
{title=Central-force scattering trajectories}
{description=Attractive and repulsive inverse-square scattering with the same incoming speed, impact parameter, and force magnitude. The incoming asymptote and closest approach are marked.}
The <hyperbola> shown has a nonzero <impact parameter>. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach <speed>.
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