= Solution
{c}
Write $q=|\dot{\mathbf r}|=\sqrt{\dot r^2+r^2\dot\theta^2}$. Differentiating the <angular momentum> and applying <Newton's second law> gives
$$
\begin{aligned}
\dot{\mathbf L}
&=m\dot{\mathbf r}\times\dot{\mathbf r}+m\mathbf r\times\ddot{\mathbf r}
=\mathbf r\times\mathbf F\\
&=\mathbf r\times(-\alpha q\dot{\mathbf r})
=-\frac\alpha m q\mathbf L.
\end{aligned}
$$
The <central force> contributes no <torque>, because it is parallel to $\mathbf r$. The <angular momentum> direction is fixed whenever it is nonzero; the motion is consequently planar. In the <plane>'s <polar coordinates>, $\mathbf L=mh\mathbf e_z$, so
$$
\boxed{\dot h=-\frac\alpha m\sqrt{\dot r^2+r^2\dot\theta^2}\,r^2\dot\theta.}
$$
For the <mechanical energy>, use the <chain rule>, <Newton's second law>, and $V'(r)=-F_c(r)$:
$$
\begin{aligned}
\dot E
&=m\dot{\mathbf r}\cdot\ddot{\mathbf r}+V'(r)\dot r\\
&=F_c(r)\dot r-\alpha q|\dot{\mathbf r}|^2+V'(r)\dot r
=-\alpha q^3.
\end{aligned}
$$
Thus
$$
\boxed{\dot E=-\alpha(\dot r^2+r^2\dot\theta^2)^{3/2}.}
$$
For a trajectory on a time interval, both $h$ and $E$ are each a <constant of motion> precisely when $\alpha=0$ or the <velocity> vanishes throughout that interval. Under the stated strictly attractive <central force> at a regular <point> $r>0$, a permanently stationary trajectory is impossible by <Newton's second law>. \b[For the stated attractive motion, both are conserved if and only if $\alpha=0$.] An isolated instant of zero <speed> only makes the <derivatives> vanish at that instant. A radial trajectory has $h=0$ even with <quadratic drag>, but its <mechanical energy> still decreases whenever it moves.
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