For the two-body central force problem, the radial equation is:
$$r(\theta)=\frac{p}{1+e\,cos(\theta - \phi)}$$
where $$\phi$$ is the phase angle, $$\theta$$ is the angle of the satellite object in its orbit, $$p\, \equiv \, \frac{L^2}{Gm^2M}$$
(L is the angular momentum, m is the mass of the satellite, M is the mass of the central body, and G is <PERSON>'s gravitational constant), and $$e$$ is the eccentricity for the orbit. For an elliptical orbit:
$$ e\, =\, \sqrt{\frac{2\,E_0\,m\,p^2}{{L_0}^2}\,+\,1}$$
Note that the angular momentum is:
$$\overrightarrow{L_{net}}=\overrightarrow{r} \times \overrightarrow{p}=$$
$$r\,\hat{r} \times m\, (\dot r \hat{r} + r\,\dot\theta \hat{\theta})=$$
$$m\,r^2\,\dot\theta\,\hat{z}$$
Note that, for strictly central force problems:
$$\dot{\overrightarrow{L_{net}}}\,=\,\tau_{net}\,=\, \overrightarrow{r} \times\sum_i\, \overrightarrow{F_i}\,=\, r\,\hat{r}\, \times\, \sum_i\, F_i\, \hat{r}\, =\, \overrightarrow{0} $$
This means that, for a central force problem, the angular momentum is constant. Thus:
$$\dot\theta\, =\, \frac{L_0}{m\,r^2}$$
where $$L_0$$ is that inital angular momentum (or any convenient time).
Thus:
$$\frac{d\theta}{dt}\,=\, \frac{L_0}{m\,r^2}$$
So:
$$\frac{m\,r^2\, d\theta}{L_0}\, =\, dt$$
$$\frac{m}{L_0}\int_{\theta_0}^{\theta_0\,+\,2\pi}\,r^2\,d\theta\,=\,\int_{0}^{T}\, dt$$
where T is the period and the limits of the integral on the left were chosen from any arbitrary angle and the upper limit was is because of the $$2\pi$$ periodicity of the radial function (at the top).
Plugging in the radial equation:
$$\frac{m}{L_0}\int_{\theta_0}^{\theta_0\,+\,2\pi}\,({\frac{p}{1\,+\,e\,cos(\theta-\phi)}})^2\,d\theta\,=\,\int_{0}^{T}\, dt$$
$$\frac{m\,p^2}{L_0}\int_{\theta_0}^{\theta_0\,+\,2\pi}\,[1\,+\,e\,cos(\theta-\phi)]^{-2}\,d\theta\,=\,T$$
Using a change of variables:
$$u \,\equiv\, (\theta\, -\, \phi)\,\,\, \Rightarrow\,\,\,du\,=\, d\theta\, $$and$$\, u_{initial}\,=\, (\theta_0\,-\,\phi) \, $$and$$u_{final}\,=\,(\theta_0\,+\,2\pi\,-\,\phi)$$
The equation then becomes:
$$\frac{m\,p^2}{L_0}\int_{(\theta_0\,-\,\phi)}^{(\theta_0\,+\,2\pi\,-\,\phi)}\,[1\,+\,e\,cos(u)]^{-2}\,du\,=\,T$$
That integral is still pretty nasty... 246