Integrating twice based on acceleration due to gravity. (Keplar could also be used, but the result is the same). m1 = mass of earth m2 = mass of object, so much smaller than mass of earth it can be ignored
gravitational constant G = 6.6743 x 10^-11
m1+m2 = 5.9722 x 10^24
r is distance between the two objects
v is combined closure rate of the two objects towards a common center of mass.
a is combined acceleration of the two objects towards a common center of mass
initial distance r0 = 76370000
final distance r1 = 6370000
$$ v = dr/dt \ a = dv/dt \ a = (dv / dr) (dr / dt) \ a = v \ dv/dr = -G (m1 + m2) / r^2\ v \ dv = -G (m1 + m2) dr / r^2 $$ Integrating both sides, with constant v = 0 at r0 $$ 1/2 v^2 = G(m1+m2)/r - G(m1+m2)/r0 \ v = -\sqrt{2G(m1+m2)/r - G(m1+m2)/r0} \ v = \frac{dr}{dt} = -\sqrt{\frac{2 \ G\ r0 \ (m1+m2)-2 \ G \ r \ (m1+m2)}{r \ r0}} \ \frac{- \sqrt{r0 \ r} \ dr} {\sqrt{2 \ G \ r0 (m1 + m2) - 2 \ G \ r \ (m1 + m2)}} = dt \ {- \sqrt{ \frac{r0}{2 \ G \ (m1 + m2)}}} \ \ \sqrt{\frac{r}{r0 - r}} \ dr = dt \ $$ Integrate again, using u for substitution. $$ u = \sqrt{\frac{r}{r0-r}} \ u^2 = \frac{r}{r0-r} \ r = \frac{r_0 u^2}{1 + u^2} = \frac{r_0 + r_0 u^2 - r_0}{1 + u^2} = \frac{ r_0(1 + u^2) - r_0}{1 + u^2} = r_0 - \frac{r_0}{1 + u^2} \ dr = \frac{2 \ r0 \ u \ du}{(1 + u^2)^2} \ when \ r = r0 = 76370000\ m, u0 = \infty \ when \ r = r1 = 6370000\ m, u1 = \sqrt{0.091} \ t = \frac{-2 {r0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \ \int_\infty^\sqrt{0.091} \frac{u^{2}}{(1+u^{2})^{2}}du \ t = \frac{-2 \ {r0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \left [ \frac{1}{2} \left (\tan^{-1}(u)-\frac{u}{1+u^{2}} \right ) \right ] _\infty^\sqrt{0.091} \ t = \frac{{-r0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \left <PHONE_NUMBER> - \frac{\pi}{2} \right ) \ t \approx 32240 \ seconds $$ 512