 Solution

A Hamiltonian system on a $2d$-dimensional symplectic manifold is Liouville integrable when it has $d$ first integrals, including the Hamiltonian, whose differentials are independent on a dense regular region and which pairwise have zero Poisson bracket. The Arnold-Liouville theorem states that a compact connected regular common level set is a $d$-torus; near it there are action-angle variables $(I,\theta )$ in which the Hamiltonian depends only on $I$ and the motion is $\dot{I}=0$, $\dot{\theta }={\nabla }_{I}H$. Without compactness, the torus conclusion needs modification.

Here $q_1$ is a cyclic coordinate, so $p_1=\ell$ is conserved and $\{H,p_1\}=0$. The Hamilton equations areThe differentials $dH,d{p}_1$ are independent away from the circular-orbit critical set, establishing integrability. Put $r=q_2$; on a fixed angular-momentum level,For a bounded noncollision orbit require $\ell \ne 0$ and $-k^2/(2{\ell }^2)<E<0$. Its two turning radii areThey are positive and $\alpha <\beta$, and $p_2^2=(-2E)(r-\alpha )(\beta -r)/r^2$. In terms of $u=1/r$ and the angular coordinate, eliminating time gives $u^{\prime \prime }+u=k/{\ell }^2$. ThusAfter an angular change of $2\pi$ the phase-space state returns to itself, and the elapsed time is finite because $r$ is bounded away from zero and infinity. This proves periodicity; a torus alone would only guarantee quasiperiodic motion. At $E=-k^2/(2{\ell }^2)$ the orbit is circular and remains periodic. For $E\ge 0$ the allowed radial motion is unbounded. The collision case $\ell =0$ reaches the excluded boundary $r=0$ and is not a periodic orbit in the given phase space.

With the usual action convention $(2\pi {)}^{-1}\oint pdq$, the angular cycle gives $I_1=\ell$, and the two radial legs giveFor completeness, substitute $r=(\alpha +\beta )/2+(\beta -\alpha )\cos \theta /2$ to evaluate the integral as $\pi [(\alpha +\beta )/2-\sqrt{\alpha \beta }]$. Therefore $I_2=k/\sqrt{-2E}-|\ell |$ and $H=-k^2/[2(I_2+|I_1|{)}^2]$. The two frequencies have equal magnitude, consistent with the closed Kepler orbits. The circular limit has $I_2=0$ and $\alpha =\beta$.