Free-fall according to Newton's gravitation law Most analysis of free-fall assume that bodies fall with constant acceleration. If however one analyses  free-fall according to Newton's gravitation law, one is lead to a differential equation which I don't know how to solve. 

The differential equation one is lead to[1] is $\frac{d^2x}{dt^2}=-\frac{1}{x^2}$. Does this equation have a closed form solution with given initial conditions $x_0,v_0$?

[1] If one picks the units so that $MG=1$, fixes one mass at $x=0$ and places the falling mass at some $x&gt;0$.
Multiply the equation by $dx/dt$ and integrate once to get
$$
\frac{1}{2}\Bigl(\frac{dx}{dt}\Bigr)^2=\frac{1}{x}+\frac{v_0^2}{2}-\frac{1}{x_0}.
$$
Then
$$
\frac{dx}{dt}=\pm\sqrt{\frac{2}{x}+v_0^2-\frac{2}{x_0}}\,,
$$
a first order equation in separated variables whose solution is
$$
\int_{x_0}^x\frac{dz}{\sqrt{\frac{2}{z}+v_0^2-\frac{2}{x_0}}}=\pm t.
$$
The integral can be computed explicitly, although I do not think you can find a closed form for $x$ in terms of $t$.
I managed to do it

NewtonGravitySolution

The final solution I had was:

$t=\sqrt{\frac {R}{2Gm} } \biggl( x \sqrt{\frac {R}{x} -1} +tan^{-1}\sqrt{\frac{R} {x}-1 }  \biggr)$

$R$ is the initial position where $v=0$

If you plug in values for Earth with Mass = Earth's Mass you can plot it.