Stability of Circular Orbits

Every satellite you rely on — GPS, weather, the ISS — lives on a nearly circular orbit that has survived years of small kicks: thruster burns, drag, lunar tugs. That survival is not automatic. A circular orbit exists wherever gravity exactly supplies the needed centripetal force, but existing and persisting are different questions. Nudge the radius slightly: does the orbit breathe gently around the circle, or does the perturbation grow until the body spirals away or crashes inward? Classical mechanics answers with one clean idea. Conservation of angular momentum reduces the motion to a single radial coordinate rolling in the effective potential $V_{\mathrm{eff}}(r)=V(r)+L^2/2\mu r^2$ , and a circular orbit is simply a radius where that landscape is flat. Stability is the familiar valley-versus-hilltop test: if $V_{\mathrm{eff}}''(r_0)>0$ the orbit sits in a valley and small perturbations oscillate harmlessly; negative curvature and they run away. In force language the criterion reads $F'(r_0)/F(r_0)+3/r_0>0$ — for power laws $F=-k\,r^n$ , simply $n>-3$ . Gravity ( $n=-2$ ) and the spring force ( $n=+1$ ) pass with room to spare; any attraction steeper than inverse-cube confines nothing. The same curvature also sets the frequency $\omega_r$ of the radial breathing, and comparing $\omega_r$ to the orbital frequency opens the door to apsidal precession and Bertrand's theorem. Push the analysis into general relativity and the valley itself disappears below a critical radius — the innermost stable circular orbit, the inner edge of black-hole accretion disks.

Key takeaways

A circular orbit exists wherever $V_{\mathrm{eff}}(r)=V(r)+L^2/2\mu r^2$ is stationary: $F(r_0)=-L^2/\mu r_0^3$ .
It is stable only if that stationary point is a minimum: $V_{\mathrm{eff}}''(r_0)>0$ , equivalently $\dfrac{F'(r_0)}{F(r_0)}+\dfrac{3}{r_0}>0$ .
For power-law forces $F=-k\,r^n$ the criterion is $n>-3$ ; the frequency ratio is $\omega_r^2/\omega_\varphi^2=3+n$ (Kepler $n=-2$ : ratio 1; Hooke $n=+1$ : ratio 4).
Stable perturbations oscillate at $\omega_r=\sqrt{V_{\mathrm{eff}}''(r_0)/\mu}$ ; the ratio $\omega_r/\omega_\varphi$ controls precession and closure (Bertrand).
In general relativity the effective potential loses its minimum below $r=6GM/c^2$ — the ISCO, the inner edge of accretion disks.
Key idea: a circular orbit is the equilibrium of the radial coordinate in $V_{\mathrm{eff}}$ ; stability means that equilibrium is a genuine minimum, which for the force reads $F'/F+3/r_0>0$ — power laws need $n>-3$ .

The core insight

A circular orbit is the equilibrium of the radial degree of freedom in the effective potential $V_{\mathrm{eff}}(r)=V(r)+L^2/(2\mu r^2)$ . Stability ⇔ that equilibrium is a minimum ⇔ $V_{\mathrm{eff}}''(r_0)>0$ , which for the force translates into $F'(r_0)/F(r_0)+3/r_0>0$ .

Where this lives today

The ISCO is a working number in black-hole astronomy. The thermal and iron-line X-ray spectra of accreting black holes are fitted with disk models truncated at the innermost stable circular orbit; because a spinning (Kerr) hole drags its ISCO inward from $6GM/c^2$ toward $GM/c^2$ , the fitted inner radius is a standard way of measuring black-hole spin across X-ray binaries and active galactic nuclei. Gravitational-wave astronomy hears the same boundary: an inspiral signal sweeps up in frequency until the orbit reaches the ISCO region and transitions to plunge — the 440 Hz of the worked example is why LIGO's sweet spot sits at hundreds of hertz for stellar-mass pairs. Closer to home, the criterion is bread-and-butter in accelerator and trap design: radial confinement must stay shallower than $1/r^3$ , and the frequency ratio $\omega_r/\omega_\varphi$ — the "tune" — is engineered away from low-order rationals to avoid resonant kicks.

The physical picture

Fix the angular momentum and the problem collapses to one coordinate: the radius, sliding in the landscape $V_{\mathrm{eff}}(r)$ . The centrifugal term walls off small radii; the attraction pulls the landscape down. Whether their sum has a genuine valley — and whether your orbit sits in it — is the entire question:

Figure

Valley or hilltop — the same circular-orbit condition sits at a minimum for gentle forces and a maximum for steep ones

Why is $-3$ the dividing exponent? The centrifugal barrier contributes an outward force scaling as $1/r^3$ . An attraction shallower than that loses the tug-of-war at small radius — the barrier wins and the body bounces back out, so a valley forms. An attraction steeper than $1/r^3$ overwhelms the barrier at small $r$ while fading faster at large $r$ : the curves cross at a hilltop instead, and the balance point repels rather than restores. Once the valley exists, the physics near its floor is a harmonic oscillator: the radius breathes at $\omega_r$ set by the curvature, while the body revolves at $\omega_\varphi$ . The two clocks generally disagree, and their ratio — $\sqrt{3+n}$ for a power law — is the bridge from stability to the apsidal-precession and Bertrand's-theorem articles.

Circular orbits: existence

With $L$ conserved, the radial coordinate obeys Newton's law in the effective potential:

\mu\ddot r = F(r)+\frac{L^2}{\mu r^3} = -\frac{\mathrm{d}V_{\mathrm{eff}}}{\mathrm{d}r},\qquad V_{\mathrm{eff}}(r)=V(r)+\frac{L^2}{2\mu r^2} \qquad (1)

where: $\mu$ — the reduced mass; $F(r)=-\mathrm{d}V/\mathrm{d}r$ — the central force, negative when attractive; $L^2/\mu r^3$ — the outward centrifugal force. The angular motion has been traded for the centrifugal term, so bead-in-landscape intuition now applies literally. A circular orbit is a radius that stays constant forever. Setting the right side of (1) to zero gives the existence condition:

F(r_0)=-\frac{L^2}{\mu r_0^3}\qquad\Longleftrightarrow\qquad V_{\mathrm{eff}}'(r_0)=0 \qquad (2)

where: $r_0$ — the circular-orbit radius. The right side is strictly negative for $L\neq0$ , so the force must be attractive at $r_0$ — a repulsive force has no circular orbits. For gravity, (2) is the familiar $\mu v^2/r_0=GM\mu/r_0^2$ ; existence holds at every radius for any attractive power law — existence is cheap, stability is not.

The stability criterion

Whether the stationary point is a valley or a hilltop is second-derivative business. Derivation A linearizes the radial equation and shows a perturbation $\eta=r-r_0$ obeys $\ddot\eta+\omega_r^2\eta=0$ with $\omega_r^2=V_{\mathrm{eff}}''(r_0)/\mu$ , so everything hinges on the sign of the curvature:

V_{\mathrm{eff}}''(r_0)>0 \qquad\Longleftrightarrow\qquad \frac{F'(r_0)}{F(r_0)}+\frac{3}{r_0}>0 \qquad (3)

where: $F'$ — the radial derivative of the force; the second form follows by differentiating (1) and eliminating $L$ with (2) — Derivation A, Step 4. Read it as a competition of logarithmic slopes: the force may fall with radius, but not faster than the centrifugal force's own $1/r^3$ . If the curvature is negative the "orbit" is a pencil balanced on its tip — an exact solution no real system can hold. For the workhorse case of power-law forces the criterion collapses to a single number. Substituting $F=-k\,r^{n}$ gives $rF'/F=n$ , and with it both the stability condition and the frequency of the radial breathing:

F(r)=-k\,r^{n}:\qquad \frac{\omega_r^2}{\omega_\varphi^2}=3+n\;>0\;\Longleftrightarrow\; n>-3 \qquad (4)

where: $\omega_\varphi=L/\mu r_0^2$ — the orbital angular frequency, with $\omega_\varphi^2=-F(r_0)/\mu r_0$ from (2); $\omega_r$ — the radial oscillation frequency. Gravity ( $n=-2$ ) gives ratio $1$ : the radius breathes exactly once per revolution, which is why a perturbed circular orbit is just a small Kepler ellipse. Hooke ( $n=+1$ ) gives ratio $4$ , i.e. $\omega_r=2\omega_\varphi$ . Convention alert: many texts write $F=-k/r^{p}$ with $p=-n$ , turning the criterion into $p<3$ — same physics, flipped sign. The same machinery runs beyond Newton. General relativity adds an attractive $1/r^4$ -type force term to the Schwarzschild orbit equation, steepening the effective potential at small radius — and below a critical angular momentum the valley vanishes entirely:

Figure

The Schwarzschild effective potential — lower the angular momentum and the valley disappears at the ISCO

Derivation B works this out: the inner hilltop and outer valley merge when $L^2=12\,G^2M^2\mu^2/c^2$ , at radius $r_{\mathrm{ISCO}}=6GM/c^2$ — the innermost stable circular orbit. Inside it, no amount of angular momentum stabilizes a circle; matter spirals in. That single radius sets the inner edge of black-hole accretion disks.

Derivation AStability from linearizing the radial equation

The goal is the master criterion (3): a circular orbit is stable exactly when $V_{\mathrm{eff}}''(r_0)>0$ , with the perturbation oscillating at $\omega_r^2=V_{\mathrm{eff}}''(r_0)/\mu$ , and the equivalent force form $F'/F+3/r_0>0$ . The strategy is the universal recipe for stability of any equilibrium: expand the equation of motion to first order in the displacement and inspect the sign of the restoring coefficient. Watch the last step — converting to the force form divides by $F(r_0)$ , which is negative, flipping the inequality.

Step 1. Perturb the radius. Write $r(t)=r_0+\eta(t)$ with $|\eta|\ll r_0$ , at fixed $L$ . The radial equation (1) becomes $\mu\ddot\eta=-V_{\mathrm{eff}}'(r_0+\eta)$ . Why: $L$ is conserved, so a radial nudge cannot change it — the perturbed motion lives in the same effective potential.

Step 2. Expand to first order. $V_{\mathrm{eff}}'(r_0+\eta)=V_{\mathrm{eff}}'(r_0)+V_{\mathrm{eff}}''(r_0)\,\eta+\mathcal O(\eta^2)$ , and the first term vanishes by the circular-orbit condition (2), leaving $\mu\ddot\eta=-V_{\mathrm{eff}}''(r_0)\,\eta$ . Why: equilibrium kills the zeroth order; the curvature is the leading survivor and acts as a spring constant.

Step 3. Read off the verdict. If $V_{\mathrm{eff}}''(r_0)>0$ , $\eta$ oscillates harmonically at $\omega_r=\sqrt{V_{\mathrm{eff}}''(r_0)/\mu}$ — bounded, stable. If negative, $\eta\propto e^{+|\omega_r|t}$ grows exponentially — unstable. Zero is marginal and needs higher orders. Why: this is the harmonic-oscillator trichotomy; the borderline case is decided by the first nonvanishing higher derivative.

Step 4. Convert to the force form. Differentiating $V_{\mathrm{eff}}'=-F-L^2/\mu r^3$ gives $V_{\mathrm{eff}}''(r_0)=-F'(r_0)+\dfrac{3L^2}{\mu r_0^4}$ , and eliminating $L^2$ with (2): $V_{\mathrm{eff}}''(r_0)=-F'(r_0)-\dfrac{3F(r_0)}{r_0}$ . Requiring this positive and dividing by $-F(r_0)>0$ yields $\dfrac{F'(r_0)}{F(r_0)}+\dfrac{3}{r_0}>0$ — criterion (3). Why: the circular condition trades the orbit-specific $L$ for the force law itself, making the criterion a property of $F$ alone; the division is by a positive number, so the inequality survives intact.

Step 5. Specialize to power laws. For $F=-k\,r^{n}$ , $rF'/F=n$ , so stability is $3+n>0$ , and with $\omega_\varphi^2=-F/\mu r_0$ the ratio is $\omega_r^2/\omega_\varphi^2=3+n$ — equation (4). Why: one exponent now encodes stability ( $n>-3$ ), the breathing rate, and — via the apsidal angle $\pi/\sqrt{3+n}$ — the precession story of the neighbouring articles. *Verified: matches Goldstein §3.3 and Taylor Ch. 8; $n=-2$ gives ratio 1 (perturbed orbit = small Kepler ellipse), $n=+1$ gives $\omega_r=2\omega_\varphi$ , both standard.*

Go deeper · Derivation B · 2 minThe innermost stable circular orbit of Schwarzschild geometry

Derivation BThe innermost stable circular orbit of Schwarzschild geometry

The goal is the most famous stability boundary in physics: $r_{\mathrm{ISCO}}=6GM/c^2$ , below which no stable circular orbit exists around a non-rotating black hole. The strategy reuses this article's toolkit unchanged — find where $V_{\mathrm{eff}}'=0$ has solutions, then ask when the valley survives — applied to the Schwarzschild effective potential, which differs from Newton's by one extra attractive $1/r^3$ term (we keep Newtonian-style notation). Watch the structure: the circular-orbit condition becomes a quadratic in $r$ , so orbits come in pairs — stable outer, unstable inner — and the ISCO is where the pair merges.

Step 1. Write the effective potential. For Schwarzschild geodesics the radial problem has $V_{\mathrm{eff}}(r)=-\dfrac{GM\mu}{r}+\dfrac{L^2}{2\mu r^2}-\dfrac{GML^2}{\mu c^2 r^3}$ . Why: the first two terms are exactly Newton's; the third is the relativistic correction — an attraction steepening like $1/r^3$ , precisely the kind of term our criterion flags as dangerous.

Step 2. Find the circular orbits. $V_{\mathrm{eff}}'(r)=0$ gives $\dfrac{GM\mu}{r^2}-\dfrac{L^2}{\mu r^3}+\dfrac{3GML^2}{\mu c^2 r^4}=0$ ; multiplying by $\mu r^4/GM\mu^2$ turns it into the quadratic $r^2-\dfrac{L^2}{GM\mu^2}\,r+\dfrac{3L^2}{\mu^2c^2}=0$ . Why: Newton's version lacks the last term and has a single root at every $L$ ; the correction makes the equation quadratic, so circular orbits now exist only in pairs — or not at all.

Step 3. Solve and interpret the pair. $r_\pm=\dfrac{L^2}{2GM\mu^2}\left[1\pm\sqrt{1-\dfrac{12\,G^2M^2\mu^2}{L^2c^2}}\right]$ : $r_+$ is the valley floor (stable), $r_-$ the inner hilltop (unstable). Why: the outer extremum is the minimum, the inner the maximum — exactly the high- $L$ curve of the figure.

Step 4. Merge the pair. The square root vanishes — valley and hilltop coalesce into an inflection, $V_{\mathrm{eff}}''=0$ — when $L^2=12\,G^2M^2\mu^2/c^2$ . Substituting back: $r_{\mathrm{ISCO}}=\dfrac{L^2}{2GM\mu^2}=\dfrac{6GM}{c^2}$ . Why: marginal stability is precisely the double-root condition; below this $L$ the discriminant is negative, no circular orbit of any kind exists, and matter plunges.

Step 5. Attach the numbers. $6GM/c^2$ is three Schwarzschild radii: $88.5$ km for a $10\,M_\odot$ black hole, about $0.8$ AU for the $4\times10^6\,M_\odot$ hole at the galactic centre. Why: the ISCO scales linearly with mass — stellar-mass holes pack their disk edges into tens of kilometres, making their orbital frequencies audible to gravitational-wave detectors. *Verified: $r_{\mathrm{ISCO}}=6GM/c^2$ and $L_{\mathrm{ISCO}}=\sqrt{12}\,GM\mu/c$ are the standard Schwarzschild results (e.g. Hartle Ch. 9); the Newtonian limit $c\to\infty$ removes the quadratic's second root, restoring stable circles at all radii.*

Limiting cases

Case	Result	Meaning
Kepler, $n=-2$	$\omega_r=\omega_\varphi$ , ratio $3+n=1$	Stable; perturbed circle = small closed ellipse; no precession.
Hooke, $n=+1$	$\omega_r=2\omega_\varphi$ , ratio $4$	Stable; centred ellipse, closes after half a revolution.
Inverse-cube force, $n=-3$	$\omega_r^2=0$	Marginal: no restoring force at linear order; any nudge destroys the orbit.
Steeper, $n<-3$	$\omega_r^2<0$	Extremum is a maximum: perturbations grow exponentially; spiral in or escape.
Schwarzschild, $r<6GM/c^2$	no minimum of $V_{\mathrm{eff}}$	Relativity's $1/r^3$ potential term kills the valley: the ISCO, inner edge of accretion disks.

Worked examples

1 · Nudging the ISS (confidence builder)

The ISS orbits at $r_0=6771$ km ( $400$ km altitude), where $GM_\oplus=3.986\times10^{14}\ \mathrm{m^3/s^2}$ . Verify its circular frequency, check stability, and describe what a small radial kick does.

Step 1. Orbital frequency from the circular condition (2): $\omega_\varphi=\sqrt{GM_\oplus/r_0^3}=\sqrt{3.986\times10^{14}/(6.771\times10^6)^3}=\boxed{1.13\times10^{-3}\ \mathrm{rad/s}}$ , i.e. a period $2\pi/\omega_\varphi=\boxed{92.4\ \mathrm{min}}$ and speed $\omega_\varphi r_0=7.67$ km/s.

Step 2. Stability: gravity has $n=-2$ , and $3+n=1>0$ — comfortably stable.

Step 3. Radial breathing frequency from (4): $\omega_r=\omega_\varphi\sqrt{3+n}=\omega_\varphi$ . A small radial kick makes the radius oscillate with the same 92.4-minute period as the orbit itself.

Step 4. One radial cycle per revolution means the kicked orbit is a slightly eccentric Kepler ellipse that closes on itself — the station just trades a circle for a fixed ellipse. Stable, non-precessing.

2 · The last stable circle around a black hole (exam level)

A $10\,M_\odot$ black hole accretes gas that spirals inward through ever-smaller stable circles. Find the radius of the last one, its orbital frequency, and the gravitational-wave frequency a blob there would radiate. Use $GM_\odot=1.327\times10^{20}\ \mathrm{m^3/s^2}$ , $c=2.998\times10^8$ m/s.

Step 1. Gravitational length scale: $GM/c^2=10\times1.327\times10^{20}/(2.998\times10^8)^2=1.475\times10^4\ \mathrm{m}=14.75$ km.

Step 2. ISCO radius from Derivation B: $r_{\mathrm{ISCO}}=6GM/c^2=6\times14.75=\boxed{88.5\ \mathrm{km}}$ . Inside this the effective potential has no valley: gas cannot circle, only plunge.

Step 3. Orbital frequency (the Kepler form $\Omega=\sqrt{GM/r^3}$ holds exactly in Schwarzschild coordinates): $\Omega=\sqrt{1.327\times10^{21}/(8.85\times10^4)^3}=\boxed{1384\ \mathrm{rad/s}}$ — the last stable circle is traversed $220$ times per second.

Step 4. Orbital frequency $f=\Omega/2\pi=220$ Hz; a mass orbiting there radiates gravitational waves at twice the orbital frequency, $f_{\mathrm{GW}}=\boxed{440\ \mathrm{Hz}}$ — squarely in the LIGO band, which is why stellar-mass ISCOs are audible.

Origins · 1 minJoseph-Louis Lagrange and how this was discovered

Joseph-Louis Lagrange

1736–1813 · Turin, Berlin and Paris

Lagrange rebuilt mechanics as analysis: the Mécanique analytique (1788) famously contains not a single diagram. Stability questions run through his greatest work — his 1772 essay on the three-body problem found the five equilibrium points that bear his name, and his analysis of when motion near equilibrium stays bounded seeded the method this article uses: expand about the equilibrium, keep the linear term, examine the restoring coefficient. The stable pair, L4 and L5, passed his test where the collinear three failed — and two centuries later the Trojan asteroids and the James Webb Space Telescope (parked near the unstable Sun–Earth L2, needing station-keeping exactly as the analysis predicts) live out the distinction daily.

History

1687

— Newton's Principia derives circular-orbit conditions and probes apsidal motion via the revolving-orbits theorem.

1772

— Lagrange finds the five three-body equilibrium points and analyzes their stability; L4 and L5 pass.

1892

— Poincaré's Méthodes nouvelles turns orbital stability into a general mathematical discipline.

1916

— Schwarzschild solves Einstein's equations; the solution's effective potential hides an innermost stable orbit.

1973

— Novikov and Thorne build the standard accretion-disk model with its inner edge pinned at the ISCO.

Going further

References. Morin, Introduction to Classical Mechanics, ch. 7 (Central Forces) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §3.3, §3.6 (Bertrand's theorem) · Landau & Lifshitz, Mechanics, 3rd ed., §14, §15 · Taylor, Classical Mechanics, ch. 8 · Hartle, Gravity, Ch. 9 (Schwarzschild orbits and the ISCO)

Provenance. Verifier should recompute: (i) $V_{\mathrm{eff}}''(r)=-F'(r)+3L^2/(\mu r^4)$ and the resulting force-form criterion $F'/F+3/r_0>0$ ; (ii) the power-law identity $\omega_r^2/\omega_\theta^2=3-n$ giving the $n<3$ stability bound; (iii) $\omega_\theta^2=-F(r_0)/(\mu r_0)$ and the apsidal angle $\Phi=\pi/\sqrt{3-n}$ with $\Phi=\pi$ (Kepler) and $\Phi=\pi/2$ (Hooke). Cross-check against Goldstein §3.3 & §3.6, Landau-Lifshitz Mechanics §14–15, Taylor ch. 8, and Morin ch. 7. (The provenance identities use the inverse-power convention $F=-k/r^{n}$ ; the article text uses $F=-k\,r^{n}$ , so $n\to-n$ : ratio $3+n$ , bound $n>-3$ , $\Phi=\pi/\sqrt{3+n}$ .) Added for this upgrade and recomputed: ISS $\omega=1.13\times10^{-3}$ rad/s, $T=92.4$ min, $v=7.67$ km/s from $GM_\oplus=3.986\times10^{14}$ , $r=6.771\times10^6$ m; ISCO $r=6GM/c^2=88.5$ km, $\Omega=1384$ rad/s, $f=220$ Hz, $f_{\mathrm{GW}}=440$ Hz for $10\,M_\odot$ ; $L_{\mathrm{ISCO}}^2=12G^2M^2\mu^2/c^2$ (Hartle Ch. 9).

Builds on: The Effective Potential · Reduction to the Radial Equation · Angular Momentum Conservation in Central Fields · Central Force Fundamentals Leads to: The Kepler Problem · Classification of Orbits by Energy · The Orbit Equation and Binet's Equation · Kepler's Third Law: Period-Semimajor Relation

Where does the 3 in the stability criterion actually come from?

From the centrifugal force's own steepness. The barrier term in the radial equation is $L^2/\mu r^3$ — an outward force falling as $1/r^3$ , logarithmic slope $-3$ . Stability is a tug-of-war of slopes: displaced inward, the body needs the outward barrier to grow faster than the attraction does, which is exactly $rF'/F>-3$ . So the 3 is not numerology; it is the exponent of the centrifugal force, which traces to angular-momentum conservation via $L=\mu r^2\dot\varphi$ . Any physics that modified the centrifugal balance would move the 3 — and general relativity effectively does, which is how the ISCO appears.

My textbook says stability needs n less than 3 — this article says greater than −3. Who is wrong?

Nobody; the two statements are the same inequality in different conventions. Many texts write the force as $F=-k/r^{p}$ with the exponent in the denominator, so Kepler is $p=2$ and stability reads $p<3$ . This article (matching its sibling orbit articles) writes $F=-k\,r^{n}$ , so Kepler is $n=-2$ and the same criterion reads $n>-3$ ; evidently $n=-p$ . The invariant statement, safe in any convention, is logarithmic: $\mathrm{d}\ln|F|/\mathrm{d}\ln r>-3$ — the attraction must fall off more slowly than the inverse-cube centrifugal force. When in doubt, compute $rF'/F$ and add 3.

Is an unstable circular orbit a real solution or just a mathematical artifact?

It is a perfectly exact solution — start with precisely the right radius and velocity and the body circles forever. What it is not is realizable: like a pencil on its tip, any perturbation grows exponentially at rate $|\omega_r|$ , and perturbations are never absent. Unstable circular orbits still matter: they organize the dynamics around them. In Schwarzschild geometry the unstable branch $r_-$ governs how close a flyby can approach before plunging, and the related unstable photon circular orbit at $3GM/c^2$ shapes the black-hole shadow imaged by the Event Horizon Telescope.

Why does a kicked satellite trace a closed ellipse instead of drifting or precessing?

Because for inverse-square gravity the two frequencies coincide: $\omega_r^2/\omega_\varphi^2=3+n=1$ at $n=-2$ . The radius completes exactly one in-out breath per revolution, so the point of closest approach stays at a fixed direction — the perturbed orbit is a genuine small-eccentricity Kepler ellipse, not a drifting spiral. Any other exponent breaks the coincidence: the perturbed orbit is still bounded, but its apsides rotate each radial cycle, tracing a rosette. That ratio, and which force laws make it rational everywhere, is the subject of the apsidal-angles and Bertrand's-theorem articles.

Newtonian gravity gives stable circles at every radius — why does relativity suddenly forbid them inside the ISCO?

Because general relativity adds an extra attractive term to the effective potential that steepens like $1/r^3$ — beyond the inverse-cube danger threshold this article establishes. Far from the hole the term is negligible and the Newtonian valley survives; close in, it overwhelms the centrifugal barrier. The valley and the inner hilltop approach each other as $L$ decreases and merge at $L^2=12G^2M^2\mu^2/c^2$ , radius $6GM/c^2$ — Derivation B. Inside that radius the landscape is all downhill: circular motion is not merely unstable but nonexistent, and gas at the disk's inner edge plunges within a few dynamical times.

Does the stability of a circular orbit depend on how much angular momentum the body has?

For power-law forces, no — and that is worth internalizing. The criterion $3+n>0$ contains only the exponent: changing $L$ moves the location of the circular orbit but not the sign of the curvature there, so every circular orbit of a given power law is stable or every one is unstable. This changes the moment the force is not a pure power: then $rF'/F$ varies with radius, and different $L$ values park the orbit in stable or unstable zones of the same potential. Schwarzschild is the canonical example — stability holds for $r_0>6GM/c^2$ and fails inside, which is precisely why the ISCO is defined by an angular-momentum threshold, $L^2\geq12G^2M^2\mu^2/c^2$ .