From the post "Space Density Compression Ratio",
\( { d }_{ s }(x)={ e }^{ -\frac { 1 }{ r } x }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }\)
where \(r\) is the radius of the planet and \({d}_{n}\) is a constant, the baseline space density of empty free space. This equation suggests that space compression above normal space about a volume of denser space decreases exponentially radially from the volume. It also suggests that since we are dealing with surface density \({d}_{s}(x\) = 0\()\), below \(r\) things can be hollow, as long as the surface density is established.
\(\frac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }} =\frac{{g}_{B}(0)}{{g}_{e}(0)}\frac { { r }_{ eo } }{ { r }_{e } }\) = 7.191e8
When surface gravity reaches 5.07071e18 ms-2 light does not escape and time speed is zero. The size of the black hole when this is achieved can vary. Gravity is proportional to the first derivative of space density not the absolute value of space density. As long as a space density profile that provides the appropriate gradient is achieved, time stops. A space density gradient that results in a surface gravity of 5.07071e18 ms-2 will stop time. One way is where one end of such a profile is open to free space. In that case, the space compression ratio on the surface will have to be 7.191e8 in order to stop time. The \(\gamma (x)\) expression is,
\(\gamma(x)=\sqrt{1-\frac{2{G}_{o}}{{c}^{2}{r}_{e}}{e}^{-\frac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}}\)
Simplifying further, because \(\frac{2{G}_{o}}{{c}^{2}{r}_{eo}}=1\) and \({G}_{o}= {g}_{o}{r}^{2}_{e}\)
\(\gamma(x)=\sqrt{1-\frac{{r}_{eo}}{{r}_{e}}{e}^{-\frac{1}{r}_{e}(x)}}\)
where \({r}_{eo}\) = 0.00886 m is the size of the black hole where the surface gravity does not allow light to escape and \({r}_{e}\) is the radius of the sphere, the surface of which has a surface gravity as a result of a space density profile along its radial line. If the surface space compression ratio is 7.191e8 it will be a black hole. Time stop inside this black hole. If the black hole is made bigger, hollow thin sphere of radius \({r}_{e}\), ie \(\frac{{r}_{eo}}{{r}_{e}} \) = 1,
\(\gamma(x)=\sqrt{1-{e}^{-\frac{1}{r}_{e}(x)}}\)
A plot of the time dilation function when \({r}_{e}\) = \({r}_{eo}\) = 3m, (1-e^(-x/3))^(0.5) is shown below,
At 14 m away time returns to normal, below the surface x < 0, inside the hollow sphere, time dilation can go to zero. All gravitational effects inside the sphere cancel, just like gravity inside Earth is solely due to the mass below, all gravitational effects of the mass above cancel out.
