Technically, the Moon

Consider space density function of the form,

${ d }_{ s }(x)=Ae^{(-\cfrac{x}{{r}_{e}})}+Be^{(-\cfrac{Orbs}{{r}_{m}}+\cfrac{x}{{r}_{m}})}+C$

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 r_e = 6371 km, Moon's radius r_m = 1737 km , Earth's gravity = 9.807 m s^-2, Moon's gravity = 1.62 ms^-2.

When x=0,g=-AD/2 r_e = 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/(2r_m) = 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?