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In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

For a central gravitational potential and nonzero specific angular momentum $L$, the positive $L^2/(2r^2)$ contribution to the radial effective potential can prevent the orbit reaching the centre at finite energy. It dominates the attractive $-\mu /r$ term at small $r$ in a Kepler orbit. A more singular attractive term may overcome it: for $\Phi =-\mu /r-\mu a/r^2$, the net inverse-square coefficient is $(L^2-2\mu a)/2$, so a central barrier exists only when that coefficient is positive.

For $V(r)=km{r}^2$, fixed nonzero angular momentum $h$ gives the effective potential $V_{\mathrm{eff}}=h^2/(2m{r}^2)+km{r}^2$. A positive-radius circular orbit exists when $k>0$, at the displayed radius, and $V_{\mathrm{eff}}^{\prime \prime }(r_0)=8km>0$ proves radial stability by the effective potential stability criterion. Its angular speed has magnitude $\sqrt{2k}$ and its small radial oscillation frequency is $\sqrt{8k}$. If $k\le 0$, no such orbit with nonzero angular momentum exists; for $h=0,k>0$, the sole minimum is at the origin, not a positive-radius circular orbit.

For unit mass with specific angular momentum $l$ in $V(r)=-e^{-r}$, the effective potential is $-e^{-r}+l^2/(2r^2)$. A positive-radius circular orbit satisfies the displayed equation. Since $r^3{e}^{-r}$ has its unique maximum $27/e^3$ at $r=3$, exactly two radii occur for $0<l^2<27/e^3$. Their radial stability follows from $V_{\mathrm{eff}}^{\prime \prime }=e^{-r}(3/r-1)$: the inner orbit is stable at fixed angular momentum, the outer unstable. The case $l=0$ has no finite-radius circular orbit despite satisfying the upper bound.

A stationary time-coordinate change $t^{\prime }=t+F(r)$ leaves both the areal radius and the time-translation Killing vector field unchanged. The conserved Killing energy and angular momentum of a null geodesic therefore yield the same radial equation in either chart, ${\dot{r}}^2+h^2(1-2\mu /r)/r^2=c^2{k}^2$. The effective potential and radial turning points agree, even when the coordinate time velocities and horizon regularity differ.

For a small, rapid horizontal displacement $a\cos \omega t$, the leading fast angular excursion is $\xi =-(a/l)\cos \Theta \cos \omega t$. Correlating that excursion with the driving torque gives the averaged force $\ddot{\Theta }=-(g/l)\sin \Theta +(a^2{\omega }^2/2l^2)\sin \Theta \cos \Theta$. Its effective potential is $-mgl\cos \Theta +m{a}^2{\omega }^2{\cos }^2\Theta /4$. The downward state loses linear stability above $a\omega =\sqrt{2gl}$ and two tilted minima appear. Small $a/l$ and slow dynamics compared with $\omega$ are required; horizontal and vertical support vibration produce different effective potentials.

In a flattened power-law gravitational potential $\Phi ={\Phi }_0{R}_0^{\beta }(R^2+{\lambda }^2{z}^2{)}^{-\beta /2}$, a magnetic surface corotating with its disk footpoint has an effective potential with meridional Hessian matrix ${\Omega }_f^2\mathrm{diag}[-(\beta +2),{\lambda }^2]$ there. A cold outward displacement along a field inclined by $\alpha$ to the vertical is downhill when ${\tan }^2\alpha >{\lambda }^2/(\beta +2)$. Equality is marginal and can depend on higher derivatives and field-line curvature; this is a local launching criterion, not a guarantee of a global escaping wind.

At exactly thirty degrees the quadratic curvature vanishes, but the specified straight-line point-mass effective potential is downhill for small positive displacement at cubic order. Thus the strict negative-curvature criterion does not exclude every nonlinear one-sided marginal launch. An exactly stationary bead remains at equilibrium until displaced.

Adding an inverse-cube force can preserve radial motion while rescaling angular motion. For potential $-A/r+B/r^2$ and angular momentum $L$, the radial effective potential equals that of $-A/r$ with effective angular momentum $\sqrt{L^2+2mB}$ when this is real.

For an equatorial null geodesic in Schwarzschild spacetime, let $k=(1-2\mu /r)dt/d\lambda$ and $h=r^{2}d\varphi /d\lambda$ be the conserved quantities for an affine parameter $\lambda$. The null metric condition gives ${\dot{r}}^2/2+V_{\mathrm{eff}}=c^2{k}^2/2$. For $h\ne 0$, the effective potential is zero at the Schwarzschild event horizon, positive outside, and tends to zero at infinity. Its unique exterior maximum is at $r=3\mu$, with value $h^2/(54{\mu }^2)$ and second derivative $-h^2/(81{\mu }^4)$. This maximum gives the unstable photon sphere.

The impact parameter of an incoming ray is $b=|h|/(ck)$ for $k>0$. Capture requires its radial energy to exceed the maximum, namely $b<3\sqrt{3}\mu$. At equality the incoming ray approaches the circular orbit asymptotically; it does not cross the horizon. For $h=0$, the potential is identically zero and the ingoing radial ray is captured.

A particle of mass $m$ constrained to the cone $z=-r$ is joined through the apex by a light inextensible ideal string to a hanging particle of mass $\sqrt{2}m$. The cylindrical radius is $r>0$ and gravitational acceleration is $g$. If the total ideal string length is $s_0$, the second particle has vertical coordinate $z_2=-s_0+\sqrt{2}r$. The ideal string and normal reaction exert no azimuthal force, so conservation of angular momentum gives $r^2\dot{\theta }=h$.

The first particle has kinetic energy $m{\dot{r}}^2+m{h}^2/(2r^2)$, while the second has kinetic energy $\sqrt{2}m{\dot{r}}^2$. Their combined gravitational potential energy is $mgr$ plus a constant. Hence conservation of energy gives the displayed equation, with effective potential $U(r)=gr+h^2/(2r^2)$. For $h\ne 0$, $U$ is strictly convex and has its unique minimum at $r=(h^2/g{)}^{1/3}$. The energy level has two radial turning points unless the motion is a circular orbit at that minimum.

The model maintains a mechanical constraint on the surface. For unilateral contact with one side of the cone, a normal reaction computed with a fixed outward sign must also have the physically allowed sign. A finite length of ideal string additionally restricts $r<s_0/\sqrt{2}$ before the hanging particle reaches the hole.

For the radial epicyclic frequency, introduce the effective potential at fixed specific angular momentum $l$: ${\Phi }_{\mathrm{eff}}=\Phi +l^2/(2r^2)$. A circular orbit at $r_c$ satisfies ${\Phi }^{\prime }(r_c)=l^2/r_c^3=r_c{\Omega }^2(r_c)$. A small radial displacement $\xi$ obeys $\ddot{\xi }=-{\Phi }_{\mathrm{eff}}^{\prime \prime }(r_c)\xi$. Differentiate at fixed $l$ before inserting the circular-orbit relation:For a flat rotation curve, $\Omega \propto r^{-1}$, so $\kappa =\sqrt{2}\Omega$.

In a perturbation proportional to $\exp [i(\int kdr+n\varphi -\omega t)]$, $n$ is the integer azimuthal mode number, giving $|n|$ arms for a nonaxisymmetric pattern. Its angular pattern speed is ${\Omega }_p=\omega /n$ for a real frequency and $n\ne 0$; $a_0$ is the unperturbed gas sound speed, or the specified effective fluid pressure-response speed. For growing patterns the angular phase speed uses $\mathrm{Re}\omega /n$. Axisymmetric modes have $n=0$: their pattern speed is undefined, and one must use $(\omega -n\Omega {)}^2\to {\omega }^2$, not set the pattern-speed expression to zero.

Define the Toomre parameter and Toomre wavenumber byThen $q=|k|/k_T$ givesFor $a_0>0$, the minimum is negative precisely when $Q<1$. Thus this condition is necessary and sufficient for an exponentially growing axisymmetric mode within the given local razor-thin fluid model; $Q=1$ is marginal. It is not a guarantee against global or nonaxisymmetric instabilities, and the collisionless stellar-disk criterion has a different pressure-response normalization.

The Toomre unstable band and fastest growth follow by minimizing ${\omega }^2$:Here the growing branch has $\omega =i{\gamma }_{\ast }$ under the stated Fourier convention. Both signs of the radial wave number have the same growth rate. The unstable band is $q_{-}<q<q_{+}$, where $q_{\pm }=2(1\pm \sqrt{1-Q^2})/Q^2$.

For the minimum Toomre parameter in an exponential flat-rotation disk with constant $a_0$, put $x=r/r_0$:The logarithmic derivative is $d\log Q/dx=1-1/x$, so the unique minimum occurs at $r=r_0$. Both $Q\to \infty$ as $r\downarrow 0$ and $Q\to \infty$ as $r\to \infty$. Hence inner and outer regions are stable within this idealized rotation law, even when an intermediate annulus is unstable. The whole disk is strictly stable in this local axisymmetric test ifEquality makes the ring at $r_0$ marginal. The central conclusion uses the assumed flat rotation curve all the way inward; real central rotation laws need not retain that limit.

Let $f(r)=1-2m/r$ and work in the exterior $r>2m$, with $G=c=1$. The affinely parametrized geodesic Lagrangian isIts polar-angle equation isThe initial data $\theta =\pi /2$, $\dot{\theta }=0$ therefore remain in that plane. Rotational symmetry allows the axes to be chosen so that any nonradial initial trajectory has those data: its conserved angular-momentum direction fixes an orbital plane through the centre. Radial trajectories can also be placed in an equatorial plane.

The coordinates $t,\varphi$ are cyclic in $L$, so their canonical momenta are constant:The corresponding Killing vector fields are ${\partial }_t$ and ${\partial }_{\varphi }$. With the displayed signature, $g({\partial }_t,\dot{x})=E$ and $g({\partial }_{\varphi },\dot{x})=-J$, explaining the sign in the second canonical momentum. Thus $E$ is conserved energy per unit rest mass for a massive particle parametrized by proper time, and $J$ is its axial angular momentum per unit rest mass. For a photon, affine rescaling rescales both constants; after the tangent is normalized to the physical four-momentum they represent its energy at infinity and axial angular momentum. Their ratio $b=J/E$ is the signed impact parameter.

In the equatorial plane, the photon satisfies the null geodesic constraintwhich givesFor $J\ne 0$, put $u=m/r$ and $\alpha =Em/J$. Thenand consequentlyThe $J=0$ radial case is not represented by $\varphi$ as an orbit parameter.

There is a circular photon orbit at $r=3m$, but it is unstable. The radial effective potential isA circular orbit requires $V^{\prime }(r)=0$ and $E^2=V(r)$, givingIts second derivative isIt is a maximum, rather than a minimum, so radial perturbations depart from the orbit. Equivalently the second-order orbit equation is $u^{\prime \prime }=3u^2-u$. Linearizing about $u_c=1/3$ gives $\delta u^{\prime \prime }=\delta u$, with modes $e^{\varphi }$ and $e^{-\varphi }$. The existence of a growing mode directly contradicts the printed stability claim. Exact initial data can remain circular, but that does not make the orbit stable.

The supplied implicit expression describes the critical-energy family with ${\alpha }^2=1/27$, as fixed by the circular orbit. DefineRationalizing once gives $F=(\sqrt{3}-s)/(2(\sqrt{3}+s))$ andDifferentiating $F(u)=A{e}^{\varphi }$ gives $F^{\prime }{u}^{\prime }=F$, soThis verifies the first-order equation for every allowed integration constant $A$ on its real branch. For $A=0$ the derivative argument is replaced by direct substitution: $u=1/3$ is constant and solves the critical equation. The displayed family does not solve the equation at an independently arbitrary value of $\alpha$.

For $A>0$, write $q=A{e}^{\varphi }$. On the physical exterior branch,As $\varphi \to -\infty$, $q\to 0^{+}$ andThus for $A=0$ the photon remains exactly on the circular orbit. For $A>0$ it approaches $r=3m$ from outside through indefinitely many revolutions in the negative-$\varphi$ limit. With increasing $\varphi$, this same branch peels away outward and reaches infinity when $q=1-\sqrt{3}/2$. Reversing the direction of motion gives an incoming critical ray asymptotically approaching the photon circle. These are separatrix trajectories, not evidence of stability.

The potential maximum and the outward-growing radial separation display the same instability in two complementary descriptions.

The assertion about every exterior radius is false for geodesics. A circular orbit here means an equatorial null geodesic at constant $r$, rather than a genuinely radial trajectory. A zero of the radial effective potential alone gives a turning point, not necessarily a circular orbit. The radial geodesic equation must hold as well.

Write $q=L-aE$ andFor an exterior circular null geodesic, the two conditions are $f(r)=f^{\prime }(r)=0$, equivalently $\mathcal{R}(r)={\mathcal{R}}^{\prime }(r)=0$ with $E,L$ fixed. One way to justify the derivative condition without dividing by $\dot{r}=0$ is to use the radial Euler-Lagrange equation. At constant $r$ it givesTogether with $g_{AB}{\dot{x}}^A{\dot{x}}^B=0$, and the fixed conserved $p_A$, this is precisely the double-root condition for $\mathcal{R}$.

NowSetting this to zero and using $f=0$ givesA nontrivial exterior null geodesic cannot have $E=0$ here: the second equation would force $q=L=0$, and then every tangent component would vanish. ThereforeBoth equations, with the same constants $E,L$, are required. They give the requested radius relations in terms of the conserved quantities, while showing why an arbitrary exterior radius is not permitted.

For a more explicit classification let $b=L/E$ be the impact parameter. The second relation gives $b-a=s{r}^{3/2}/\sqrt{M}$ with $s=\pm 1$. Substituting into the first relation yieldsFor $0<a<M$, $s=+1$ gives the prograde equatorial circular Kerr photon orbit and $s=-1$ the retrograde one. The two allowed exterior radii areBoth are strictly greater than $r_{+}$ in the subextremal range. Squaring the signed radius equation gives $r(r-3M{)}^2=4M{a}^2$, but introduces an additional inner root and loses the sign distinguishing the two senses of rotation. At $a=0$ the two exterior radii coincide at the photon sphere $r=3M$, with $b=\pm 3\sqrt{3}M$.

For a direct counterexample to the all-radii claim take $a=0$ and $r_0=4M>r_{+}=2M$. Choose $E=1$ and $L=4\sqrt{2}M$ so that $f(r_0)=0$. The radial geodesic equation nevertheless givesIt is a turning point, not a circular orbit. Alternatively, constant-$r$ null directions exist throughout the exterior because the two angular-velocity roots of $g_{tt}+2\Omega g_{t\varphi }+{\Omega }^2{g}_{\varphi \varphi }=0$ are real; almost all of their integral curves require acceleration and fail the radial geodesic equation. This distinction explains how an all-radii assertion can arise.

Finally, $r=r_{+}$ itself has the null geodesic generators of its Killing horizon, proportional to ${\chi }_{+}={\partial }_v+{\Omega }_{+}{\partial }_{\psi }$. They are a separate horizon case. On the horizon ${\chi }_{+}$ is normal to every tangent, so its contractions with ${\partial }_v$ and ${\partial }_{\psi }$ both vanish: $E=L=0$ for a generator tangent. Its affine parameter can be chosen by rescaling that tangent. The exterior calculation dividing by $E$ does not include it. The presence of these generators does not supply circular null geodesics at the intervening arbitrary exterior radii.

The unit vectors of plane polar coordinates obey ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$, as follows by differentiating their Cartesian sine/cosine components. Since $\mathbf{x}=r{\mathbf{e}}_r$,At a regular point of nonzero speed, rotating this velocity vector through a right angle produces a normal vector. Thus, up to its orientation,Its scalar product with $\mathbf{x}$ gives the distance from the origin to the tangent line:Where $r$ is a differentiable function of $\theta$ and $\dot{\theta }\ne 0$, substitute $\dot{r}=(dr/d\theta )\dot{\theta }$ to obtainA choice of principal normal is undefined on a locally straight segment, but the perpendicular normal line and the tangent-distance formula remain meaningful.

For a central force, the torque vanishes. Use physical signed angular momentum $h=m{r}^2\dot{\theta }$, with $h\ne 0$. The preceding geometric formula implies $|h|=mpv$, where $v=|\dot{\mathbf{x}}|$. Conservation of energy gives $E=m{v}^2/2+V(r)$. Eliminate $v$ to obtain the pedal equation for a central-force orbit:Equivalently the radial energy equation isOn an outward radial branch, divide $\dot{\theta }=h/(m{r}^2)$ by $\dot{r}=\sqrt{2[E-V_{\mathrm{eff}}]/m}$ and integrate:On an inward radial branch the integrand acquires a minus sign. At a turning point the two branches must be joined; the quoted positive-root integral is a local branch representation, not a single global time-parametrized orbit.

For $V=c/r^2$ with $c>0$, defineA finite moving orbit has $E>0$. In the radial integral, set $s=r_{\min }/r$. Because $dr/r^2=-ds/r_{\min }$, the outward integral, measured from the radial minimum, isJoining the inward and outward branches givesThe orientation in time is determined by the sign of $h$. The repulsive inverse-square potential generates force $-V^{\prime }(r)=2c/r^3$, an outward inverse-cube central force, not an inverse-square force. The orbit approaches infinity along two asymptotic directions, turns once at $r_{\min }$ and escapes again. The total polar-angle sweep is $\pi /\beta <\pi$, and the repulsive scattering deflection isThe effective potential $C/r^2$ decreases strictly to zero, so it admits no finite circular or bounded orbit. The sketch shows this scattering in a repulsive inverse-square potential. In the radial case $h=0$, the formulas dividing by $h$ are inapplicable; the direction is fixed, the turning radius is $\sqrt{c/E}$ and the particle reverses on the same ray.

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.

For a circular mid-plane orbit, radial force balance in the Newtonian gravitational potential is $r{\Omega }^2={\Phi }_{,r}$. HenceAxisymmetry conserves the specific angular momentum $j=r^2\dot{\varphi }$. At fixed $j$, radial motion obeys $\ddot{r}=-{\partial }_r{\Phi }_{\mathrm{eff}}$, where the effective potential is ${\Phi }_{\mathrm{eff}}=\Phi +j^2/(2r^2)$. Linearize around a circular radius $r_0$, writing $r=r_0+\xi$. Since $j^2=r_0^3{\Phi }_{,r}(r_0,0)$ there,This is the radial epicyclic frequency; equivalently ${\kappa }^2=r^{-3}d(r^4{\Omega }^2)/dr$.

Reflection symmetry gives ${\Phi }_{,z}(r,0)=0$ and ${\Phi }_{,rz}(r,0)=0$, so vertical and radial displacements decouple at linear order. The vertical equation is $\ddot{z}=-{\Phi }_{,zz}(r_0,0)z$, giving the vertical epicyclic frequencyPositive squared frequencies give oscillatory stability; a negative squared frequency gives exponential instability rather than a real oscillation frequency.

The imposed constraint now fixes $\dot{\varphi }=\omega$, while a torque allows $h=r^2\omega$ to vary as the particle moves radially. The meridional equations become $\ddot{r}=-{\Phi }_r+{\omega }^{2}r$ and $\ddot{z}=-{\Phi }_z$. Hence $\boxed{{\Phi }_2=\Phi -\frac{1}{2}{\omega }^2{r}^2}$ is the appropriate effective potential. One must not substitute a variable h into the fixed-angular-momentum potential and then differentiate as though h were constant.

For a point mass and nonzero $\omega$, the stationary point is $(r_c,0)$ with $r_c=(GM/{\omega }^2{)}^{1/3}$. Its meridional Hessian matrix isIt is a saddle, at ${\Phi }_2=-3GM/(2r_c)$. Near it the contours are hyperbolas, with the saddle contour locally tangent to $z=\pm \sqrt{3}(r-r_c)$, as shown in the right panel. On the equatorial plane the potential tends downward both toward the point mass and at large cylindrical radius; at fixed r it increases as $|z|$ increases. The radial perturbation obeys $\delta \ddot{r}=3{\omega }^2\delta r$, while the vertical perturbation obeys $\delta \ddot{z}=-{\omega }^2\delta z$. Thus the radial motion has growing and decaying exponential solutions and the vertical motion is oscillatory. For $\omega =0$ no such stationary point exists.

This saddle is the basis of magnetocentrifugal acceleration. A sufficiently strong, open poloidal magnetic field can enforce approximate corotation with its disc footpoint while allowing motion along the field. If its outward tangent makes angle $\vartheta$ with the vertical, a small distance l along it has $\delta r=l\sin \vartheta$, $\delta z=l\cos \vartheta$. Its leading potential change at a Keplerian rotation footpoint isThe thirty-degree magnetocentrifugal launching criterion follows: cold local launching has a downhill quadratic direction when $\boxed{\vartheta >30^{\circ }}$ from the vertical. Equality is marginal at quadratic order and depends on higher terms and field-line geometry. The field transmits the torque needed to increase the material's angular momentum; a freely moving particle has no such constraint and instead sees the stable minimum in part (a). The local criterion does not by itself prove global escape or determine the mass loading of a complete magnetohydrodynamic wind.

For a particle constrained to $z=h(x,y)$, the kinetic energy and potential energy areIn the stipulated gentle-surface small-slope approximation, the second kinetic term is higher order. The leading Euler-Lagrange equations for $T-V$ therefore give the small-slope particle motion on a height graph:This is also the small-slope component of gravity along the surface, with $\sin \varphi \simeq \tan \varphi$. Geometric acceleration terms must remain negligible at the speeds considered; the approximation is not asserted for arbitrarily fast motion on a sharply curved surface.

For $h=\alpha x^2$, the equations are $\ddot{x}+2g\alpha x=0$, $\ddot{y}=0$. If $\alpha >0$, put $\omega =\sqrt{2g\alpha }$. For arbitrary initial data the motion isIt is simple harmonic motion across the trough and uniform motion along it. If $\alpha =0$, both coordinates move uniformly; if $\alpha <0$, the same formula uses $\cosh (\sqrt{-2g\alpha }t)$ and $\sinh (\sqrt{-2g\alpha }t)/\sqrt{-2g\alpha }$ instead.

For the radial height profile, $h(r)=(1-r)/r^2$ is positive for $r<1$, zero at $1$, and negative for $r>1$. Its derivative is $(r-2)/r^3$, so the minimum is $h(2)=-1/4$. It diverges to positive infinity as $r\downarrow 0$ and tends to $0$ from below as $r\to \infty$. The requested profile sketch is shown below.

The equations in polar coordinates in the planar approximation areMultiplying the second equation by $r$ gives $d(r^2\dot{\theta })/dt=0$. Thus the specific angular momentum $L=r^2\dot{\theta }$ is conserved, andAt release with negligible radial velocity, the initial specific kinetic energy is indeedIt is not the total energy. Put $A=L^2+2g$. The conserved specific total energy and effective potential areThe radial turning points for an inverse-square height well follow fromAs a function of $r$, this expression tends to $-\infty$ at zero, has its maximum at $r=A/g$, and tends to $2\mathcal{E}$ at infinity. Only regions where it is nonnegative are dynamically accessible. If $\mathcal{E}<0$, there are two positive turning radii,and the radial coordinate is confined between them. They merge at $r_0=A/g$, the circular orbit. If $\mathcal{E}=0$, the particle escapes to arbitrarily large radius with radial speed tending to zero; if $\mathcal{E}>0$, it escapes with positive limiting radial speed. Consequently the bounded-orbit condition, including the circular case, isThe circular release is at $r_0=2+L^2/g$. For $A/(2g)<r_0<A/g$, the release point is the smaller turning radius; for $r_0>A/g$, it is the larger one. These statements concern the prescribed planar small-slope model. The literal surface is steep near its central singularity, so that approximation must be valid along the particular orbit under consideration.

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.