\({ d }_{ s }(x)=Ae^{(-\cfrac{x}{{r}_{e}})}+Be^{(-\cfrac{Orbs}{{r}_{m}}+\cfrac{x}{{r}_{m}})}+C\)
where \({r}_{e}\) and \({r}_{m}\) are radius of the bodies, Orbs is their distance apart (surface to surface) and A, B are constants to be determined. C disappears after differentiation.
Then we let the inverse relationship between time speed squared \({v}^{2}_{t}\)and space density, \({ d }_{ s }(x)\) to be,
\({v}_{t}^{2}=F-D{ d }_{ s }(x)\) where \(F, D\) are constants.
Differentiating this with respect to x,
\(2{v}_{t}\cfrac{d{v}_{t}}{dx}=-D\cfrac{d({d}_{s}(x))}{dx}\)
From the energy conservation equation,
\({v}^{2}_{t}+{v}^{2}_{s}={c}^{2}\)
Differentiating with respect to time and since total energy does not change with time,
\(2{v}_{t}\cfrac{d{v}_{t}}{dt}+2{v}_{s}\cfrac{d{v}_{s}}{dt}=0\)
\({v}_{t}\cfrac{d{v}_{t}}{dx}\cfrac{dx}{dt}+{v}_{s}g=0\), since \(g=\cfrac{d{v}_{s}}{dt}\) and \({v}_{s}=\cfrac{dx}{dt}\)
\(g=-{v}_{t}\cfrac{d{v}_{t}}{dx}\)
From previously, \({v}_{t}\cfrac{{v}_{t}}{dx}=-\cfrac{D}{2}\cfrac{d({d}_{s}(x))}{dx}\)
\(g=\cfrac{D}{2}\cfrac{d({d}_{s}(x))}{dx}\)
And so,
\(g=\cfrac{D}{2}(-\cfrac{A}{{r}_{e}}e^{-\cfrac{x}{{r}_{e}}}+\cfrac{B}{{r}_{m}}e^{(-\cfrac{Orbs}{{r}_{m}}+\cfrac{x}{{r}_{m}})})\)
We know that the distance between the Moon and Earth is 384400 km, Earth's radius re = 6371 km, Moon's radius rm = 1737 km , Earth's gravity = 9.807 m s-2, Moon's gravity = 1.62 ms-2
When x=0,g=-AD/2 re = 9.807
\(g=-{g}_{e}.e^{-\cfrac{x}{{r}_{e}}}+{g}_{m}.e^{(-\cfrac{Orbs}{{r}_{m}}+\cfrac{x}{{r}_{m}})}\)
When x = Orbs = 384400-6371-1737 = 376292, g = BD/(2rm) = 1.62 all higher negative power of e are ignored. And so we have an expression for gravity between Earth and the Moon
g(x)=-9.807*e^(-x/63.71)+1.62*e^((x-3762.92)/17.31)
This graph has been scaled on the x-axis by 10 km. It illustrate the gravity between Earth and the Moon. But still, why wouldn't the Moon falls?
