= Solution
{c}
With $\alpha=0$, <4/a> gives constant $h$, and the <polar coordinates> <basis> satisfies $\dot{\mathbf e}_r=\dot\theta\mathbf e_\theta$, $\mathbf e_r\times\mathbf e_z=-\mathbf e_\theta$. Also $\ddot{\mathbf r}=(F_c/m)\mathbf e_r$. Differentiating the proposed <Laplace-Runge-Lenz vector> therefore gives
$$
\begin{aligned}
\dot{\mathbf A}
&=m^2h\ddot{\mathbf r}\times\mathbf e_z-mk\dot{\mathbf e}_r\\
&=-mhF_c(r)\mathbf e_\theta-mk\dot\theta\mathbf e_\theta\\
&=-mh\left(F_c(r)+\frac{k}{r^2}\right)\mathbf e_\theta.
\end{aligned}
$$
For this to vanish for arbitrary $h$, including nonzero $h$, the <central force> must satisfy
$$
\boxed{F_c(r)=-\frac{k}{r^2},\qquad V(r)=-\frac{k}{r}+\text{constant}.}
$$
Attraction requires $k>0$. The condition “for any value of $h$” matters: with $h=0$ alone, the <Laplace-Runge-Lenz vector> is constant on each radial ray for any <central force>, so that special case would not determine the force law.
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