Flux It,

The derivation for force in this post is wrong, although the conclusion that the force is zero at infinity is true.  Please refer to posts "Opps! Lucky Me" and "Flux It Too" dated 25 May 2015.

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.

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.