# Differential equation free fall in gravitational field

For a physics problem I was told to set up a differential equation for the free fall in the gravitational field of the earth. The equation (via Newton) I've got is following:

$$\ddot{r} = - G M \frac 1 {r^2}$$

where $G$ is the gravitational constant, $M$ is the mass of the earth and $r$ is the distance to earths center of gravity. I was told to use this equation to find the velocity $\dot{r}$ but I have no idea how to do that, and it seems almost impossible to solve this differential equation analytically at all?

• Are you sure this is correct? The LHS is certainly force accoridng to Newton's law, but the RHS looks like a potential, not a force. The gravitational force is $1/r^2$. Oct 15, 2015 at 18:19
• @RonGordon Of course, you are right, thanks! Oct 15, 2015 at 18:28

After your edit of the question

Hint: Try to multiply with $\dot r$ and integrate, $$\int \ddot r\dot r\,dt=-GM\int \frac{\dot r}{r^2}\,dt.$$

• Thank you for your answer, that is a really neat trick, I'm gonna try that. Oct 15, 2015 at 18:29

The way you framed the problem, it looks like you're talking about rectilinear motion - i.e. straight up or down. Rewrite it as $$\frac{dr}{dt} = v, \quad \frac{dv}{dt} = -\frac{GM}{r^2}.$$

Since the motion is straight up or down, you can treat the velocity $$v$$ as a function of the height $$r$$, instead of as a function of time $$t$$ ... at least until you reach a stationary point, if you're not going up at or beyond escape velocity.

For infinitesimal change in time, $$dt$$, you have $$dr = v dt, \quad dv = -\frac{GM}{r^2} dt.$$ Therefore $$\frac{dv}{dr} = \frac{-GM/r^2}{v},$$ or $$v dv = -\frac{GM}{r^2}dr.$$ or $$d\left(\frac{v^2}{2}\right) = d\left(\frac{GM}{r}\right).$$ Therefore, $$v^2 = \frac{2GM}{r} + \text{ constant}.$$

The simplest case is escape velocity, where $$v → 0$$ as $$r → ∞$$, because then, the integration constant is $$0$$, and we could write $$\frac{dr}{dt} = v = \sqrt{\frac{2GM}{r}}. \tag{Escape Velocity}$$ Then, you have $$\sqrt{r} dr = \sqrt{2GM} dt,$$ or $$d\left(\frac{2}{3}r^{3/2}\right) = d\left(\sqrt{2GM}t\right).$$ Thus $$r^{3/2} = \frac{3}{2} \sqrt{2GM} \left(t - t_0\right),$$ or $$r^3 = \frac{9}{2}GM \left(t - t_0\right)^2,$$ where the integration constant has, without loss of generality, been written as $$t_0$$.

Compare this with Kepler's Third Law: for elliptical orbits, period-squared goes as mean orbital radius cubed.