Time Speed Field

If we look at the expression for gravity

$g=-\cfrac { 1 }{ 2 } .\cfrac { d{ v }_{ t }^{ 2 } }{ dx }$                  

Integrating both sides

$-2\int { { g } } dx=\int { d{ v }_{ t1 }^{ 2 } } ={ v }_{ t1 }^{ 2 }$ ----(1)

The left hand side is negative given that g decreases with increasing x.

We have seen that g is of the form,

$g=-{ G }_{ o }.\cfrac { 1 }{ { (x+{ r }_{ e }) }^{ 2 } }$

where the negative sign suggests that g, a vector is pointing towards x=0.  Integrating,

$\int { g } dx={ G }_{ o }.\cfrac { 1 }{ (x+{ r }_{ e }) }+A$ ----(2)

Combining the above equations (1) and (2) and using constant C=-2A

${ v }_{ t }= \sqrt { C-\cfrac { 2{ G }_{ o } }{ (x+r_{ e }) }  }$

Since we know that at no gravity, space is not compressed where $x\rightarrow \infty$, ${ v }_{ t }=c$, we have $C={ c }^{ 2 }$, the normal time speed.  Therefore,

${ v }_{ t }= \sqrt {{ c }^{ 2 }-\cfrac { 2{ G }_{ o } }{ (x+r_{ e }) }  }$

This expression is time speed at various distance x, at x = 0 space is densest.