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I'm learning about orbits in classical mechanics at the moment. I'm not understanding something about the evenness of the distance component and oddness of the angle component.

I see that the radial motion equation tells $\frac{1}{2}\dot{r}^2+V(r)+\frac{L^2}{2r^2}=E$ with $V(r)$ any sort of potential energy per unit mass and $E$ the total energy per unit mass of the system. The radial motion equation $L=r^2\dot{\theta}$ can be used to obtain $\theta$ from $r$ since the angular momentum per unit mass $L$ is constant.

Apparently this should imply that $r(t)$ is an even function and $\theta(t)$ is an odd function if you define them such that $t=0$ is at the farthest point (or closest I guess, as long as it's an extreme). This should hold for any potential energy function $V(r)$ too. I'm not understanding why. I see that integrating $\dot{\theta}$ gives $\theta(t)=-\frac{2L}{r^3(t)}$, but if $r(t)$ is even, that would mean that $\theta(t)$ is even too, which makes no sense. $\theta(r)$ would be odd but that doesn't mean anything since $r>0$.

So my question is, why is $r(t)$ even and why is $\theta(t)$ odd? For the case of an attractive inverse square field like gravity, the orbits should be closed too, which is related to this. Don't understand that implication either.enter image description here

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Consider $r(t)$ and $\theta(t)$ close to $t=0$ where we have set the time so it is an extreme of $r(t)$.

If we Taylor expand, $r(t)=r(0) + r''(0)t^2/2! + r'''(0)t^3/3!\ldots$ since $r'(0)=0$. So unless $r''(0)=0$ the function will be even(ish) close to $t=0$.

Meanwhile $\theta(t)=\int_0^t L/r(\tau) d\tau$, which for small $t$ $\approx (L/r(0)) t$, nicely odd.

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