= Solution
A <central force> has zero <torque>, so $h=r^2\dot\theta$ is constant and the <angular momentum> is $mh$. The <areal velocity> is $h/2$. Put $u=1/r$ and let primes denote angular <derivatives>. Then $\dot r=-hu'$, $\ddot r=-h^2u^2u''$, and $r\dot\theta^2=h^2u^3$. The radial equation with inward force magnitude $f(u)$ is $m(\ddot r-r\dot\theta^2)=-f(u)$, giving the <Binet equation>
$$
\boxed{u''+u=\frac{f(u)}{mh^2u^2}}.
$$
Choose the angular origin along a diameter of a radius-$R$ circle through $O$. Its polar equation is $r=2R\cos\theta$, so $u=(2R)^{-1}\sec\theta$ on $-\pi/2<\theta<\pi/2$. Since $(\sec\theta)''+\sec\theta=2\sec^3\theta$, substitution gives $u''+u=8R^2u^3$ and hence
$$
\boxed{f(u)=cu^5,\qquad c=8mh^2R^2>0}.
$$
The force is \b[attractive]. For fixed force coefficient $c$ and mass, $R|mh|=\sqrt{mc/8}$, so $R$ is inversely proportional to the magnitude of the <angular momentum>. The time along the circle is
$$
\boxed{T=\int_{-\pi/2}^{\pi/2}\frac{r^2}{|h|}\,d\theta=\frac{2\pi R^2}{|h|}=4\sqrt2\,\pi\sqrt{\frac mc}R^3}.
$$
This <circular orbit through an inverse-fifth-power singularity> reaches the force center at the two endpoints of this angular interval. The integral is finite; continuation through the singular collision is not specified by the force law alone.
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