\({ d }_{ s }(x)=A{ e }^{ -bx }+B\) 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,
\({ d }_{ s }(x)={ e }^{ -bx }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }\)
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,
\({v}_{t}^{2}={c}^{2}-\cfrac{ 2{ G }_{ o }({ d }_{ s }(x)-{d}_{n}) }{r_{ e }({ d }_{ e } -{d}_{n})}\)
So,
\(\gamma (x)^{ 2 }=1-\cfrac{ 2{ G }_{ o }({ d }_{ s }(x)-{d}_{n}) }{{c}^{2}r_{ e }({ d }_{ e } -{d}_{n})}\)
Comparing this with the equation obtained from gravity,
\(\gamma (x)^{ 2 }=1-\cfrac { 2{ G }_{ o } }{ { c }^{ 2 }{ r }_{ e } } { e }^{ -\cfrac { { g }_{ o }{ r }_{ e } }{ { G }_{ o } } (x) }\)
\({ e }^{ -\cfrac { { g }_{ o }{ r }_{ e } }{ { G }_{ o } } (x) }=\cfrac{({ d }_{ s }(x)-{d}_{n}) }{({ d }_{ e } -{d}_{n})} \)
But, \({ d }_{ s }(x)={ e }^{ -bx }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }\)
\({ e }^{ -\cfrac { { g }_{ o }{ r }_{ e } }{ { G }_{ o } } }={ e }^{ -b }\)
So, \(b=\cfrac { { g }_{ o }{ r }_{ e } }{ { G }_{ o } }=\cfrac{1}{{r}_{e}}\)
because \({g}_{o}=\cfrac{{G}_{o}}{{r}^{2}_{e}}\)
We have for the case of earth,
\( { d }_{ s }(x)={ e }^{ -\cfrac { 1 }{ { r }_{e } } x }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }\)
Similarly, for black hole
\({ d }_{ sB }(x)={ e }^{ -\cfrac {1 }{ { r }_{eo } } x }({ d }_{ B }{ -d }_{ n })+{ d }_{ n }\)
Differentiating both expressions
\(\cfrac{d({ d }_{ s }(x))}{dx} = -\cfrac { 1 }{ { r }_{e } }{ e }^{ -\cfrac { 1 }{ { r }_{e } } x }({ d }_{ e }{ -d }_{ n })\)
\(\cfrac{d({ d }_{ sB }(x))}{dx} = -\cfrac { 1 }{ { r }_{ eo } }{ e }^{ -\cfrac { 1 }{ { r }_{eo } } x }({ d }_{ B }{ -d }_{ n })\)
Since both are proportional to gravity,
\(\cfrac{{g}_{e}(x)}{{g}_{B}(x)} = \cfrac { { r }_{ eo } }{ { r }_{e } }\cfrac{{ e }^{ -\cfrac { 1 }{ { r }_{e } } x }({ d }_{ e }{ -d }_{ n })}{{ e }^{ -\cfrac { 1 }{ { r }_{eo } } x }({ d }_{ B }{ -d }_{ n })}\)
and so,
\(\cfrac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }} =\cfrac{{g}_{B}(x)}{{g}_{e}(x)}\cfrac { { r }_{ eo } }{ { r }_{e } } { e }^{ (\cfrac { 1 }{ { r }_{eo }}-\cfrac { 1 }{ { r }_{e } }) x} \)
when \(x\) = 0,
\(\cfrac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }} =\cfrac{{g}_{B}(0)}{{g}_{e}(0)}\cfrac { { r }_{ eo } }{ { r }_{e } }.1 \)
Therefore, the relative amount of space compression around a black hole is,
\(\cfrac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }}\) = (5.07071e18/9.80665)*(0.00886/6371000) = 7.191e8
