Here are several phase-space orbits for the Kepler problem:
The Hamiltonian for the Kepler problem, and hence the energy, is
\[ H = \frac{p^2}{2m} + \frac{L^2}{2mr^2} - \frac{k}{r} = E \]The Hamilton equations are
\[ \dot{r} = \frac{p}{m} \hspace{5em} \dot{p} = -\frac{L^2}{mr^3} + \frac{k}{r^2} \]so that the phase space has one critical point at \( \displaystyle \left[ \frac{L^2}{mk}, 0 \right] \), which explains why the orbits are symmetric with respect to the r-axis.
The upper limitation on angular momentum comes from the value for a circular orbit at the greatest negative energy, which is found by setting the eccentricity equal to zero:
\[ e = \sqrt{ 1 + \frac{2EL^2}{mk^2}} \hspace{2em} \rightarrow \hspace{2em} L = \sqrt{ \frac{mk^2}{2(-E)} } \]For the purposes of this visualization, \( m = k = 1 \) suffices. The endpoints of the orbits on the r-axis are where \( p = 0 \).
Complete code for this example: