What exactly is meant by 'Integrate the Equation of Motion'?

Question

Short Question: if a question says to 'integrate the equation of motion', what does it mean?

Long Question:

Question: Take a planet of mass $M$ and place a satellite at rest at a distance $R$ from the planet, where $R$ is much greater than the planet's radius. How long does the satellite take to hit the surface of the planet?

Part 1 of the question asks the reader to perform dimensional analysis. This yields

$$\textrm{Time taken }T=C\sqrt{\frac{R^3}{GM}}$$

Part 2 - integrate the equation of motion of the satellite to show that $C=\pi /2\sqrt{2}$.

As far as I'm aware, the equation of motion for the satellite is

$$\ddot{r}=-\frac{GM}{r^2}.$$

I've tried solving this differential equation, but to no avail ($r=\frac{9}{2}GMt^{\frac{2}{3}}$ is a particular solution, but I have no idea how to find the more general case; substituting the dimensionless quantity $\kappa=\frac{1}{GM}r^3t^{-2}$ almost worked but not quite). I also tried using the potential $V=\frac{-GMm}{r}$ to form the equation

$$T=\int_R^0{\frac{dr}{2\sqrt{\frac{GM}{r}-\frac{GM}{R}}}}.$$

However, this integral doesn't look as if it's going to give the right answer. The $\pi$ in the given expression for $C$ seems to suggest that we're going to get an integral involving a $\sin$ substitution.

So - what does 'integrate the equation of motion' mean? Integrate which equation? And with respect to what?

Answer 1

Regarding the question in the title: 'to integrate the Equation of Motion', or in general, to 'integrate a differential equation' just means to solve it. The expression alludes to the fact that you want to find $r(t)$, the position of the particle as a function of time ("the motion"), but originally you have an equation that relates the motion and its derivatives; so, to "integrate the equation" means, roughly, to get rid of the derivatives ($\dot{r}$, $\ddot{r}$, ...) and find explicitly $r(t)$

Answer 2

Multiply both sides of your equation of motion for $\ddot{r}$ by $2 \dot{r}$ then use you knowledge of the chain rule on the right hand side. You will end up integrating a square root but fear not you will be heading in the right direction.

Answer 3

As you said, the equation of motion is $$ \ddot{r}=-\frac{GM}{r^2}, $$ and the total energy $E=\dot{r}^2/2 - GM/r$ is conserved (and equal to $-GM/R$), letting you write $$ \dot{r}=-\sqrt{2GM}\sqrt{\frac{1}{r}-\frac{1}{R}} $$ directly; or, in terms of $u=r/R$, $$ \dot{u}=-\sqrt{\frac{2GM}{R^3}}\sqrt{\frac{1-u}{u}}. $$ So $$ T=\sqrt{\frac{R^3}{GM}}\frac{1}{\sqrt{2}}\int_{0}^{1}du\sqrt{\frac{u}{1-u}}. $$ The integral can be looked up, or you can use the trig substitution $u=\sin^2\theta$ to yield $$ C =\frac{1}{\sqrt{2}}\int_{0}^{1}du\sqrt{\frac{u}{1-u}}=\frac{1}{\sqrt{2}}\int_{0}^{\pi/2}2\sin^2\theta d\theta=\frac{\pi}{2\sqrt{2}}. $$