= Solution
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 imply
$$
\boxed{-\frac{Q^2}{2h^2}\le E<0\quad\text{for bounded orbits}.}
$$
At 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 $E\ge0$ does not give a radially bounded orbit.
Write $\varepsilon=|E|$ and $D=Q^2-2\varepsilon h^2\ge0$. At each <turning point> $E=U(r)$, so
$$
\varepsilon r^2-Qr+\frac{h^2}{2}=0.
$$
The smaller and larger roots give the <periapsis distance> and <apoapsis distance>:
$$
\boxed{r_{\min}=\frac{Q-\sqrt D}{2\varepsilon},\qquad
r_{\max}=\frac{Q+\sqrt D}{2\varepsilon}.}
$$
They coincide for the circular orbit. Their sum and product are
$$
\boxed{r_{\min}+r_{\max}=\frac Q{|E|}},\qquad
r_{\min}r_{\max}=\frac{h^2}{2|E|}.
$$
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> is
$$
v_{\min}+v_{\max}
=h\left(\frac1{r_{\max}}+\frac1{r_{\min}}\right)
=h\frac{Q/|E|}{h^2/(2|E|)}=\frac{2Q}{h}.
$$
Hence $\boxed{v_{\min}+v_{\max}=2Q/h}$. Individually, $v_{\min}=(Q-\sqrt D)/h$ and $v_{\max}=(Q+\sqrt D)/h$, also valid when the orbit is circular.
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