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Write $h=r^2\dot{\theta }$ for the conserved specific angular momentum. Zero specific orbital energy in the inverse-square Newtonian gravitational field givesDefine $r_0=h^2/(2GM)$, the periapsis radius for a nonradial parabolic trajectory. ThenOn the outgoing branch, integrate from the periapsis time $t_0$ and put $s=\sqrt{r-r_0}$:The incoming branch has the opposite sign. Squaring gives the radius-time relation for a parabolic Kepler orbit, a version of the Barker equation:Here $r_0$ and $t_0$ are fixed by angular momentum and the time origin at periapsis. The relation applies on both branches, with $r\ge r_0$; it is not necessary to solve for the polar angle.

Let $M_{\ast }$ be the central mass, so ${\Omega }^2=G{M}_{\ast }/r^3$ and the specific orbital energy is $\epsilon =-G{M}_{\ast }/(2r)$. Define the inward accretion rate $\dot{M}>0$. The rate at which the gas acquires binding energy in moving from $b$ to $a$ isThe vertically integrated viscous heating per unit horizontal area, including the entire column, is $\bar{\nu }\Sigma (r{\Omega }^{\prime }{)}^2$. Since $r{\Omega }^{\prime }=-3\Omega /2$, the dissipated power in the annulus isHere the column integral already includes dissipation associated with both disk faces; adding another factor of two would double-count it. Performing the elementary integrals givesFor the whole disk, take $a=r_{\mathrm{in}}$ and $b\to \infty$. Both expressions giveIf $a\gg r_{\mathrm{in}}$, the bracket $1-\sqrt{r_{\mathrm{in}}/r}$ is uniformly close to one throughout the annulus. Comparing the positive integrals showswith the correction negative. Thus far from the inner edge the viscous dissipation is approximately three times the local binding-energy acquisition.

There is no violation of conservation of energy because the viscous torque in an accretion disk also transports mechanical energy. The outward viscous torque in an accretion disk isand its outward energy flux is $\Omega \mathcal{G}=G{M}_{\ast }\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$. The explicit expressions satisfyEnergy transported from the inner disk supplies the extra dissipation at larger radii. The torque power vanishes at the zero-torque inner boundary condition and tends to zero at infinity, leaving global balance. This is torque transport of energy in a steady accretion disk.

Write the gravitational potential with spherical symmetry as $\Phi (r)$ and let $\mathbf{n}$ be the unit vector normal to a circular orbit, oriented in the direction of its angular velocity. Put $\mathbf{\Omega }=\Omega \mathbf{n}$ and denote the specific angular momentum by $\mathbf{h}=h\mathbf{n}$. Circular force balance and the specific orbital energy giveDifferentiating along the family of circular orbits,A change in orbital orientation contributes $hd\mathbf{n}$ to $d\mathbf{h}$, but $\mathbf{n}\cdot d\mathbf{n}=0$. ConsequentlyThis circular-orbit energy gradient accounts for radial and inclination changes together; spherical symmetry makes the orbital energy independent of the plane.

For either variable-mass particle, $\mathbf{H}=M\mathbf{h}$ and $E=M\epsilon$. Thus $d\mathbf{H}=Md\mathbf{h}+\mathbf{h}dM$, soConservation of mass and conservation of angular momentum impose $d{M}_2=-d{M}_1$, $d{\mathbf{H}}_2=-d{\mathbf{H}}_1$. Adding the two particle contributions provesTo interpret the signs, label the inner orbit 1 and the outer orbit 2. For $f(r)=\epsilon -\Omega h$ the first identity gives $f^{\prime }=-h{\Omega }^{\prime }$. The assumed outward decrease of angular velocity therefore makes $f_1<f_2$, so the mass-exchange part lowers the energy when $d{M}_1>0$: mass is transferred inward. In a coplanar, co-rotating pair, ${\Omega }_1>{\Omega }_2$ and an outward angular momentum transfer has $d{\mathbf{H}}_1=-dH\mathbf{n}$ with $dH>0$. Its contribution is $-({\Omega }_1-{\Omega }_2)dH<0$.

The vector difference also identifies the inclination-reducing exchange. For a fixed small magnitude of admissible angular momentum exchange, its energy contribution is most negative whenby the Cauchy-Schwarz inequality. This gives $d{E}_{\mathrm{angular}}=-s|{\mathbf{\Omega }}_1-{\mathbf{\Omega }}_2{|}^2$. Let $q={\mathbf{n}}_1\cdot {\mathbf{n}}_2=\cos \iota$ and $H_i=|{\mathbf{H}}_i|$, taking fixed particle masses for this exchange. Projecting $d{\mathbf{H}}_i$ perpendicular to its own direction givesThereforeThe downhill exchange reduces the relative orbital inclination. Its inner angular momentum component is ${\mathbf{n}}_1\cdot d{\mathbf{H}}_1=-s({\Omega }_1-{\Omega }_{2}q)<0$; in the aligned limit the outer orbit receives that angular momentum directly. For strongly inclined orbits, part of the exchange instead cancels differently directed angular momenta as the planes align. Together these gradients explain the tendencies toward inward mass transfer, outward angular momentum transport and coplanarity.

The marginal circular orbit has conserved specific angular momentum and specific orbital energyAt fixed $j_0$, the derivative of the effective potential factors asAn inward displacement therefore produces inward radial acceleration; no inner circular equilibrium or centrifugal barrier halts the plunge. In the infinitesimal-displacement limit, energy conservation givesSelecting the inward branch and the original positive sense of rotation yields the Paczyński-Wiita ballistic plungeThe azimuthal speed remains positive while the radius decreases, so the trajectory spirals inward. From any fixed starting radius below $3a$, the time to reach $a$ is finite: $1/|v_r|$ behaves as $(r-a{)}^{1/2}$ near $a$, which is integrable.

The exactly marginal circular state with exactly zero displacement remains a circular solution. A finite inward release from rest changes the energy by a term cubic in the displacement; the displayed expression is its limiting separatrix with $E=E_0$. On that separatrix $|v_r|\propto (3a-r{)}^{3/2}$ near $3a$, so departure from exactly $3a$ takes an infinite limiting time. This distinguishes the infinitesimal-release formula from spontaneous motion of the unperturbed marginal orbit.

One integral can occur in an autonomous system with a fully chaotic region and no additional continuous orbit constraints. The specific orbital energy is conserved in a time-independent gravitational potential. If a chaotic stellar orbit is dense in a connected region of its energy hypersurface, every continuous integral of motion is constant on that region. Thus it cannot supply a new independent constraint there, and energy is the only independent isolating integral. A generic nonaxisymmetric potential can contain such regions; not every orbit in such a potential is necessarily chaotic.

Zero stationary isolating integrals are possible when time dependence removes conservation of energy and no other symmetry or invariant survives. For instance, a generic time-dependent nonsymmetric gravitational potential need not have a stationary integral. This is not possible in an ordinary time-independent galactic gravitational potential: $E=\frac{1}{2}{v}^2+\Phi$ is already an isolating integral on a regular specific orbital energy surface. An explicitly time-dependent constant that encodes an initial condition does not contradict the stationary count.

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).

More than three occur when symmetries or frequency degeneracies reduce the orbit's invariant closure below a generic three-dimensional invariant torus. For a nonradial orbit in a spherical gravitational potential, specific orbital energy and the three components of specific angular momentum are generically four independent isolating integrals. Their values fix the orbital plane, angular momentum magnitude and radial specific orbital energy, but a nonclosed rosette still fills a two-dimensional invariant torus. Thus more than three does not by itself imply closure. Orbital resonances can further reduce the number of independent orbital phases. The count concerns independent integrals, not arbitrarily many functions of the same invariants.

One isolating integral occurs when specific orbital energy is the only independent stationary invariant. An autonomous chaotic orbit that densely explores a connected five-dimensional specific orbital energy surface is the standard case. Every continuous integral of motion must then be constant on that surface, so it is locally a function of specific orbital energy rather than an additional independent integral. Irregular motion need not mean zero integrals; specific orbital energy still isolates the orbit from other specific orbital energy surfaces.

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 ${\mu }_g=\gamma m$ for the gravitational parameter of the relative Kepler orbit, and use specific orbital energy and momenta per unit reduced mass. The Hamiltonian isIf physical momenta rather than these specific momenta are used, multiply the actions by the reduced mass. This only introduces a constant factor into the phase space measure and leaves the eccentricity fractions unchanged. The formulas in the question adopt the specific-action convention.

Here is an explicit local canonical transformation. A separated complete integral of the Hamilton-Jacobi equation isChoose branches along an orbit and take $p_j=\partial S/\partial q_j$, $l=\partial S/\partial L$, $g=\partial S/\partial G$, $h=\partial S/\partial H$. Thenwhose exterior derivative proves preservation of the canonical symplectic form. Thus these variables are canonical, and the absolute Jacobian determinant is one. The angular separation gives $G=|\mathbf{r}\times \mathbf{v}|$ and $H=G\cos i$, including the sign of the angular momentum along the reference axis.

For clarity, the radial action can be evaluated rather than assumed. Let $r=a(1-e\cos u)$, with $u$ the eccentric anomaly. Then $p_r=\sqrt{{\mu }_g/a}e\sin u/(1-e\cos u)$, andIn the integral, use $e^2{\sin }^{2}u/(1-e\cos u)=1+e\cos u-(1-e^2)/(1-e\cos u)$ and ${\int }_0^{2\pi }(1-e\cos u{)}^{-1}du=2\pi /\sqrt{1-e^2}$. The polar action is $J_{\theta }=G-|H|$, while the signed azimuthal action is $J_{\varphi }=H$. The separated angular coordinates have the usual Delaunay variables interpretation: $g$ is the argument of periapsis, and $h$ is the longitude of ascending node. In particular,with a suitable origin at periapsis, so $l$ is the mean anomaly. We have therefore recoverednot merely assigned names to the new coordinates.

For nondegenerate bound Kepler orbits, the ranges areThe boundaries describe circular, coplanar or radial degeneracies, where some angles cease to be defined; they have zero measure in the present calculation. The actions are canonical coordinates with nested bounds, rather than three independently unrestricted Cartesian coordinates. The square of $H$ alone would miss the distinction between prograde and retrograde inclinations.

By the canonical transformation, the population measure becomesAt fixed $L$, the condition $e^{\prime }<e$ is $L\sqrt{1-e^2}<G<L$. Integrating all three angles and then $H$ givesConsequently the Delaunay phase-volume proof of the thermal eccentricity distribution yieldsThe PDF's $e^2$ dependence must be interpreted as a cumulative distribution function, not an eccentricity probability density function or the value of the original six-dimensional phase-space distribution function. A finite value of $F(E)$ at every energy also does not suffice to make the total population finite: $F=1$ on all bound energies gives the divergent integral ${\int }_0^{\infty }{L}^{2}dL$. The fractions are well defined at each fixed energy, over a finite allowed energy range, or for an energy distribution with a finite total integral.

A Boltzmann distribution is therefore unnecessary. Any normalizable energy-only distribution has the same eccentricity marginal, although such a distribution is a collisionless steady state by Jeans theorem. The marginal alone does not even prove stationarity: for $0<|\eta |<1$,is a positive time-dependent solution of the Collisionless Boltzmann equation and has exactly the same eccentricity marginal after integration over $l$. An observed $e^2$ cumulative law establishes neither thermal relaxation nor the age of the system or Universe.

For an equatorial circular orbit in the Paczyński-Wiita potential, radial force balance givesIts specific angular momentum is $j=R^2\Omega =\sqrt{GM}{R}^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the radial epicyclic frequency squared ${\kappa }^2=R^{-3}d(j^2)/dR$. ThereforeCircular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$, joining a plunging region of a black-hole accretion disk inside the innermost stable circular orbit. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.

Take the inward mass accretion rate $\dot{M}$ as positive, and write the positive outward viscous torque in an accretion disk asSteady conservation of angular momentum makes the net inward flux $C=\dot{M}j-\mathcal{G}$ independent of radius. Impose a zero-torque inner boundary condition at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. ThenThis is the steady accretion disk with arbitrary rotation law angular-momentum balance.

The local viscous dissipation integrated through the full disk thickness is $D=\nu \Sigma (R{\Omega }^{\prime }{)}^2=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Sincesubstitution gives the two-face dissipation of a Paczyński-Wiita disk:In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total accretion luminosity is therefore $L={\int }_{3R_S}^{\infty }2\pi RDdR$, with no additional factor of two.

For a direct evaluation put $x=R/R_S$. ThenThe first integral is $7/4$, using $(3x-1)/(x-1{)}^3=3/(x-1{)}^2+2/(x-1{)}^3$. For the second,so it equals $2/[\sqrt{3}(3-1)]=1/\sqrt{3}$. HenceThis is the radiative efficiency of black-hole accretion within this pseudo-Newtonian model.

Independently, the specific orbital energy of a circular particle isThus $E(\infty )=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated accretion luminosity. The Paczyński-Wiita circular orbit reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.

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.