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Set $\mu =GM>0$ and interpret $L$ in the displayed equation as specific angular momentum, $L=r^2\dot{\psi }$, rather than total angular momentum. The gravitational potential is per unit mass and $a$ has dimensions of length. Spherical symmetry fixes the orbital plane and conserves the vector angular momentum, while the autonomous specific orbital energy is also conserved.

For the orbit equation for combined inverse-square and inverse-cube attraction, defineA regular nonradial bounded oscillation requires $L\ne 0$ and $L^2-2\mu a>0$. Solving the linear Binet equation then givesFor a noncircular bound orbit $0<e<1$; its specific orbital energy is $E={\mu }^2(e^2-1)/[2(L^2-2\mu a)]$. The effective potential $[L^2-2\mu a]/(2r^2)-\mu /r$ has a centrifugal barrier and a minimum in this case. If $L^2-2\mu a\le 0$, the oscillator frequency is zero or imaginary and there is no regular finite-radius bound oscillation: the attractive centre lacks a centrifugal barrier. Unbound branches with $e\ge 1$ are not closed ellipselike trajectories.

Successive pericentres differ in angle by $2\pi /\nu$. The full phase space state returns after $p$ radial oscillations and $q$ revolutions precisely whenwhere $p,q$ are coprime positive integers; the second formula applies when $a\ne 0$. Thus $a>0$ requires $0<p/q<1$ and $L^2>2\mu a$, whereas $a<0$ requires $p/q>1$. An irrational $\nu$ advances the apsidal direction through a dense set of angles and gives a nonclosed rosette, with only the four regular spherical invariants needed to label its two-dimensional closure.

To see the additional resonant label explicitly, let $u_c=\mu /(L^2-2\mu a)$ and $w=u-u_c-i{u}^{\prime }/\nu$. Then $w=C{e}^{i\nu (\psi -{\psi }_0)}$, so on a family with fixed resonant $L$,Its argument fixes the remaining apsidal phase modulo the symmetry of the closed curve; its magnitude is already determined by specific orbital energy and $L$. Choosing reference axes in the fixed orbital plane makes this single-valued under $\psi \mapsto \psi +2\pi$. This is the fifth independent orbit label on the resonant family described by closed orbits with an inverse-cube perturbation.

For $a\ne 0$, $\nu$ varies with $L$. The orbital resonance condition holds only on selected angular momentum surfaces, so this construction is not a fifth global smooth integral on an open set of arbitrary nearby $L$. If the question asks for a maximally superintegrable Hamiltonian system on an open region of noncircular bound orbits, the required case is $a=0$: then $\nu =1$ for every $L$, and the conserved Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{v}\times \mathbf{L}-\mu \hat{\mathbf{r}}$ fixes the apsidal direction globally. The relations $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2={\mu }^2+2E{L}^2$ leave five independent constants. For the usual orbit-wise interpretation, noncircular bound orbits with rational $\nu$ have five isolating labels; irrational $\nu$ has four. Circular orbits are closed for any allowed $\nu$ but have no independent apsidal phase, so the generic noncircular count cannot simply be applied to them.

For completeness, regular unbound branches are different. When $e\ge 1$ and $\nu >0$, positive radius restricts the radial phase to $|\nu (\psi -{\psi }_0)|<\arccos (-1/e)\le \pi$. The apsidal invariant on an unbound central-force branch isThe principal argument is continuous on this scattering branch because it never reaches an apocentre across the negative-real-axis cut. Thus the periapsis direction can be a fifth single-valued scattering label for any positive $\nu$, including irrational $\nu$; the rationality condition applies to the noncircular bound case. For plunging or collision branches with $L^2-2\mu a\le 0$, local trajectory labels can still be defined away from the singular centre, but the preceding periodic-orbit criterion is inapplicable and no smooth continuation through $r=0$ is asserted.

Set $\theta =0$ initially and choose its positive sense to agree with the initial tangential motion. The inward radial component is $\dot{r}(0)=-4\cos (\pi /3)=-2$, while the tangential speed is $4\sin (\pi /3)=2\sqrt{3}$. HenceSubstitution in the Binet equation gives $12(u^{\prime \prime }+u)=3+9u$, or $u^{\prime \prime }+u/4=1/4$. This is an orbit equation for combined inverse-square and inverse-cube attraction. Solving the constant-coefficient equation and applying both initial conditions givesOn $0\le \theta \le 2\pi$, the sine is nonnegative, so $r$ stays finite and positive. At $\theta =2\pi$, $u=1$, hence $r=1$ and the particle is back at its original spatial position after one revolution. But $u^{\prime }(2\pi )=-1/\sqrt{3}$, so its radial velocity is now $+2$, rather than the original $-2$. The return is not a periodic return in position and velocity.

After that revolution, the first zero of $u$ is at $\theta =8\pi /3$, where $\sin (\theta /2)=-\sqrt{3}/2$. The physical branch therefore has $r\to \infty$ as $\theta \uparrow 8\pi /3$; it must not be continued into negative $u$. Since $dt/d\theta =1/(l{u}^2)$ and $u$ has a simple zero, reaching infinity takes infinite time. As a check, the specific potential energy is $\Phi (r)=-3/r-9/(2r^2)$ and the conserved energy per unit mass is $4^2/2+\Phi (1)=1/2$, giving escape speed $1$ at infinity, also obtained from $-l{u}^{\prime }(8\pi /3)=1$.

Radial energy and effective potential. For a central force with potential energy $U(r)$, angular momentum conservation fixes $J=m{r}^2\dot{\varphi }$ and confines motion to a plane. The kinetic energy separates into radial and angular parts, givingThe second term is the centrifugal barrier. It converts the radial motion to a one-dimensional energy problem: allowed radii have $E\ge U_{\mathrm{eff}}$, turning points have equality, and a circular orbit occurs at a stationary point of $U_{\mathrm{eff}}$. Stability of that radial orbit requires a local minimum.

The first-order orbit equation. Write $A=GMm$. Since $f=-dU/dr=-A/r^2-A\gamma /r^3$, choose zero potential at infinity to getAlso $\dot{r}=(dr/d\varphi )J/(m{r}^2)$. Substitute this into the energy equation and multiply by $2m{r}^4/J^2$:Here $\gamma$ has units of length. A regular bound orbit with a centrifugal barrier requires $J^2>mA\gamma$ as well as the appropriate negative energy.

Solving in inverse radius. Let $u=1/r$ and defineDividing the first-order equation by $r^4$ givesDifferentiating, or directly using the Binet equation, gives $u^{\prime \prime }+qu=b$. The continuation through turning points follows from this smooth second-order equation. Its noncircular bound solution, with angular origin at a pericentre, isSubstitution into the first-order equation fixes the amplitude through this $e$. Therefore the requested representation holds withsince ${\alpha }_1/r-{\alpha }_2=\cos ({\alpha }_3\varphi )$. A general angular origin replaces $\varphi$ by $\varphi -{\varphi }_0$. This is the orbit equation for combined inverse-square and inverse-cube attraction. The circular limit $e=0$ is regular in $u=b/q$ but is not represented by finite ${\alpha }_1,{\alpha }_2$ in the source's normalized cosine form.

Precession and the inferred length. The last expression immediately givesSuccessive pericentres occur after angular advance $2\pi /{\alpha }_3$, so the apsidal precession per orbit isFor ${\alpha }_3$ near one, $1-{\alpha }_3^2\simeq \Delta \varpi /\pi$. Inferring an inverse-cube perturbation from apsidal precession consequently gives $\gamma \simeq [J^2/(mA)]\Delta \varpi /\pi$. For the stipulated nearly circular approximation, $J^2/(mA)\simeq r_0$, and Kepler's third law givesusing $P=88(86400)\mathrm{s}$, $M=2.0\times 10^{30}\mathrm{kg}$ and $G=6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. A century contains about $36525/88=415.1$ orbits. Convert the observed advance to radians per orbit before using it:HenceThe positive sign strengthens the attraction and gives prograde precession. The fractional perturbation $\gamma /r_0\simeq 1.6\times 10^{-7}$ validates the small-deviation expansion. This is the parameter fitting the proposed phenomenological force law under the specified small-eccentricity approximation, not a derivation of that law as the physical explanation of Mercury's anomaly.