= Solution
For $k=\kappa>0$, the same asymptotic data give $E=v^2/2$ and $|h|=bv$. At the <central-force radial turning point>,
$$
\frac12v^2=\frac{b^2v^2}{2\widetilde p^2}+\frac\kappa{\widetilde p},\qquad v^2\widetilde p^2-2\kappa\widetilde p-b^2v^2=0.
$$
The positive root and the tangential <speed>, again with $q=\kappa/v^2$, are
$$
\boxed{\widetilde p=\sqrt{b^2+q^2}+q,\qquad\widetilde u=\frac{bv}{\widetilde p}=\frac v b\left(\sqrt{b^2+q^2}-q\right).}
$$
For $b>0$, $\widetilde p>b$ and $\widetilde u<v$. The <repulsive inverse-square force> bends the trajectory away from the origin, as in the right-hand sketch. Its <polar equation of a repulsive inverse-square orbit> is $r=h^2/[\kappa(e\cos\phi-1)]$, with $e=\sqrt{1+b^2v^4/\kappa^2}>1$ and $\phi$ measured from closest approach.
For $b=0$, the regular radial turning point is $\widetilde p=2\kappa/v^2$ and its <speed> is zero; the particle reverses direction there. The formula $bv/\widetilde p$ still gives zero, while its rationalized form containing $1/b$ should be interpreted only by a limit.
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