The simple inverse (a correction)

Plus bonus near and far field formulas!

I need to make a short but crucial correction. I’m human (so they say), and subject to the same force of habit as everybody else. I’ve been wrong about the curvature of the potential energy field. It curves as 1/x, not 1/x². If the force of gravity follows an inverse square law, then the curvature that causes it must be the simple inverse. Duh. Why did it take me this long to notice? I had to start running the numbers to see that I kept coming up with an inverse cube force to see the problem. Now, all my diagrams are wrong.

Symmetry at work: The classical gravitational force (inverse square) is the derivative of the curvature (inverse) of the potential energy field. The gravitational potential is the integral of the force. That means the classical gravitational potential is the curvature of the potential energy field is the simple inverse.

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Here’s a bonus from the paper I’m working on. The force felt by a particle of fixed radius r in the potential energy field at distance x from a stationary particle of mass M also of fixed radius r is the tangent of the gradient across the diameter of that particle:

\(-M(\frac{\frac{1}{x-r}-\frac{1}{x+r}}{2r}) = -M\frac{1}{x^{2}-r^{2}}, |x| \geq 2r\)

The integral of this function (the gravitational potential) is:

\(M \frac{ln(|x|-r) - ln(|x|+r)}{2r} , |x| \geq 2r\)

The function bounds are restricted because the stationary particle’s gradient is zero inside |x| ≤ r. In that region (the “near field”), the force function becomes:

\(-M(\frac{1-\frac{1}{x+r}}{2r}), |x|\le 2r\)

The potential in the “near field” region becomes:

\(M \frac{- ln(|x|+r)}{2r} , |x| \le 2r\)

QED.