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I have been given these two equations in polar coordinates:

$m(r''− rθ'^2) =−f,$

$m(r''θ + 2r'θ') = 0$

And have been told that I need to differentiate to show the angular momentum $L=mr^2θ'$ is conserved. And then to use this and the substitution $r=1/u$ to turn the first equation into Binet's equation.

I am not sure how to do this as all of my attempts have gotten to something different.

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You need to show that the angular momentum $L = m r^2 \theta'$ is conserved. This means that $L$ always has the same value, i.e. that its value does not change in time. Therefore, you need to show that $L' = 0$.

What is $L'$? Given that $L = m r^2 \theta'$, we see that \begin{equation} L' = (m r^2 \theta')' = m r^2 \theta'' + 2 m r r' \theta' = m r( r \theta'' + 2 r' \theta') \tag{1} \end{equation} by the product and chain rules. You can now try to use the given equations to argue that $(1)$ must equal zero (check if the second equation you've written down is correct!).

For the second problem, it is given that you should substitute $r = \frac{1}{u}$. This means that \begin{equation} r' = -\frac{1}{u^2} u' \end{equation} by the chain rule, and therefore (differentiating once more) \begin{equation} r'' = -\frac{u''}{u^2} + 2 \frac{1}{u^3} (u')^2. \end{equation} Substituting these into the given equations should get you closer to the Binet equation. For more information, see Wikipedia.

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$$\frac 1r\frac{d}{dt}(r^2 \dot{\theta})=\frac 1r(2r\dot{r}\dot{\theta}+r^2\ddot{\theta})=0$$ Hence $L$ is constant

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  • $\begingroup$ Would that just be 0? $\endgroup$
    – user295542
    Feb 17 '16 at 13:08

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