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Friday, August 8, 2014

Gravity Exponential Form Again

Consider s space density function of the form,

ds(x)=Aebx+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)=ebx(dedn)+dn ----(1)

Then we let the inverse relationship between time speed squared  v2t  and space density,  ds(x)   to be,

v2t=CDds(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=c2D(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=bebx(dedn) subsitute into the above we have,

g=D(dedn).bebx

when  x=0,    g=go

go=Gor2e=D(dedn).b

g=goebx=Gor2eebx

And if  we consider

0gdx=Gore=0Gor2eebxdx

Gore=Gor2e[0ebx1b]

re=1b

b=1re

So,

g=goexre=Gor2eexre

This expression is based on compression of free space, in a exponential manner.