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

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?

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I asked a similar question. I think the best you can get is velocity with respect to height: $r' = r'(r)$. math.stackexchange.com/questions/114610/… – Javier Apr 13 '12 at 10:37
@Javier - your question was asking for a full solution to the differential equation. I suspect that there is an easier way to do this question, and my main reason for asking it is to find out what 'integrate the equation of motion' means. Thanks for pointing it out, though. – Donkey_2009 Apr 13 '12 at 10:39
@Donkey_2009: "Integrate the equation" is a slightly old-fashioned way of saying "solve the (differential) equation". – Zhen Lin Apr 13 '12 at 11:42

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 $r(t)$

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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.

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Thanks - if you look at my question, you'll see that I actually got to exactly that integral of a square root by another method; however, I don't think that that's what the question's asking for, as the integral looks hard to solve. – Donkey_2009 Apr 13 '12 at 10:46
Is the question focussed on another aspect, ie, can you find the integral by looking at a table of integrals in the university library rather than the concentrating too long on the method of integration? – Autolatry Apr 13 '12 at 10:50
As I said, the question is focused on finding the constant $C$ in the equation $T=C\sqrt{\frac{R^3}{GM}}$, by 'integrating the equation of motion'. I suspect that what it's asking for is somewhat simpler than solving that differential equation. – Donkey_2009 Apr 13 '12 at 10:52
@Donkey, no. As Zhen said, you should solve the differential equation, as that's what the problem is indeed asking, but cast in old-fashioned language. – J. M. May 13 '12 at 13:24

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 $$C =\frac{1}{\sqrt{2}}\int_{0}^{1}du\sqrt{\frac{u}{1-u}}=\frac{1}{\sqrt{2}}\int_{0}^{\pi/2}2\sin^2\theta d\theta=\frac{\pi}{2\sqrt{2}}.$$

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