= Solution
At infinity the <potential energy> tends to zero, so the conserved <mechanical energy> is $E=\tfrac12v_\infty^2$. The <impact parameter> $b$ is the perpendicular distance from the origin to the incoming asymptotic line. Conservation of <angular momentum> gives $|h|=bv_\infty$.
For $b>0$, a <turning point> of the radial motion satisfies
$$
\frac12v_\infty^2=\frac{b^2v_\infty^2}{2r^2}-\frac Qr,
\qquad
v_\infty^2r^2+2Qr-b^2v_\infty^2=0.
$$
Only the positive root is physical, giving
$$
\boxed{r_{\min}=\frac{\sqrt{Q^2+b^2v_\infty^4}-Q}{v_\infty^2}.}
$$
For $Q=0$ this reduces to the straight-line closest distance $b$.
For nonzero $Q$, put $\eta=bv_\infty^2/|Q|\ll1$. If $Q<0$, the numerator is $|Q|(\sqrt{1+\eta^2}+1)$, so
$$
r_{\min}=\frac{2|Q|}{v_\infty^2}\left(1+\frac{\eta^2}{4}+O(\eta^4)\right).
$$
If $Q>0$, rationalize the numerator instead:
$$
r_{\min}=\frac{b^2v_\infty^2}{\sqrt{Q^2+b^2v_\infty^4}+Q}
=\frac{b^2v_\infty^2}{2Q}\left(1-\frac{\eta^2}{4}+O(\eta^4)\right).
$$
Thus the requested leading distances are
$$
\boxed{r_{\min}\sim\frac{2|Q|}{v_\infty^2}\quad(Q<0),\qquad
r_{\min}\sim\frac{b^2v_\infty^2}{2Q}\quad(Q>0).}
$$
Repulsion gives a finite barrier even for zero <impact parameter>; attraction allows much closer passage. At $b=0$, the attractive formula has the limiting closest distance zero, corresponding to collision with the singular centre rather than a positive <turning point>.
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