Speculative physics: gravitational force as 1/r³

Problem

Speculative physics: $1/r^3$ instead of $1/r^2$. Imagine that the gravitational force scaled as $1/r^3$ instead of $1/r^2$. Discuss the qualitative consequences for planetary orbits.

Solution

With the real law $F \propto 1/r^2$, bound orbits are closed ellipses (Kepler’s first law): after one revolution the planet passes exactly through the same point again. This property is exceptional.

Bertrand’s theorem. The only central radial forces for which all bound orbits are closed are two: the one $\propto 1/r^2$ (gravitation, Coulomb) and the one $\propto r$ (harmonic oscillator). Any other exponent produces orbits that do not close: the planet traces a rosette that precesses, or spirals.

The $1/r^3$ case. Let’s analyse stability with the effective potential. For a force $F \propto 1/r^3$ the potential goes as $U \propto -1/r^2$, and the angular-momentum centrifugal term also goes as $+L^2/(2mr^2) \propto 1/r^2$. The two terms have the same dependence $1/r^2$: no minimum forms in the effective potential to trap the planet in a stable orbit. $U_\text{eff}(r) = -\frac{k}{r^2} + \frac{L^2}{2m\,r^2} = \frac{1}{r^2}\!\left(\frac{L^2}{2m} - k\right)$ The sign of the bracket decides everything: either the planet spirals inward towards the star (if gravity wins), or it spirals outward to infinity (if the centrifugal term wins). No intermediate stable closed orbits exist.

Consequence. A Solar System like the one we know — planets on stable, recurring orbits for billions of years — could not exist. The stability of the planetary system depends crucially on the $-2$ exponent of the law of gravitation.

$\ev{\text{with } F \propto 1/r^3:\ \text{no stable closed orbit} \Rightarrow \text{spiral fall or escape}}$

Links

Topics: Gravitation Concepts: Law of universal gravitation · Orbits Skills: Analysis of limiting cases and speculative physics