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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 $\dot{r}=-h{u}^{\prime }$, $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$, and $r{\dot{\theta }}^2=h^2{u}^3$. The radial equation with inward force magnitude $f(u)$ is $m(\ddot{r}-r{\dot{\theta }}^2)=-f(u)$, giving the Binet equationChoose 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 $(\sec \theta {)}^{\prime \prime }+\sec \theta =2{\sec }^3\theta$, substitution gives $u^{\prime \prime }+u=8R^2{u}^3$ and henceThe force is 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 isThis 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.