top of page

A New Take on the Inverse-Square Law

With the Sun at one of the foci, Newton's inverse-square law states that the product of the acceleration at a point on the tangent to the curve, and the square of the distance from that point to the source is constant.

If a curve C is the boundary of a convex region with one of the foci of the conic in its interior, then the its polar C* is the boundary of the polar body. The two curves are intrically related and what we shall show is that the acceleration a* at P* on C* and the square of the radius vector from point P on C to the source S is what is constant.

The distance from S to P* is r* and it cut the tangent line emanating from the point P normally. Continuing SP to the tangent line of the polar curve, it too will cut it at right angles. If C is a conic, so, too, must its polar, C*, be. By duality, (C*)*=C The pedal p is the disnce from S to the tangent line of C, so the pedal p* is the distance from S to P*. By reciprocity we have r*=1/p, and r=1/p*.

The product of the curvatures satisfies

rho rho*= (r/p)^3. (1)

This easily follows from Newton's definition of the radius of curvature

rho = (1+(r'/r)^2)^3/ rho*.

Multipling the numerator through,

rho rho*= {(dr/dt)^2+r^2(d phi/dt)^2]^3/(r d phi/dt)^3 = (v/r dphi/dt)^3 (2)


rho* =(1/r){1+(r'/r)^2-(r'/r)'}

and the prime stands for differentiation with respect to the angle phi, the support of the curve C*. Likewise, phi* will be the support angle of C. On the strength of the conservation of angular momentum, h=h*, expression (1) is the same as (2) since p v= r^2 d phi/dt= p*v*=r*^2 d phi*/dt=h*. These allow us to write

r^2r*2/p p*= v^2 v*2/phi phi*.

Since v=h r* and v*=h r, the above expression can be written as

rho rho* =(1/h^2) v^2 v*^2/(r d phi/dt)(r* d phi*/dt) (3)

We can split this into two expressions

rho = v^2/r* d phi*/dt h and rho* =v*^2/r d phi/dt h.

For if we introduce rho=v/phi* into the first term in the first expression we get v=h r*.

As for the second expression we can write it as

rho* = a* r^2/h^2 (4)

where we multiplied numerator and denominator by r^2 in the second expression.

Expression for is the result we anticipated: If the curvature of the dual is constant, the product of the acceleration at P* on the polar curve C* and the square of the distance from P on the tangent line to the curve C to the source S must be constant. This is based on the invarancy of the angular momenta h and h*.

So it is not the acceleration a=v^2/r*, at P and the square of the distance r^2 that is constant, but rather the acceleration a*=v*2/r at P* and the sqaure of the distance that is constant. By duality, it also follows that the acceleration at P and the square of the distance SP*, r*2 that is constant. Thus, it is the acceleration at one of the tangent points and the square of the product of the distance to the other that remain constant, and are a statement of Newton's inverse-square law.

This can easily be extended to higher time derivatives for we have

rho rho*= (v/r d phi/dt)^3= (a/v d phi/dt)^3=(j/a dphi/dt)^3 (5)

which asserts that the acceleration a=v^2/r. Introducing the dual jerks, j= a^2/v* and j*=a*^2/v, the invariancy of the angular momenta in velocity space, v*2 d phi*/dt=v*^2 d phi*/dt, whereby the velocity vector vector sweeps out equals in velocity space in equal times

that leads to the invariancy of the jerk and the corresponding area in velocity space, i.e.

j* v^2=j v*^2,

and the curvature is rho=a/d phi*/dt and rho*= a*/d phi/dt.

This resolves the confusion in an earlier blog which stated incorrectly that the accelerations and jerks can be determined solely from the velocities and accelerations of either the curve C or its polar dual, C*. For, in fact, these kinematic variables require both the curve and is dual.

10 views0 comments

Recent Posts

See All


bottom of page