= Radial turning points for an inverse-square height well
{title2=$\dot r^2=2\mathcal E+2g/r-(L^2+2g)/r^2$}
For the height profile $h(r)=r^{-2}-r^{-1}$, set $A=L^2+2g$ and use the planar small-slope model. Its <effective potential> per unit mass is $V_{\rm eff}=A/(2r^2)-g/r$. Release with negligible radial speed at $r_0$ gives $\mathcal E=A/(2r_0^2)-g/r_0$ and the displayed radial-speed law. A bounded trajectory requires $\mathcal E<0$, equivalently $r_0>A/(2g)$. The two positive turning radii are $r_0$ and $Ar_0/(2gr_0-A)$; they coincide at the circular solution $r_0=A/g$. Zero energy is marginal escape, not bounded motion.
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