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$$m\frac{d^2r}{dt^2}=-G\frac{Mm}{r^2}, \qquad r(0)=R ,\qquad \dot{r}=v_0>0$$

describes the free fall of an object of mass $m$. $r(t)$ is the distance of m to the center of the earth which is $R$ at $t=0$. $G$ is the gravitational constant.

I don't even know how to start with it. I thought about integrating it twice, but it was apparently wrong. And once I get $r(t)$ I am supposed to find the velocity a rocket should have to escape the gravitational area of the earth, and to do that I'm supposed to assume that the velocity of the rocket at $r(t)=\infty$ is zero.

Qmechanic
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1 Answers1

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Here is one way to solve it. You can multiply both terms by $\dot r$. $$ m\frac{d^2 r}{dt^2}\frac{dr}{dt} + \frac{GMm}{r^2}\frac{dr}{dt} = 0 \quad\Longrightarrow\quad \frac{d}{dt}\left[\frac{m}{2}\left(\frac{dr}{dt}\right)^2\right] - \frac{d}{dt}\frac{GMm}{r} = 0 $$

Basically $\dot r$ is the integrating factor. We now recognize the potential and kinetic energies over there. Also, we see, $E = T + V$ is a dynamical invariant. Which means: $$ E = \frac{m}{2}\left(\frac{dr}{dt}\right)^2 - \frac{GMm}{r}, \quad\quad \frac{dE}{dt} = 0, \quad\quad E = cte. $$

Now all you have to do is to isolate $\dot r$, separate $r$-terms and $t$ terms, and integrate to explicitly find $\Delta t$, then compute the inverse to find $r(t)$ for a given energy $E$.

Tip: you don't need to solve that differential equation completely and plug initial conditions to realize the escape velocity of a rocket. You can work with conservation of energy directly to find the escape velocity of your rocket (that way you will save a lot of paper).

Physicist137
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