Consider s space density function of the form,
\({ d }_{ s }(x)=A{ e }^{ -bx }+B\) where A, B and b are constant to be determined.
At \({ d }_{ s }(0)={ d }_{ e }\) space is compressed, its space density is \({d}_{s}\), on surface of earth. And
\({ d }_{ s }(x\rightarrow \infty )={ d }_{ n }\) where at long distance, space is relaxed and the corresponding space density is \({d}_{n}\).
We have,
\({ d }_{ s }(x)={ e }^{ -bx }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }\) ----(1)
Then we let the inverse relationship between time speed squared \({v}^{2}_{t}\)and space density, \({ d }_{ s }(x)\) to be,
\({v}_{t}^{2}=C-D{ d }_{ s }(x)\) where \(C, D\) are to be determined.
We know that as \(x\rightarrow \infty, { d }_{ s }(x)\rightarrow{ d }_{ n }\)where space is relaxed and time speed is \(c\),
\({v}_{t}^{2}={c}^{2}\), \({ d }_{ s }(x)={ d }_{ n }\)
\(C={c}^{2}+D{d}_{n}\) then,
\({v}_{t}^{2}={c}^{2}-D({ d }_{ s }(x)-{d}_{n})\)
If we then consider the time dilation equation from previously, (The use of this equation is still valid because a linear approach can be consider as a first approximate to an exponential one. Consider the power series expansion of \(e\).)
\({ v }_{ t }= \sqrt {{ c }^{ 2 }-\cfrac { 2{ G }_{ o } }{ (x+r_{ e }) }}\)
when \(x=0\)
\({ v }^{2}_{ t }={ c }^{ 2 }-\cfrac { 2{ G }_{ o } }{ (r_{ e }) }\)
\({v}_{t}^{2}={c}^{2}-D({ d }_{ e } -{d}_{n})\) where \({ d }_{ s }(0)={ d }_{ e }\)
we have,
\(\cfrac { 2{ G }_{ o } }{ (r_{ e })}=D({ d }_{ e } -{d}_{n})\)
therefore,
\(D=\cfrac{ 2{ G }_{ o } }{r_{ e }({ d }_{ e } -{d}_{n})}\)
Hence,
\({v}_{t}^{2}={c}^{2}-\cfrac{ 2{ G }_{ o }({ d }_{ s }(x)-{d}_{n}) }{r_{ e }({ d }_{ e } -{d}_{n})}\)
Differentiating both side with respect to time
\(2{v}_{t}\cfrac{d{v}_{t}}{dt}=-\cfrac{2{G}_{o}}{{r}_{e}({d}{e}-{d}{n})}\cfrac{d({d}_{s}(x))}{dx}\cfrac{dx}{dt}\)
\({v}_{t}\cfrac{d{v}_{t}}{dt}=-\cfrac{{G}_{o}}{{r}_{e}({d}{e}-{d}{n})}\cfrac{d({d}_{s}(x))}{dx}{v}_{s}\) ----(2)
But from the energy equation,
\({v}^{2}_{t}+{v}^{2}_{s}={c}^{2}\) differentiating it
\(2{v}_{t}\cfrac{d{v}_{t}}{dt} + 2{v}_{s}\cfrac{d{v}_{s}}{dt}=0\), since \(\cfrac{d{v}_{s}}{dt}=g\)
\(2{v}_{t}\cfrac{d{v}_{t}}{dt} = -2g.{v}_{s}\)
\(g.{v}_{s}=-{v}_{t}\cfrac{d{v}_{t}}{dx}\) substitute into (2)
and so,
\(g=\cfrac{{G}_{o}}{{r}_{e}({d}_{e}-{d}_{n})}.\cfrac{d({d}_{s}(x))}{dx}\)
From (1), differentiating with respect to x,
\(\cfrac{d({d}_{s}(x))}{dx}=-b{e}^{-b(x)}({d}_{e}-{d}_{n})\) subsitute into the above we have,
\(g=\cfrac{{G}{o}}{{r}_{e}}.(-b).{e}^{-b(x)}\)
But we know that at \(x=0\),
\(g={g}_{o}=-b\cfrac{{G}_{o}}{{r}_{e}}\), therefore,
\(b=\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}=\cfrac{1}{{r}_{e}}\) \(\because {g}_{o}=\cfrac{{G}_{o}}{{r}^2_{e}}\)
And so we have an expression for gravity,
\(g=-{g}_{o}{e}^{-\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}\) the negative sign indicates the direction of g.
This expression is based on compression of free space, in a exponential manner.
