= Solution
{c}
By <Newton's second law>, the <central force> obeys
$$
m\ddot{\mathbf r}=-\Phi'(r)\frac{\mathbf r}{r}.
$$
Differentiating the <cross product> gives
$$
\dot{\mathbf n}=\dot{\mathbf r}\wedge\dot{\mathbf r}+\mathbf r\wedge\ddot{\mathbf r}=\mathbf0,
$$
since the <acceleration> is parallel to $\mathbf r$. Hence $\mathbf n=\mathbf n(0)$ is constant. Since $\mathbf r\cdot\mathbf n=0$, the entire motion lies in the fixed <plane> through the origin with nonzero <normal vector> $\mathbf n(0)$.
Choose <polar coordinates> in that <plane>, orienting $\theta$ so that its positive <normal vector> points along $\mathbf n(0)$. The <velocity> is
$$
\dot{\mathbf r}=\dot r\,\mathbf e_r+r\dot\theta\,\mathbf e_\theta,
\qquad
\mathbf r\wedge\dot{\mathbf r}=r^2\dot\theta\,\mathbf e_z.
$$
Therefore
$$
\boxed{r^2\dot\theta=h=|\mathbf n(0)|>0.}
$$
Multiplication by $m$ identifies this with <conservation of angular momentum>; geometrically, the swept <area> per unit time is $h/2$.
Taking the <dot product> of <Newton's second law> with the <velocity> gives
$$
\frac{d}{dt}\left(\frac m2|\dot{\mathbf r}|^2+\Phi(r)\right)
=m\dot{\mathbf r}\cdot\ddot{\mathbf r}+\Phi'(r)\dot r=0.
$$
Thus
$$
\boxed{\frac m2\left(\dot r^2+r^2\dot\theta^2\right)+\Phi(r)=E.}
$$
This expresses <conservation of mechanical energy>, the sum of <kinetic energy> and <potential energy>.
Finally, for $u=1/r$ and a prime denoting a <derivative> with respect to $\theta$, the <chain rule> gives
$$
\dot\theta=hu^2,\qquad
\dot r=-u^{-2}u'\dot\theta=-hu',\qquad
r\dot\theta=hu.
$$
Substitution into the conserved <energy> yields
$$
\boxed{\frac{mh^2}{2}\left[\left(\frac{du}{d\theta}\right)^2+u^2\right]+\Phi(1/u)=E.}
$$
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