= Solution
A <central force> has no angular component, so the supplied polar acceleration formula and <Newton's second law> give $d(r^2\dot\theta)/dt=0$. Thus $l=r^2\dot\theta$ is the conserved <specific angular momentum>. For a nonradial orbit, $l\ne0$ allows $\theta$ to be used as a parameter. With $u=1/r$ and a prime denoting $d/d\theta$,
$$
\dot r=-u^{-2}u'\dot\theta=-lu',\qquad\ddot r=-lu''\dot\theta=-l^2u^2u''.
$$
Also $r\dot\theta^2=l^2u^3$. The radial equation $\ddot r-r\dot\theta^2=-f(r)$ therefore becomes the <Binet equation>
$$
\boxed{l^2u^2(u''+u)=f(1/u).}
$$
For a purely radial trajectory $l=0$, the angular parameter is unavailable, and one instead uses the direct radial equation. The orbit in part (b) is nonradial.
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