= Solution
{c}
Write $\mu=GM$. In <polar coordinates>, <Newton's second law> per unit <mass> gives the radial and transverse <equations of motion>
$$
\ddot r-r\dot\theta^2=-\frac\mu{r^2},\qquad
r\ddot\theta+2\dot r\dot\theta=0.
$$
The transverse <equation> implies
$$
\frac d{dt}(r^2\dot\theta)=r(r\ddot\theta+2\dot r\dot\theta)=0,
\qquad\boxed{h=r^2\dot\theta\text{ is constant}.}
$$
This is <conservation of angular momentum> per unit <mass>. For the nonradial <trajectory> in the next part, choose <orientation> so $h>0$. With $u=1/r$ and primes denoting <derivatives> with respect to $\theta$, the <chain rule> gives
$$
\dot\theta=hu^2,\qquad \dot r=-hu',\qquad \ddot r=-h^2u^2u''.
$$
Substituting into the radial <equation> yields $-h^2u^2(u''+u)=-\mu u^2$, hence
$$
\boxed{u''+u=\frac{GM}{h^2}.}
$$
The change of independent variable requires $h\ne0$; a purely radial <trajectory> instead has constant $\theta$.
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