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The radial energy equation is ${\dot{r}}^2=2(E-U(r))$, so motion requires $U(r)\le E$. For $Q>0$ and $h>0$, the effective potential minimum and its limiting value zero at infinity implyAt the lower endpoint the particle stays at $r=h^2/Q$, giving a circular orbit. Strictly between that endpoint and zero, two positive turning points confine the radial motion. For $E$ below the minimum there is no allowed motion, while $E\ge 0$ does not give a radially bounded orbit.

Write $\epsilon =|E|$ and $D=Q^2-2\epsilon h^2\ge 0$. At each turning point $E=U(r)$, soThe smaller and larger roots give the periapsis distance and apoapsis distance:They coincide for the circular orbit. Their sum and product are

The speed satisfies $v^2=2(E+Q/r)$, which decreases strictly with $r$ because $Q>0$. Thus its maximum is at the periapsis and its minimum at the apoapsis. At either endpoint the radial velocity vanishes, so $v=h/r$. Therefore the apsidal speed sum for an attractive inverse-square orbit isHence $\boxed{v_{\min }+v_{\max }=2Q/h}$. Individually, $v_{\min }=(Q-\sqrt{D})/h$ and $v_{\max }=(Q+\sqrt{D})/h$, also valid when the orbit is circular.