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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 2

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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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    $\begingroup$ 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. $\endgroup$
    – Yrogirg
    Commented Aug 17, 2012 at 7:15
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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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    $\begingroup$ +1 although he asked for steps, not the whole solution $\endgroup$ Commented Aug 17, 2012 at 14:12

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