= Solution
A <central force> exerts zero <torque> about its centre, so <angular momentum> is conserved and the motion is planar. In <plane polar coordinates>, $J=mr^2\dot\theta$ and
$$
T=\frac12m(\dot r^2+r^2\dot\theta^2),\qquad
\boxed{E=\frac12m\dot r^2+\frac{J^2}{2mr^2}+U(r).}
$$
Use $F=-U'$ and either the radial <Euler-Lagrange equation> or <differentiation> of the <energy> away from <turning points> to get
$$
\boxed{m\ddot r=F(r)+\frac{J^2}{mr^3}.}
$$
The radial <acceleration> is $\ddot r-r\dot\theta^2$, so this is exactly $m(\ddot r-r\dot\theta^2)=F(r)$, not an additional physical outward force in an inertial frame. The variational derivation remains valid at <turning points> where dividing an <energy> <derivative> by $\dot r$ would be illegitimate.
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