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.
