= Solution
For an attractive <central force>, the transverse <acceleration> vanishes. In <polar coordinates>,
$$
2\dot r\dot\theta+r\ddot\theta=0
\quad\Longrightarrow\quad
\frac d{dt}(r^2\dot\theta)=0.
$$
Thus the <specific angular momentum> $h=r^2\dot\theta$ is constant. Assume $h\ne0$, so angle can parameterize the <orbit>, and put $u=1/r$. Primes denote angle <derivatives>. The <chain rule> gives
$$
\dot\theta=hu^2,\qquad
\dot r=-hu',\qquad
\ddot r=-h^2u^2u'',\qquad
r\dot\theta^2=h^2u^3.
$$
The outward radial component of the attractive <force> is $-f(u)$. Its <Newton's second law> equation is therefore $m(\ddot r-r\dot\theta^2)=-f(u)$, giving the <Binet equation>
$$
\boxed{u''+u=\frac{f(u)}{mh^2u^2}.}
$$
The sign reflects $f$ being the positive inward force magnitude. Zero <angular momentum> produces radial motion and is outside this angular parametrization.
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