= 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 $\dot I=0$, $\dot\theta=\nabla_IH$. 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> are
$$
\dot q_1=\frac{p_1}{q_2^2},\quad\dot q_2=p_2,\quad\dot p_1=0,\quad\dot p_2=\frac{p_1^2}{q_2^3}-\frac{k}{q_2^2}.
$$
The differentials $dH,dp_1$ are independent away from the circular-orbit critical set, establishing integrability. Put $r=q_2$; on a fixed angular-momentum level,
$$
p_2^2=2E+\frac{2k}{r}-\frac{\ell^2}{r^2}.
$$
For a bounded noncollision orbit require $\ell\ne0$ and $-k^2/(2\ell^2)<E<0$. Its two turning radii are
$$
\boxed{\alpha=\frac{k-\sqrt{k^2+2E\ell^2}}{-2E},\qquad\beta=\frac{k+\sqrt{k^2+2E\ell^2}}{-2E}.}
$$
They 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 $u''+u=k/\ell^2$. Thus
$$
r=\frac{\ell^2}{k[1+e\cos(q_1-q_*)]},\qquad e^2=1+\frac{2E\ell^2}{k^2}<1.
$$
After 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 $E\geq0$ 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 $(2\pi)^{-1}\oint p\,dq$, the angular cycle gives $I_1=\ell$, and the two radial legs give
$$
\boxed{I_1=p_1,\qquad I_2=\gamma\int_\alpha^\beta\frac{\sqrt{(r-\alpha)(\beta-r)}}r\,dr,\qquad\gamma=\frac{\sqrt{-2E}}\pi.}
$$
For 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 $\alpha=\beta$.
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