Mistake in proof of virial theorem for gravity I'm having trouble proving the virial theorem for gravity. I get an extraneous term, but I think my work is correct. 
Starting with the Lagrangian: $$\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2+\frac{GMm}{r}$$
I found $$L=mr^2\dot{\theta}^2\qquad\text{and}\qquad m\ddot{r}=mr\dot{\theta}^2-\frac{GMm}{r^2}.$$ Where $L$ is angular momentum. 
From this I got Binet's equation $$\frac{1}{r}=\frac{GMm^2}{L^2}+B\cos{(\theta-\theta_0)}$$ Where $B$ is an as yet to be determined constant, but is related to the eccentricity of the orbit by geometric arguments. 
I subbed $A=\frac{GMm^2}{L^2}$ for convenience and $\theta_0=0$ since it doesn't effect average values. 
$V=\frac{-GMm}{r}$ so $$<V>=-GMmA=\frac{-L^2A^2}{m}$$ since the cosine averages to zero over a full cycle. 
Differentiating the expression for $1/r$ and subbing in the angular momentum, the result: $\dot{r}=\frac{BL}{m}\sin{\theta}$ and $\frac{m}{2}\dot{r}^2=\frac{B^2L^2}{2m}\sin^2{\theta}$ is the kinetic energy due to radial motion. There is also a kinetic energy term due to angular motion:
$$\frac{L^2}{2mr^2}=\frac{L^2}{2m}(A^2+2AB\cos{\theta}+B^2\cos^2{\theta})$$
So average kinetic energy $$<T>=\frac{L^2(A^2+B^2)}{2m}.$$
So $$\frac{2<T>}{<V>}=\frac{-(A^2+B^2)}{A^2}.$$ 
The Virial Theorem says that ratio should be $-1$, but that's only the case if $B=0$, which implies circular and not elliptical motion. 
It looks like I"m missing a term, but I'm not sure from where. Any  thoughts? 
 A: The virial theorem $\langle U\rangle_t =-2\langle T\rangle_t$ is for time-averages $\langle f\rangle_t=\frac{1}{T}\int_0^T\! \mathrm{d}t~f(t)$, while OP considers angular averages $\langle f\rangle_{\theta}=\frac{1}{2\pi}\int_0^{2\pi}\! \mathrm{d}\theta~f(\theta)$. These averages will in general be different because of the larger (smaller) angular velocity at the perigee (apogee), respectively. Of course, when the eccentricity $e\propto B$ is zero, the angular velocity is constant, and the distinction doesn't matter.
A: I wanted to elaborate in Qmechanic's answer. 
The angular average for $f(\theta)$ is $\langle f(\theta)\rangle_\theta= \frac{1}{2\pi}\int_0^{2\pi} f(\theta) d\theta$. The time average of $f(\theta)$ is $\langle f(\theta)\rangle_t=\frac{1}{\tau}\int_0^{2\pi} \frac{f(\theta)}{\dot{\theta}} d\theta$. 
Where $1/\dot{\theta}=\frac{m}{L(A+B\cos{\theta})^2}$ and $\tau=\int_0^{2\pi} \frac{d\theta}{\dot{\theta}}$. 
This results in $\langle\cos {\theta} \rangle=-\epsilon$.
Relevant integrals can be evaluated by setting $I=\int_0^{2\pi}\frac{d\theta}{A+B\cos{\theta}}=\int_\lambda \frac{-idz}{z[1+\epsilon(\frac{z+1/z}{2})]}$ and using The Residue Theorem. Then the other integrals for calculating the averages can be determined by derivatives of I. Finally $\langle \cos{\theta}\rangle=-\epsilon=-B/A$. Ultimately this yields the expected result from the Virial Theorem, $2\langle T\rangle/\langle V \rangle=-1.$
