ds(x)=Ae−bx+B where A, B and b are constant to be determined.
At ds(0)=de space is compressed, its space density is ds, on surface of earth. And ds(x→∞)=dn where at long distance, space is relaxed and the corresponding space density is dn.
We have,
ds(x)=e−bx(de−dn)+dn ----(1)
Then we let the inverse relationship between time speed squared v2t and space density, ds(x) to be,
v2t=C−Dds(x) where C,D are to be determined.
We know that as x→∞,ds(x)→dn where space is relaxed and time speed is c,
v2t=c2, ds(x)=dn
So,
C=c2+Ddn then,
v2t=c2−D(ds(x)−dn)
Differentiating with respect to time,
2vtdvtdt=−Dd(ds(x))dxdxdt=−Dd(ds(x))dxvs ---(2)
But from the energy equation,
2v2t+v2s=c2 differentiating it
4vtdvtdt+2vsdvsdt=0, since dvsdt=g
2vtdvtdt=−g.vs substitute into (2)
and so,
g=D.d(ds(x))dx
From (1), differentiating with respect to x,
d(ds(x))dx=−be−bx(de−dn) subsitute into the above we have,
g=−D(de−dn).be−bx
when x=0, g=−go
go=Gor2e=D(de−dn).b
g=−goe−bx=−Gor2ee−bx
And if we consider
∫∞0gdx=Gore=∫∞0Gor2ee−bxdx
Gore=Gor2e[∞0−e−bx1b]
re=1b
b=1re
So,
g=−goe−xre=−Gor2ee−xre
This expression is based on compression of free space, in a exponential manner.