Let's assume that the speed of time is some terminal velocity in space of density \({ d }_{ s }\) and that all factors are constant except for \({ d }_{ s }\). We have
\({ v }_{ t }^{ 2 }\propto \cfrac { 1 }{ { d }_{ s } } \) this is an assumption.
But, if we let,
\( { v }_{ t }^{ 2 }=\quad \cfrac { A }{ { d }_{ s } }\) where A is a constant
then
\(g=-\cfrac { 1 }{ 2 } .\cfrac { d{ v }_{ t }^{ 2 } }{ dx } =-\cfrac { 1 }{ 2 } .\cfrac { d }{ dx } (\cfrac { A }{ { d }_{ s } } )=\cfrac { A }{ 2 } .\cfrac { 1 }{ { d }_{ s }^{ 2 } } .\cfrac { d{ d }_{ s } }{ dx }\)
assume further that space is compressed linearly, where \(x=0\) has the highest compression \({ a }_{ 0 }-a{ r }_{ 0 }\), and the equation is valid up to \({ d }_{ s }=0\) or some residue value \({ d }_{ o }\) not negative.
\({ d }_{ s }=-a(x+{ r }_{ 0 })+{ a }_{ 0 },\quad \cfrac { d{ d }_{ s } }{ dx } =-a\)
then
\(g=-G.\cfrac { 1 }{ { (-a(x+{ r }_{ 0 })+{ a }_{ 0 }) }^{ 2 } }\) where all constants have been grouped into G. This equation shows that gravity is directed at x=0 where space is most compressed and that the inverse square law effect applies assuming that space compresses linearly. Space will be compressed around a massive body or be artificially compressed by moving space towards one end.
