# Newton's Law of Gravitation in Velocity Space

Updated: Oct 6

We have alluded to the fact that the laws of physics repeat themselves in the same form as we go to higher spaces. Here we will show that gravitational acceleration decays as the inverse of the square of the velocity in an analogous way that it decays as the inverse square of the distance from the source in configuration space. And just as the radial vector sweeps out equal areas in equal times (Kepler's II), so, too, does the velocity vector sweep out equal areas in equal time intervals. It is these conservation laws that are behind the almost identical nature of the physical laws as we go from radial vector to velocity vector, and then to acceleration vector.

Newton's expression for the radii of curvature is given in terms of his slope, z=(1/r)r', where the prime denotes differentiation with respect to the angular variable, phi, as

rho rho* =(1+z^2)^3/2, (1)

where the star denotes the polar reciprocal curve. It was Newton's intuition that led him to suppose that rho*=const. is the condition for his inverse square law.

The radius of curvature of the polar reciprocal is

rho* = (1/r){ 1+ z^2 -(r/2) d (1+z^2)/dr}/ (2)

And since

vr/h= (1+z^2)^1/2

(2) becomes

rho* = - r^2 (v^2/r)/h^2= a r^2/h^2, (3)

since a=v^2/r. Hence, the condition that (3) be constant, implies that the acceleration decreases as the inverse square of the distance from the source.

Since

v^2= (dr/dt)^2+ r^2(d phi/dt)^2,

Eq (1) can be written as

rho rho* = (v/r dphi/dt)^3,

where the tangential component has a role comparable to the normal pedal, p, in the classical expression

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

Continuing we write the square of the acceleration as

a^2 = (dv/dt)^2 + v^2(d phi/dt)^2

and write the radii of curvature in the same way as (1), but now z=(1/v)v;. However, what interests us occurs at the next order; the square of the jerk is

j^2 = (da/dt)^2 + a^2(d phi/dt)^2.

Defining a new Newtonian slope as *z* =(1/a)a',

rho rho* = (1+*z*^2)^3/2

or equivalently,

rho rho* = (j/a d phi/dt)^3.

That radius of curvature goes from rho = v/ d phi/dt to rho= j/d phi/dt, and the Newtonian slope

av/*h* = (1+4 z^2)^1/2 = (1+*z*^2)^1/2,

where we used the fact that *z=- 2z, *or da/dv=-2a/v, and *h* =v^2 d phi/dt, the conserved angular momentum in velocity space.

The Newtonian slopes for velocity and configuration space are the same so that either (1/v)v' or (1/r)r' is tan phi. This follows from the fact that either r or v--but not both simultaneously-- can be varied in the expression for the angular momentum,

v r cos phi = h.

taking the variation with respect to phi, we get

(1/v)v' + (1/r)r' = tan phi.

This is only possible if either the velocity or the radial distance is held constant.

The radius of curvature of the polar reciprocal is

rho* = (1/a){ 1+*z*^2 -(a/2)d/da (av/*h*)2

or equivalently

rho* =+ (1/2) a v^2/*h*^2. (4)

The condition that the curvature of the polar reciprocal remain constant is that the acceleration decrease as the inverse square of the velocity. The analogy between (3) and (4) is manifestly evident. Note the change in sign of the radius of curvature.