= Solution
A <central force> has no component in the angular direction. The angular component of the <acceleration in polar coordinates> therefore gives $2\dot r\dot\theta+r\ddot\theta=0$, or
$$
\frac d{dt}(r^2\dot\theta)=0.
$$
For unit <mass>, $h=r^2\dot\theta$ is the signed <angular momentum> perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the <torque> $\mathbf r\times\mathbf F$ vanishes, which also fixes that plane.
The given <potential energy> is $k/r$, whose radial <force> is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The <kinetic energy> in <polar coordinates> is $(\dot r^2+r^2\dot\theta^2)/2$, so <conservation of energy> gives
$$
\boxed{E=\frac12\dot r^2+V_{\rm eff}(r),\qquad V_{\rm eff}(r)=\frac{h^2}{2r^2}+\frac kr.}
$$
For $h\ne0$ and $k>0$, the <effective potential> is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa<0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa)$, and has its unique minimum at
$$
\boxed{r_c=\frac{h^2}{\kappa},\qquad V_{\rm eff}(r_c)=-\frac{\kappa^2}{2h^2}.}
$$
It then approaches zero from below. The minimum corresponds to a stable <circular orbit>. If $h=0$, the centrifugal barrier disappears; the attractive <effective potential> is simply $-\kappa/r$ and has no minimum.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-4-effective-potentials.png]
{title=<Effective potentials> and turning points of central-force orbits}
{description=Effective potentials with unit angular momentum and k equal to plus or minus one. The attractive minimum and zero are marked.}
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