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For equal $|k|=\kappa$ in the two cases, their closest distances are conjugate roots expressed with the same $q=\kappa /v^2$. ConsequentlyAt either central-force radial turning point, the velocity is tangential and conservation of angular momentum gives $pu=\tilde{p}\tilde{u}=bv$. ThusThese 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.

For $k=\kappa >0$, the same asymptotic data give $E=v^2/2$ and $|h|=bv$. At the central-force radial turning point,The positive root and the tangential speed, again with $q=\kappa /v^2$, areFor $b>0$, $\tilde{p}>b$ and $\tilde{u}<v$. 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 $r=h^2/[\kappa (e\cos \varphi -1)]$, with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$ and $\varphi$ measured from closest approach.

For $b=0$, the regular radial turning point is $\tilde{p}=2\kappa /v^2$ and its speed is zero; the particle reverses direction there. The formula $bv/\tilde{p}$ still gives zero, while its rationalized form containing $1/b$ should be interpreted only by a limit.

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 isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ 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 $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

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.

Now $E=u^2/2-k/R<0$, so the orbit is bound. Since $L=ub$, the initial energy isAt the farthest central-force radial turning point $q$, the radial speed vanishes and the transverse speed is $L/q$, givingEquating these expressions and multiplying by $2q^2$ yieldsThus $A=L^2/b^2-2k/R$, $B=2k$, and $C=-L^2$.