\({ d }_{ s }(x)=A{ e }^{ -bx }+B\) 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 \(c\),
\({v}_{t}^{2}={c}^{2}\), \({ d }_{ s }(x)={ d }_{ n }\)
So,
\(C={c}^{2}+D{d}_{n}\) then,
\({v}_{t}^{2}={c}^{2}-D({ d }_{ s }(x)-{d}_{n})\)
Differentiating with respect to time,
\(2{v}_{t}\cfrac{d{v}_{t}}{dt}=-D\cfrac{d({d}_{s}(x))}{dx}\cfrac{dx}{dt}=-D\cfrac{d({d}_{s}(x))}{dx}v_s\) ---(2)
But from the energy equation,
\(2{v}^{2}_{t}+{v}^{2}_{s}={c}^{2}\) differentiating it
\(4{v}_{t}\cfrac{d{v}_{t}}{dt} + 2{v}_{s}\cfrac{d{v}_{s}}{dt}=0\), since \(\cfrac{d{v}_{s}}{dt}=g\)
\(2{v}_{t}\cfrac{d{v}_{t}}{dt} = -g.{v}_{s}\) substitute into (2)
and so,
\(g=D.\cfrac{d({d}_{s}(x))}{dx}\)
From (1), differentiating with respect to x,
\(\cfrac{d({d}_{s}(x))}{dx}=-b{e}^{-bx}({d}_{e}-{d}_{n})\) subsitute into the above we have,
\(g=-D({d}_{e}-{d}_{n}).b{e}^{-bx}\)
when \(x=0\), \(g=-g_o\)
\(g_o=\cfrac{G_o}{r^2_e}=D({d}_{e}-{d}_{n}).b\)
\(g=-g_o{e}^{-bx}=-\cfrac{G_o}{r^2_e}{e}^{-bx}\)
And if we consider
\(\int^{\infty}_0{g}dx=\cfrac{G_o}{r_e}=\int^{\infty}_0{\cfrac{G_o}{r^2_e}{e}^{-bx}}dx\)
\(\cfrac{G_o}{r_e}=\cfrac{G_o}{r^2_e}[^{\infty}_0-{e}^{-bx}\cfrac{1}{b}]\)
\(r_e=\cfrac{1}{b}\)
\(b=\cfrac{1}{r_e}\)
So,
\(g=-g_oe^{-\cfrac{x}{r_e}}=-\cfrac{G_o}{r^2_e}{e}^{-\cfrac{x}{r_e}}\)
This expression is based on compression of free space, in a exponential manner.
