= Circular orbit through an inverse-fifth-power singularity
{title2=$c=8mh^2R^2,\quad T=2\pi R^2/|h|$}
For an attractive <central force> of magnitude $c/r^5$, a circle of radius $R$ through the force center has polar equation $r=2R\cos\theta$. With <specific angular momentum> $h$, substitution into the <Binet equation> gives $c=8mh^2R^2$. Consequently $R$ is inversely proportional to the magnitude of the <angular momentum>, and the angular-time integral gives $T=2\pi R^2/|h|\propto R^3$. This is the finite travel time along the punctured circle from one approach to the singular center to the next; the force law alone does not specify continuation through the collision.
Back to article page
