= Solution
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.
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