I am trying to describe the position of a free falling ball by gravity: if
- $x$ is the time in seconds,
- $y$ is the position of the falling ball,
- $y^{\prime\prime}(x)$ is its acceleration
then $$ F=G\frac{M_1\cdot M_2}{y(x)^2}=M_2 y^{\prime\prime}(x), $$ and since for the Earth $G\cdot M_1=4\cdot10^14$, we just need to solve for $y$ the following equation: $$ y(x)^2\cdot y^{\prime\prime}(x)=4\cdot 10^{14} $$ I have put it in Mathematica by using the following command:
DSolve[{y[x]^2*y''[x]==4*10^14},y[x],x]
but I got no answer: how can I solve for $y$ and plot it?
Here's an analytic solution. The ODE and initial conditions are:
$$y'' = -\frac{G M_1}{y^2(t)}$$ $$y(0)=y_0$$ $$y'(0)=v_0$$
Define dimensionless variables as follows: $$t_0 \equiv \sqrt{\frac{y_0^3}{2 G M_1}}$$ $$T \equiv t/t_0$$ $$Y \equiv y/y_0$$ $$V_0 \equiv v_0 t_0/y_0$$
Then the ODE and initial conditions can be written as: $$2Y'' = -\frac{1}{Y^2}$$ $$Y(0) = 1$$ $$Y'(0) = V_0$$
To solve, multiply both sides of the ODE by $Y'$ and integrate $$2 Y' Y'' = -\frac{Y'}{Y^2}$$ $$ Y'^2 = C_1 + \frac{1}{Y}$$
Use the initial condition to eliminate $C_1$ then solve for $Y'$ $$ V_0^2 = C_1 + 1$$ $$ Y'^2 - V_0^2 = \frac{1}{Y} - 1$$ $$ Y' = \pm\sqrt{V_0^2 + \frac{1}{Y} - 1}= \pm\sqrt{W + \frac{1}{Y}}$$
where $W \equiv V_0^2 - 1$ and the sign is chosen to match $sign(v_0)$. Separating variables gives $$dT = \pm \frac{1}{\sqrt{W+\frac{1}{Y}}} = \pm \frac{Y}{\sqrt{WY^2+Y}} dY$$
Case 1. When $W<0$ then $M_2$ has a velocity less than the escape velocity; if $M_2$ is initially rising ($v_0>0$), gravity will decelerate it enough to cause it to reverse direction and fall.
To integrate, we will manipulate the term under the radical to a form $1-z^2$. Multiply top and bottom of the RHS by $2\sqrt{-W}$
$$dT = \pm \frac{2\sqrt{-W}Y}{\sqrt{-4W^2Y^2-4WY}}dY = \pm \frac{2\sqrt{-W}Y}{\sqrt{-4W^2Y^2-4WY-1+1}}dY = \pm \frac{2\sqrt{-W}Y}{\sqrt{1-(1+2WY)^2}} dY$$
Now introduce $U\equiv 1+2WY; dU=2W dY$. Then
$$2 (-W)^{3/2} dT = \pm \frac{U-1}{\sqrt{1-U^2}} dU$$
Integrate
$$2 (-W)^{3/2} T = C_2 \pm \left[-\sqrt{1-U^2} - \arcsin{U} \right]$$ $$2 (-W)^{3/2} T = C_2 \pm \left[-\sqrt{1-(1+2WY)^2} - \arcsin(1+2WY) \right]$$
From the initial conditions $$0 = C_2 \pm \left[-\sqrt{1-(1+2W)^2} - \arcsin(1+2W) \right]$$
Subtracting and rearranging gives $$ T = \pm \left[-\left(\sqrt{\frac{Y^2}{W}+\frac{Y}{(-W)^2}} - \sqrt{\frac{1}{W}+\frac{1}{(-W)^2}}\right) - \frac{\arcsin(1+2WY) - \arcsin(1+2W)}{2 (-W)^{3/2}} \right]$$
If the motion is initially upward $M_2$ will rise to a height $Y=-1/W$ before beginning to fall.
For the special case when $M_2$ is initially stationary, $v_0=0$, $W=-1$ so $$ T = \left[\left(\sqrt{-Y^2+Y}\right) + \frac{1}{2}\left(\arcsin(1-2Y) - \arcsin(-1)\right) \right]$$
Case 2. When $W>0$ then $M_2$ has a velocity greater than the escape velocity; if $M_2$ is initially rising ($v_0>0$), it will decelerate but never fall.
The steps for solving this are to Case 1, but the integral includes $\sqrt{z^2-1}$ instead of $\sqrt{1-z^2}$. After integration the result is
$$ T = \pm \left[\left(\sqrt{\frac{Y^2}{W}+\frac{Y}{W^2}} - \sqrt{\frac{1}{W}+\frac{1}{W^2}}\right) - \frac{1}{2 W^{3/2}} \ln{\frac{1+2WY + 2\sqrt{W^2Y^2+WY}}{1+2W + 2\sqrt{W^2+W}}} \right]$$
Case 3. The solution for the degenerate case $W=0$ is straightforward. $$dT = \pm \sqrt{Y} dY$$ $$T = C_2 \pm \frac{2}{3}{Y}^{3/2}$$
Initial condition $$0 = C_2 \pm \frac{2}{3}$$ $$T = \pm \frac{2}{3}\left[{Y}^{3/2} - 1 \right]$$
Hope that helps.