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In the gravitational potential $-\mu (1/r+a/r^2)$, a regular noncircular bound orbit requires $L^2>2\mu a$ and has $u=\mu (L^2-2\mu a{)}^{-1}[1+e\cos \nu (\psi -{\psi }_0)]$ with $0<e<1$. Each radial period advances the angle by $2\pi /\nu$. The full position and velocity return after finitely many radial periods precisely when $\nu$ is rational. Irrational $\nu$ gives a dense rosette on the two-frequency invariant torus. Rationality permits a fifth orbit label fixing the apsidal phase on that resonant angular momentum family, but does not give global superintegrability for all nearby angular momenta. When $a=0$, every regular bound Kepler orbit is closed and the Laplace-Runge-Lenz vector supplies the globally defined additional integral.

For an attractive Kepler orbit with gravitational parameter $GM$, velocity $\mathbf{v}$ and specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$, the eccentricity vector isIts magnitude is the orbital eccentricity, and it points toward periapsis for a noncircular orbit. It is a normalized attractive Laplace-Runge-Lenz vector; it remains constant under the pure inverse-square force.

An isolating integral is a regular single-valued integral of motion whose fixed value restricts accessible phase space to a lower-dimensional level set. Functional independence of several such integrals means their differentials are linearly independent on the region considered. A central potential has specific orbital energy and the three components of specific angular momentum as four generically independent integrals. A Kepler orbit has a fifth independent integral supplied by its fixed apsidal direction, encoded by the Laplace-Runge-Lenz vector; its vector components obey two constraints with energy and angular momentum. Circular and radial orbits are degenerate cases of this count.

For $H=p^2/2-\kappa /r$, the angular momentum and Laplace-Runge-Lenz vector obey $\{K_i,K_j\}=-2H{ϵ}_{ijk}{L}_k$. On negative or positive energy regions use $K/\sqrt{|2H|}$ to obtain the compact or Lorentz Lie algebra, respectively. At zero energy the algebra on the characteristic orbit quotient is the Euclidean motion algebra. Global group actions require complete generators or collision regularization; simply replacing $H$ by a constant inside a Hamiltonian-vector-field derivative is not valid on the unreduced energy hypersurface.

For the signed inverse-square force $\mathbf{F}=-\kappa \mathbf{r}/r^3$ on a particle of mass $m$, with Hamiltonian $H=p^2/(2m)-\kappa /r$, the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}-m\kappa \hat{\mathbf{r}}$ is conserved. Attraction has $\kappa >0$ and repulsion has $\kappa <0$. It obeys $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2=m^2{\kappa }^2+2mH{L}^2$. For unit mass in a Kepler orbit, $\kappa =GM$.

More than three integrals occur when additional symmetry or degeneracy restricts motion beyond a generic three-frequency invariant torus. In a time-independent central gravitational potential, energy and the three components of specific angular momentum are four generically independent isolating integrals. They fix the orbital plane as well as the radial-motion parameters. A nonclosed planar rosette typically has two-dimensional orbit closure, rather than a three-dimensional invariant torus.

The Kepler orbit is a stronger example. In addition to $E$ and $\mathbf{L}$, the conserved Laplace-Runge-Lenz vectorfixes the apsidal direction. Its constraints $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2={\mu }^2+2E{L}^2$ leave five independent integrals among $E,\mathbf{L},\mathbf{A}$. It is a maximally superintegrable Hamiltonian system, with closed nondegenerate bound orbits.

More than three integrals need not form a set of Poisson-commuting functions; the three-integral limit applies to an independent commuting set, rather than to every conserved orbit constraint. At a nonstationary point in six-dimensional phase space, integral gradients annihilate the nonzero flow vector, so at most five are independent. This explains the isolating integrals and orbital closure dimension count. A resonance on one isolated orbit does not by itself prove an additional smooth global integral on a neighborhood.

Five independent isolating orbit labels describe a regular closed, nonstationary orbit in three-dimensional space: its phase space closure is one-dimensional. Five is the maximum at a nonzero flow vector, since every conserved differential annihilates that vector in the six-dimensional tangent space. A global superintegrable Hamiltonian system example is the Kepler orbit, where specific orbital energy, angular momentum and the Laplace-Runge-Lenz vector have two vector/scalar relations leaving five independent invariants. A circular or radial orbit requires a separate treatment of any rank-degenerate formula for these invariants. Five integrals do not force an escaping trajectory to be periodic: unbound Kepler orbits also have the conserved apsidal vector. The closure criterion concerns recurrent bound motion.

Closure of one resonant orbit need not imply five smooth independent integrals on an open neighborhood of all nearby orbits. The orbit-wise count and global superintegrability must be distinguished; the distinction matters in part (b).

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.

Write $\mu =G{M}_{\odot }$, and interpret the specified radial force field as acceleration, or force per unit test-particle mass. Since it is a central force, the specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ is conserved: $\dot{\mathbf{h}}=\mathbf{r}\times \mathbf{F}=0$. For $h=|\mathbf{h}|\ne 0$, motion is confined to the fixed plane perpendicular to $\mathbf{h}$. Choose the orientation of the polar angle so that $h=r^2\dot{\theta }>0$.

To derive the Binet equation here, let $u=1/r$ and differentiate with respect to $\theta$. Then $\dot{r}=-h{u}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{u}^2(u^{\prime \prime }+u)$. Equating this to the radial acceleration givesPut $K=h^2+C$ and $\alpha =\sqrt{K/h^2}$. For $K>0$, the solution isThe radial potential energy per unit mass is $V(r)=-\mu /r+C/(2r^2)$, so the conserved specific orbital energy isSubstitute the displayed $u$ and $\dot{r}=-h{u}^{\prime }$ to obtain $\mathcal{E}={\mu }^2(e^2-1)/(2K)$. Therefore defineFor regular bound motion, $K>0$ and $-{\mu }^2/(2K)\le \mathcal{E}<0$, giving $a>0$ and $0\le e<1$. Since $K/\mu =a(1-e^2)$, choose the angular origin at a pericentre, so ${\theta }_0=0$, and obtainThe turning radii are $r_{\mathrm{p}}=a(1-e)$ and $r_{\mathrm{a}}=a(1+e)$. Thus $a$ is the arithmetic mean of the turning radii; when the orbit precesses it should not be interpreted as the semimajor axis of a fixed ellipse. If $K\le 0$, the centrifugal barrier needed for a regular two-turning-point bound orbit is absent, and this bound-orbit parametrization does not apply.

For $\alpha =1$, one has $C=0$ and the orbit is a closed Kepler orbit: an ellipse of eccentricity $e$ with the Sun at a focus. In addition to $\mathcal{E}$ and the three components of $\mathbf{h}$, the Laplace-Runge-Lenz vectoris conserved. Indeed,so $\dot{\mathbf{A}}=0$. It points toward the fixed pericentre. The constraints $\mathbf{A}\cdot \mathbf{h}=0$ and $A^2={\mu }^2+2\mathcal{E}{h}^2$ leave five generically independent isolating integrals. One may takechoosing a component $A_j$ whose differential is independent of the other four on the local region under consideration. The extra integral fixes the apsidal direction in the already fixed orbital plane.

For $\alpha \ne 1$, successive pericentres are separated by $2\pi /\alpha$ in polar angle. The generic bound orbit is a rosette in that same plane. Its perihelion advance per radial cycle isFor $\alpha <1$ this is prograde; for $\alpha >1$ it is retrograde. If $\alpha$ is rational, the orbit closes after a finite number of radial cycles; if it is irrational and $0<e<1$, its successive radial phases densely fill the accessible annulus. A generic independent set of single-valued isolating integrals isThe apsidal direction is no longer fixed. For irrational frequency ratio, any continuous integral of motion must be constant on the whole two-dimensional orbital torus, because one orbit is dense there; there is no further independent continuous integral specifying a fixed apsidal direction. Exceptional closed resonant orbits do not provide a fifth global integral on an open region of phase space. Circular and radial orbits are degenerate cases of these generic counts.

The plotted precession is enlarged for visibility. Both examples have $e=0.5$ and the same turning radii; the second has $\alpha =\sqrt{0.9}$.

For the precession rate, it is useful to derive the radial period rather than assume an unperturbed angular period. Set $r=a(1-e\cos \chi )$ in the radial energy equation. On a complete radial cycle,Hence $n=2\pi /P_r=\sqrt{\mu /a^3}$. Also $K=\mu a(1-e^2)$ and $h^2=K-C$, so ${\alpha }^{-1}=\sqrt{1-C/K}$. For $|C|\ll K$, inverse-cube perturbation of Kepler apsidal motion givesThere is a dimensional issue with the printed parameter: $C$ has units $L^4/T^2$ and $\mu$ has units $L^3/T^2$, so $\eta =aC/\mu$ has units $L^2$. Retaining that definition literally, the requested expression isA dimensionless perturbation coefficient would instead be $\tilde{\eta }=C/(\mu a)=\eta /a^2$. In that convention the rate is $-n\tilde{\eta }/[2(1-e^2)]$. These two conventions must not be conflated.

Use the stated residual precession and period. The advance in one radial cycle isWith $e=0.206$, the small-perturbation estimate becomesThe negative sign is required: an additional attractive inverse-cube acceleration advances perihelion, whereas the positive-$C$ term would make it regress. If a length-unit estimate of the printed $\eta$ is desired, $P_r=0.24\mathrm{yr}$ and the solar value $\mu \simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ give $a\simeq 5.78\times 10^{10}\mathrm{m}\simeq 0.386\mathrm{au}$ through the radial-period law. ThusThe dimensionless value is the estimate obtainable directly from the supplied period, eccentricity and advance; the dimensional value additionally uses the solar gravitational parameter to set the length scale.

With $\psi =0$, the angular momentum becomes $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$ and is constant. The identity $\mathbf{L}\cdot \mathbf{r}=0$ confines the trajectory to the plane through the origin perpendicular to $\mathbf{L}$ when $\mathbf{L}\ne 0$. If $\mathbf{L}=0$, the velocity is radial and a central force preserves the radial line; this degenerate motion is also planar.

For the proposed Laplace-Runge-Lenz vector convention, differentiate $\mathbf{K}=\mathbf{L}\times \dot{\mathbf{r}}-\varphi (r)\mathbf{r}$. Since $\ddot{\mathbf{r}}=-{\varphi }^{\prime }(r)\hat{\mathbf{r}}/m$ and $\mathbf{L}$ is constant, the vector triple product givesThereforeIt vanishes for general motion precisely when $(r\varphi {)}^{\prime }=0$, givingThis is an inverse-square potential, attractive when $C<0$ and repulsive when $C>0$; $C=0$ is the free-particle case. An additive constant in $\varphi$ leaves the force unchanged but would change this particular $\mathbf{K}$ by a constant multiple of $\mathbf{r}$, so its definition requires the displayed choice of energy zero.

Put $\mu =GM$ and use energy and angular momentum per unit particle mass. The Kepler orbit equation is $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$. HenceThus $E$ and $\mathbf{L}=\mathbf{r}\times \dot{\mathbf{r}}$ are integrals of motion. Splitting the velocity into radial and transverse parts, $|\dot{\mathbf{r}}{|}^2={\dot{r}}^2+L^2/r^2$, gives the Kepler radial energy equationHere $\dot{r}$ denotes the derivative of the scalar radius, unlike $\dot{\mathbf{r}}$.

For the Laplace-Runge-Lenz vector, differentiate directly:The vector triple-product identity yieldsThe terms cancel, proving every component of $\mathbf{A}$ is conserved.

Write $q=\bar{r}(r)$ and $\tau =\bar{t}$ for a monotone radial orbit time transformation. The chain rule givesImpose $q^{\prime }=(r/q)\dot{r}$, equivalentlyMultiplying the radial energy equation by $r^2/q^2$ then givesThis is a transformation along each trajectory; the new clock is obtained by integrating its radius-dependent rate.

For the Kepler–harmonic radial duality, take the old inverse-radius term as the new constant energy, $\bar{E}=\mu r/q^2$. On a positive-radius branch $\bar{E}>0$, soThe remaining term is minus the new Newtonian gravitational potential:For $E<0$ this is an attractive isotropic harmonic oscillator with ${\bar{\omega }}^2=-2E(\bar{E}/\mu {)}^2$. For $E>0$ it is an inverted oscillator, and for $E=0$ the Newtonian gravitational potential vanishes. Calling it a confining oscillator therefore requires a bound original Kepler orbit.

For the angular reconstruction of a radial orbit transformation, choose the new angle by $d\bar{\theta }/d\tau =L/q^2$. The same $L$ then supplies its centrifugal term. From $d\theta /dt=L/r^2$, the transformation is $d\bar{\theta }/d\theta =(r/q)(dq/dr)$; in the harmonic case this gives $\bar{\theta }=\theta /2$ up to a constant. Keeping the old angle unchanged would not in general preserve the angular momentum used in the displayed transformed radial equation.

For the second transformation use the explicit displayed split of the energy terms. It assigns weight $\lambda$ to $E$ and $1-\lambda$ to $\mu /r$; the surrounding printed prose interchanges these weights. Define the transformed quantities consistently byTheir difference is exactly $(r^2/q^2)(E+\mu /r)$, so they give the same transformed radial energy equation.

Let $a=(1-\lambda )\mu$, $h=\lambda E$, and $\epsilon =\bar{E}$. Constancy of the new energy imposesFor the standard regular isochrone branch take $h>0$, $\epsilon >0$, and $a\ge 0$. The positive-radius root isFor $q>0$ it is monotone. The corresponding clock is also regular there:With $\alpha =\sqrt{\epsilon /h}$ and $s=\sqrt{q^2+b^2}$, rationalization gives $r/q^2=\alpha /(s+b)$. Eliminate $r^2/q^2$ using $h{r}^2/q^2=\epsilon -ar/q^2$:where one possible identification isThis is the Kepler–isochrone radial transformation to the spherical isochrone model, up to the additive energy constant $k$.

The attractive branch of the Kepler–isochrone transformation requires a suitable parameter domain. The displayed square root needs $\lambda E\bar{E}>0$, and the positive-root branch used above needs $\lambda E>0$, $\bar{E}>0$. An attractive regular isochrone also needs $b\ge 0$ and $\bar{M}>0$. For bound original orbits a convenient domain is $E<0$, $\lambda <0$, $\bar{E}>0$: it satisfies all these requirements and gives $\bar{E}-k=\bar{E}/\lambda <0$. For $E>0$ one may instead take $1/2<\lambda \le 1$, with $\bar{E}>0$; these map to unbound attractive isochrone orbits. Other parameter choices can produce a repulsive Newtonian gravitational potential or a different branch, so no unrestricted real-parameter statement is justified. The singular algebraic case $\lambda =0$ is the harmonic transformation already handled separately. The angular reconstruction $d\bar{\theta }/d\tau =L/q^2$ likewise completes the isochrone radial solution into a central-force orbit.

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $\mathbf{v}=\dot{\mathbf{r}}$. Relative to the centre of mass, the two positions are ${\mathbf{r}}_1=-(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=(M_1/M)\mathbf{r}$. Their velocities have the same mass factors. Adding their kinetic energies and angular momenta gives, with reduced mass $\mu =M_1{M}_2/M$,These are the internal orbital quantities, excluding any uniform centre-of-mass motion. Newton's law of universal gravitation givesFor constant masses, differentiating $E$ makes the gravitational work cancel the derivative of the potential energy, so $\dot{E}=0$. Also $\dot{\mathbf{h}}=\mathbf{r}\times \ddot{\mathbf{r}}=0$, since the force is central; hence conservation of angular momentum gives $\dot{\mathbf{J}}=0$.

The eccentricity vector, the normalized attractive Laplace-Runge-Lenz vector, isUsing $\dot{\hat{\mathbf{r}}}=\mathbf{v}/r-\mathbf{r}(\mathbf{r}\cdot \mathbf{v})/r^3$ andshows that its derivative also vanishes. Adding a relative perturbing acceleration $\mathbf{f}$ changes these cancellations only by the perturbing terms. The perturbed Kepler-orbit conservation laws are therefore

For the distant Newtonian gravitational potential, expand each source term about the binary's centre of mass:The dipole vanishes because ${\sum }_i{M}_i{\mathbf{r}}_i=0$. With $\mathbf{a}=\mathbf{r}$, the quadratic mass moment is ${\sum }_i{M}_i{\mathbf{r}}_i{\mathbf{r}}_i=\mu \mathbf{aa}$, so the quadrupole potential of a circular binary isOn the inner binary's fast timescale, the distant companion's position is approximately fixed. Averaging over the inner circular orbit in the common plane gives $⟨{\cos }^2\theta ⟩=1/2$ and henceThis averaged potential function is independent of both time and azimuth. Consequently the companion's energy and axial angular momentum are conserved in the averaged problem. The exact motion has small fast variations; the statement is a secular approximation for a hierarchical system away from an orbital resonance.

For a circular outer orbit, force balance givesso the circumbinary orbital-period correction isConservative binary mass transfer leaves $M$ unchanged, so it does not change the leading monopole force. It changes $\mu$ and $a$, and therefore the quadrupole correction. If the inner orbital angular momentum is conserved, $J_{\mathrm{in}}=\mu \sqrt{GMa}$ gives $a\propto {\mu }^{-2}$ and $Q=\mu a^2\propto {\mu }^{-3}$. The correction is thus largest as the binary becomes strongly unequal in mass and widens, within the hierarchy $a\ll x$.

One must also allow the outer radius to respond. For slow axisymmetric evolution its specific angular momentum $h_3$ remains constant, while its energy need not: the averaged potential now depends on time. A nearly circular outer orbit has $h_3^2=GMx+3GQ/(4x)$ and $P_3=2\pi x^2/h_3$. At fixed $M,h_3$, differentiation givesThus mass exchange inside the binary can produce a small measurable outer-period change even without total mass loss. Nonconservative mass loss additionally changes the leading monopole term.

The repulsive force is $\mathbf{F}=k\mathbf{r}/r^3=-\nabla (k/r)$, so the Hamiltonian isThe Hamilton's equations give $\dot{\mathbf{r}}=\mathbf{p}/m$ and $\dot{\mathbf{p}}=k\mathbf{r}/r^3$. Consequentlyso the angular momentum is conserved. Moreover,It follows that the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}+mk\hat{\mathbf{r}}$ is conserved. By the Hamiltonian conservation law, these time derivatives are exactly $\{\mathbf{L},H\}$ and $\{\mathbf{A},H\}$, so both Poisson brackets vanish.

The integrals of motion are the energy $H$, the components of $\mathbf{L}$, and the components of $\mathbf{A}$, subject toTaking the dot product of $\mathbf{A}$ with $\hat{\mathbf{r}}$ givesTherefore the polar equation of a repulsive inverse-square orbit iswhere $\lambda =L^2/(mk)\ge 0$ and $e=A/(mk)\ge 0$. For a nonradial physical scattering orbit $H>0$, the identity above gives $e>1$, as expected for a hyperbolic Kepler orbit.

Taking the dot product of the Laplace-Runge-Lenz vector with $\hat{\mathbf{r}}$ gives $A\cos \theta =L^2/r+mk$. Thus

A superintegrable Hamiltonian system with $n$ degrees of freedom has more than $n$ functionally independent integrals of motion, including an integrable set of $n$ in involution. At a regular nonzero Hamiltonian vector field, at most $2n-1$ independent integral differentials can annihilate the flow. A maximally superintegrable system attains that bound. The three-dimensional Kepler orbit has five independent integrals through energy, angular momentum and the Laplace-Runge-Lenz vector. A closed orbit on one isolated resonance is weaker than maximal superintegrability on an open phase-space region.