# First integral of an equation of motion: $\mu\ddot r=-\frac{k}{r^2}$

I've got an equation of motion (EOM), which is

$\mu\ddot r=-\frac{k}{r^2}$

How do I find the first integral of this EOM? I'd appreciate it if someone could show me the steps involved. I should get

$\frac{1}{2}\mu\dot r^2=-k \left( \frac{1}{R}-\frac{1}{r} \right)$

but I'm not sure how to proceed.

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## 2 Answers

Multiply the left- and the right-hand sides of the equation by $\dot r$, and there will be full differentials at both sides.

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one can put it in a more physical setting. First integral is energy, so you need to get energy conservation from Newton's second law. To do it you multiply both sides by $\dot r$ to get the power produced by the force. Integrating resulting equations gives that energy is conserved. – Yrogirg Aug 17 '12 at 7:15

Let $\dot{r}$ be $p$

$${d^2 r \over dt^2} = {dp \over dt } = {dp \over dr}\times {dr \over dt} = p {dp \over dr} \hspace{2 cm} (1)$$

Then we have $$\mu p {dp \over dr} = {-\frac{k}{r^2}}$$ $$\mu \int p \, dp = - k \int {dr \over r^2}$$ $$\mu {p^2 \over 2} = {k \over r} + C \hspace{2 cm} (2)$$ Assuming $\large \dot{r}(0) = 0$ when $r = R$, you get $$C = - {k \over R} \hspace{2 cm} (3)$$ Hence from $(2)$ and $(3)$, we have $$\mu {\dot{r}^2 \over 2} = -k \left [ {1 \over R}- {1 \over r}\right ]$$

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+1 although he asked for steps, not the whole solution – Rody Oldenhuis Aug 17 '12 at 14:12