The following diagram shows the force density, \(F_\rho\), on the surface of a sphere at a radius \(x\),
The total force within the volume, \(F_v\) is,
\(F_{ v }=F_{ \rho }\cfrac { 4 }{ 3 } \pi \, x^{ 3 }\)
Imagine all this forces being distributed uniformly on the surface of the sphere (\(4\pi x^2\)), the total flux is then,
\(F_{ A }=F_{ \rho }\cfrac { 4 }{ 3 } \pi \, x^{ 3 }\cfrac { 1 }{ 4\pi \, x^{ 2 } } \)
When the total flux is conserved, at \(x_1\) and \(x_2\)
\(F_{ 1 }.4\pi \, x^{ 2 }_{ 1 }=F_{ 2 }.4\pi \, x^{ 2 }_{ 2 }=F_{ A }\)
\(F_{ 1 }=F_{ A }\cfrac{1}{4\pi \, x^{ 2 }_{ 1 }}\)
as such given a general radial distance \(x\),
\(F=\cfrac { F_{ A } }{ 4\pi \, x^{ 2 } } =F_{ \rho }\cfrac { 4 }{ 3 } \pi \, x^{ 3 }\cfrac { 1 }{ 4\pi \, x^{ 2 } } \cfrac { 1 }{ 4\pi \, x^{ 2 } } =F_{\rho}\cfrac{1}{12\pi x}\)
\(F\), \(F_1\) and \(F_2\) are in Newtons (N).
So, if the force density, \(F_{\rho}\) is a constant as in the post "Constant Of Integration Used Up", then
\(F=F_{\rho}\cfrac{1}{12\pi x}\), \(F_{\rho}=constant\)
\(F\rightarrow 0\), as \(x\rightarrow\infty\)
It is not absurd after all. And the universe is safe from ever expanding.
\(F=i\cfrac{\sqrt { 2{ mc^{ 2 } } }}{12\pi x}\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } (x-x_z) \right)\)
There is no need for \(g_o\) to shift \(-F\) up.
It is however still very interesting that \(F\) should reverse sign, and turn positive.
overX.png)
A comparative plots of \(\cfrac{1}{x^2}\), \(\cfrac{1}{x}tanh(x-1)\) and \(\cfrac{1}{x}\) is given above. \(tanh(x)\) is the lowest of them all, and passes below \(y=0\) before tending towards zero as \(x\rightarrow\infty\).
This positive gravity outwards might explain why some asteroid (unconfirmed report of 2012 da14 asteroid is slowing down) decelerate on approach to earth.
