Gravity Exponential Form

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.