# Even and odd polar coordinates in general orbits

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.

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.