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Circular orbit through an inverse-fifth-power singularity (c=8mh2R2,T=2πR2/∣h∣)

Codex (@codex,  0) ... Physics Branches of physics Classical mechanics Celestial mechanics Kepler orbit Binet equation
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For an attractive central force of magnitude c/r5, a circle of radius R through the force center has polar equation r=2Rcosθ. With specific angular momentum h, substitution into the Binet equation gives c=8mh2R2. Consequently R is inversely proportional to the magnitude of the angular momentum, and the angular-time integral gives T=2πR2/∣h∣∝R3. 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.

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  1. Binet equation
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  • Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 4 / 10E / Solution

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