= Solution
{c}
Take $\beta>0$. At a radial <turning point>, $\dot r=0$, and the <effective potential> gives
$$
E=\frac{L^2}{2mr^2}-\frac\beta r,\qquad
\boxed{Er_{\min}^2+\beta r_{\min}-\frac{L^2}{2m}=0.}
$$
For $L\ne0$, the smaller positive radial <turning point> is conveniently written without division by $E$:
$$
\boxed{r_{\min}=\frac{L^2}{m\beta+\sqrt{m^2\beta^2+2mEL^2}}.}
$$
This includes $E=0$, where $r_{\min}=L^2/(2m\beta)$. For $E<0$ there is also an outer radial <turning point>; the two coincide for the <circular orbit> at $E=-m\beta^2/(2L^2)$. Smaller <energies> are inaccessible at that $L$. For $E>0$ there is just one positive radial <turning point>. If $L=0$, there is no <centrifugal barrier> and an inward trajectory can reach the singular centre; its limiting closest distance is zero.
\b[The requested omission of $\beta$ is not possible.] For example, at $E=0$ and fixed nonzero $L,m$, doubling $\beta$ halves $r_{\min}$. Thus $E,L,m$ alone cannot determine closest approach.
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