top of page

Derivation of Inverse-Square Law from Epicycles of Constant Angular Velocity, not Angular Momentum

We have shown [Seeing Gravity, p. 236] that Ampere's law leads to geodesics for constant angular velocity. There should be an analogous relation for Newton's law of gravitation. Yet, Newton found it necessary to use Kepler's areal law to derive his inverse square law. Can closed elliptical orbits still be obtained when the conservation of angular momentum is replaced by the conservation of angular velocity?

Newton's starting point was the law of centrifugal force

F cosq=v^2/p (1)

where q is the angle between the point, P, on the point where the instantaneous circle of curvature makes contact with the point on the trajectory. The radius of curvature, p, was found earlier, by Newton, to be

p=r/cos^3q [1+z^2-z'],

where z is the slope of the curve, (1/r)dr/dq, and cos q=1/[1+z^2]^1/2. The prime denotes differentiation with respect to the angle q. The radius of curvature has all the earmarking of making the angular velocity the limiting velocity.

It was Newton's intention to replace "the motion along an incremental arc of the general curve at point P with uniform circular motion along an incremental arc of the circle at the same point." [Brackenridge, "Curvature in Newton's dynamics"] This is supposedly encapsuled in his equation (1). Under the condition of constant angular velocity w=rdq/dt, his equation can be written simply as


If we convert to the inverse variable, u=1/r, this simply reduces to


A constant force, F, will therefore lead to a conic section,

u=A(1+e cos q),

where the amplitude A=F/w^2, and e is the eccentricity.

This clearly shows that we can get a closed or open conic section, depending on the value of e<1 or e>1, respectively, without invoking Kepler's areal law, or the conservation of angular momentum, h=r^2 dq/dt. Why is this?

The slope z is the ratio of dr/dt to r dq/dt. The latter is the angular velocity, and if this is introduced into Newton's force law, it is seen that it is the maximum velocity attainable. This can be thought of as the velocity of the epicycle as the planet, whose period is that of uniform motion, as it move on the deferent. The deferent we have shown to be a conic section, one of constant force.

This is entirely compatible with Ptolemy's vision of epicycles: The center of the epicycle moves around the deferent at constant angular velocity. The planet also moves around the epicycle at constant angular angular velocity.

If this is so, where does Newton's inverse square come in? We can write the conserved energy as

U=(1/r^2){1+(dr/dt/w)^2}. (2)

The negative derivative with respect to r is the force

F_N=(w^2/r){2+(dr/dt/w)^2-r d^2r/dt^2/w^2}=F/r^2,

where F is our constant force. So it is Newton's inverse square law that can be defined from a conservative energy, (2), which also shows that the angular velocity is the limiting velocity.


So Newton didn't need Kepler's second law, and could have arrived at his force by considering epicycles of constant angular velocity!

Newton's slope z has incorporated in it either the conservation of angular momentum, dr/dt=v sin q, or angular velocity w=v sin q, where v^2=w^2+(dr/dt)^2. That is

cos^2 q=(1+z^2)^(-1)=w^2/v^2,


sin^2 q=z^2/(1+z^2)=(zw)^2/v^2=(dr/dt)^2/v^2.

The former can be written as the inner product


while the later can be written as cross product,

(dr/dt)= (r x v)/r=h/r,

where h is the angular momentum per unit mass.

24 views0 comments

Recent Posts

See All

Putting "words" in Newton's "mouth"

Given the nature of Kepler's elliptical orbits of the planets, what was unknown at Kepler's time was the nature of the force producing such orbits. Newton referred to this as the "inverse" problem, na


bottom of page