Solution
= Solution
The transverse equation of motion is $r^{-1}d(r^2\dot\theta)/dt=0$, so
$$
l=r^2\dot\theta
$$
is constant. With $u=1/r$,
$$
\dot r=\frac{dr}{d\theta}\dot\theta
=-u^{-2}u'\,lu^2=-lu'.
$$
Differentiating once more and substituting into the radial equation $\ddot r-r\dot\theta^2=-f(r)$ gives the <Binet equation>
$$
\boxed{l^2u^2(u''+u)=f(1/u).}
$$