For equal $|k|=\kappa$ in the two cases, their closest distances are conjugate roots expressed with the same $q=\kappa/v^2$. Consequently
$$
\boxed{p\widetilde p=(\sqrt{b^2+q^2}-q)(\sqrt{b^2+q^2}+q)=b^2.}
$$
At either <central-force radial turning point>, the <velocity> is tangential and <conservation of angular momentum> gives $pu=\widetilde p\widetilde u=bv$. Thus
$$
\boxed{u\widetilde u=\frac{b^2v^2}{p\widetilde p}=v^2.}
$$
These <attractive and repulsive inverse-square closest approaches> pair a shorter, faster attractive passage with a longer, slower repulsive one. The speed-product conclusion assumes $b>0$; at $b=0$, the attractive orbit hits the singular centre with unbounded <speed>, so the formal product with the repulsive zero <speed> is not defined.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 10A / b / iii / Solution

{title2=$p_-=\sqrt{b^2+q^2}-q,\quad p_+=\sqrt{b^2+q^2}+q$}

For unit <mass>, incoming <speed> $v>0$, <impact parameter> $b>0$, and <potential energy> $\mp\kappa/r$, put $q=\kappa/v^2>0$. At closest approach, <conservation of energy> and <conservation of angular momentum> give the displayed radii and tangential <speeds> $u_\mp=bv/p_\mp$. Consequently $p_-p_+=b^2$ and $u_-u_+=v^2$. These relations compare equal force magnitudes. The attractive radial case $b=0$ is a singular collision, so the speed-product statement does not extend to that endpoint.

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 Incoming links: Attractive and repulsive inverse-square closest approaches