Bertrand's Theorem

In 1873 Joseph Bertrand published a three-page note in the Comptes Rendus asking a question nobody had thought to ask sharply: among all possible central force laws, which ones bend every bound orbit into a closed curve? Newton had shown the inverse-square law gives perfect ellipses; Hooke's linear force does too. Bertrand proved something far stronger — these two are the only central forces for which every bound orbit, at every energy and angular momentum, closes on itself. Everything else, however close to inverse-square, produces open rosettes that precess forever. The theorem turns the two most important forces in physics into geometrically privileged objects, with a practical edge: because closure is so fragile, any observed precession is a measurement of a deviation from the pure force law. Mercury's stray 43 arcseconds per century is exactly a Bertrand violation — general relativity's small correction breaking inverse-square perfection. The proof runs in two stages, and this article walks both. First the near-circular filter: closure near every circular radius forces the apsidal angle $\Delta\varphi=\pi/\beta$ to one fixed rational number, which integrates to a power law — infinitely many candidates survive. Then the amplitude test: closure at finite eccentricity collapses the survivors to $\beta=1$ (inverse-square) and $\beta=2$ (Hooke) alone. The first stage we derive completely; the second we sketch honestly, marking where the heavy bookkeeping lives.

Key takeaways

Only the inverse-square force $F=-k/r^2$ and the linear force $F=-kr$ close every bound orbit — Kepler with apsidal angle $\pi$ , Hooke with $\pi/2$ .
The proof filters twice: near-circular closure forces a power law $F=-c\,r^{n}$ with $\beta^2=3+n$ rational and constant; finite-eccentricity closure leaves only $\beta=1$ ( $n=-2$ ) and $\beta=2$ ( $n=1$ ).
The master formula is the near-circular apsidal angle $\Delta\varphi=\pi\,[3+rF'/F]^{-1/2}$ — $\pi$ times the ratio of orbital to radial frequency.
Kepler ellipses have the force centre at a focus; Hooke ellipses have it at the centre, closing after half a revolution.
Any perturbation — general relativity, oblateness, a third body — reopens the orbit: measured precession quantifies the perturbation.
Key idea: closure of all bound orbits requires the apsidal angle to be a fixed rational multiple of $\pi$ for every orbit; matching this constant across near-circular and eccentric orbits forces the inverse-square or linear force law — nothing else.

The core insight

Closure of all bound orbits requires the apsidal angle $\Delta\varphi$ to be a fixed rational multiple of $\pi$ for every orbit; matching this constant across both near-circular and highly eccentric orbits forces $V(r)\propto-1/r$ or $V(r)\propto r^2$ — nothing else.

Where this lives today

Bertrand's theorem is used today mostly through its contrapositive: real orbits do not close, and every measured precession rate is a clean probe of whatever spoils the pure force law. The star S2, tracked through pericentre around Sagittarius A*, precesses by the Schwarzschild prediction — the GRAVITY collaboration's headline measurement, now sharpened by GRAVITY+ to hunt for an extended dark-mass component that would add its own Bertrand violation. Binary-pulsar periastron advance plays the same game at higher precision; in exoplanet systems the apsidal precession of eccentric hot Jupiters is read backwards to infer tidal deformation. And the Hooke survivor runs the modern laboratory: ion traps approximate isotropic harmonic confinement precisely because Bertrand guarantees closed, non-dephasing orbits — anharmonicity announces itself as amplitude-dependent precession, the diagnostic trap designers actually monitor.

The physical picture

Watch the radius as a bead oscillating in the effective-potential well while the whole picture turns at $\dot\varphi=L/mr^2$ . During one radial half-cycle the azimuth advances by the apsidal angle $\Delta\varphi$ : rational multiple of $\pi$ means the orbit eventually retraces itself; anything else precesses forever. Bertrand's question: which force laws make the rational value come out the same for every orbit?

Figure

One rosette and the only two survivors — a generic force precesses while inverse-square and Hooke close their ellipses

Near a circular orbit the picture simplifies to two clocks: the radius wobbles harmonically at $\omega_r$ , set by the curvature of the effective-potential well, while the angle turns at $\omega_\varphi$ . Demanding their ratio be the same rational number at every circular radius already kills every force law that is not a power law. The second demand is harsher: as the orbit grows eccentric, the apsidal angle of a generic power law drifts with amplitude, so an orbit that closes when nearly circular fails when eccentric. Only two force laws hold the apsidal angle fixed at every amplitude — the same two whose closed orbits carry hidden conserved quantities (the Laplace–Runge–Lenz vector and the Fradkin tensor; see the hidden-symmetry article).

From effective potential to apsidal angle

Conservation of $E$ and $L$ compresses the planar motion into a radial problem, and the closure question then lives in one integral — the angle swept as the radius runs from minimum to maximum:

\Delta\varphi=\int_{r_{\min}}^{r_{\max}}\frac{(L/r^2)\,\mathrm{d}r}{\sqrt{2m\,[E-V_{\mathrm{eff}}(r)]}},\qquad V_{\mathrm{eff}}=V(r)+\frac{L^2}{2mr^2} \qquad (1)

where: $\Delta\varphi$ — apsidal angle, perihelion to aphelion; $r_{\min},r_{\max}$ — the turning points, where $E=V_{\mathrm{eff}}$ ; $m$ — the (reduced) mass; $L$ — angular momentum. It comes from dividing $\dot\varphi=L/mr^2$ by $\dot r$ from the energy equation. Successive perihelia sit $2\Delta\varphi$ apart, so the orbit closes iff $\Delta\varphi=\pi\,p/q$ : after $q$ radial oscillations and $p$ revolutions the trajectory bites its tail. For the second stage of the proof we will need the orbit equation itself. Trading time for angle and radius for $u=1/r$ turns Newton's radial equation into Binet's form:

\frac{\mathrm{d}^2u}{\mathrm{d}\varphi^2}+u=J(u),\qquad J(u)\equiv-\frac{m}{L^2u^2}\,F\!\Big(\frac1u\Big) \qquad (2)

where: $u=1/r$ ; $F(r)$ — the central force, negative for attraction; $J(u)$ — the force term as a function of inverse radius. A circular orbit is a fixed point $u_0=J(u_0)$ ; everything about closure is encoded in how oscillations about that fixed point behave. Throughout we write $F(r)=-c\,r^{n}$ — force exponent $n$ , Kepler at $n=-2$ , Hooke at $n=+1$ — for which $J(u)=(mc/L^2)\,u^{-(n+2)}$ is itself a pure power, the fact that makes the whole analysis tractable. (The potential goes as $r^{n+1}$ ; the hidden-symmetry article's potential-exponent formula $\pi/\sqrt{n_V+2}$ is the same result with $n_V=n+1$ .)

The near-circular filter

Perturb a circular orbit slightly and the radius oscillates harmonically in the effective well while the angle advances at the circular rate. The apsidal angle becomes $\pi$ times the frequency ratio, computed from the force law alone — Derivation A builds it and then squeezes it:

\Delta\varphi=\pi\,\frac{\omega_\varphi}{\omega_r}=\pi\left[\,3+\frac{r_0F'(r_0)}{F(r_0)}\,\right]^{-1/2}\equiv\frac{\pi}{\beta} \qquad (3)

where: $r_0$ — the circular radius; $\beta^2=3+rF'/F$ — the squared frequency ratio $\omega_r^2/\omega_\varphi^2$ . Stability needs $\beta^2>0$ ; closure needs $\beta$ rational. But the decisive demand is uniformity: the same rational $\beta$ at every circular radius, since $\beta$ varying continuously cannot stay rational. A constant $\beta$ is a differential equation for $F$ , and it integrates to a pure power law:

\beta^2=3+\frac{rF'}{F}=\text{const}\;\Longrightarrow\; F(r)=-\frac{c}{r^{\,3-\beta^2}}=-c\,r^{\,n},\quad \beta^2=3+n \qquad (4)

where: $c>0$ — the force strength; $n=\beta^2-3$ — the force exponent. Read the survivors off: $\beta=1$ gives $n=-2$ with $\Delta\varphi=\pi$ ; $\beta=2$ gives $n=+1$ with $\Delta\varphi=\pi/2$ . But so far $\beta=3$ ( $n=6$ ), $\beta=\tfrac32$ and every other rational survive too — the near-circular filter narrows infinitely many force laws to a one-parameter family, and can do no more:

Figure

The apsidal angle across all power-law forces — a smooth curve on which only two points survive the full theorem

The amplitude test

To kill the remaining rationals, push beyond the harmonic approximation. Write $u=u_0+x$ in the Binet equation and keep terms through third order in $x$ ; the oscillation is no longer exactly harmonic, and its frequency — hence the apsidal angle — picks up a correction proportional to the squared amplitude:

\Delta\varphi(a)=\frac{\pi}{\beta}\Big[1+\gamma(\beta)\,a^2+\mathcal O(a^4)\Big] \qquad (5)

where: $a$ — the dimensionless oscillation amplitude ( $\propto$ eccentricity for small $e$ ); $\gamma(\beta)$ — the anharmonic frequency-shift coefficient, computable from the Taylor coefficients of $J(u)$ . Closure for all amplitudes demands $\gamma=0$ : otherwise an orbit closing at one eccentricity precesses at another. Working out $\gamma$ for the power law (Derivation B) gives the celebrated selection rule

\beta^2\,(\beta^2-1)\,(\beta^2-4)=0 \qquad (6)

where: the root $\beta^2=0$ is the unstable $n=-3$ boundary, leaving $\beta=1$ (Kepler, $n=-2$ ) and $\beta=2$ (Hooke, $n=+1$ ). This is Bertrand's theorem; both survivors are checked exactly after the derivations below.

Derivation AConstancy of the apsidal angle forces a power law

We derive the near-circular apsidal angle $\Delta\varphi=\pi[3+rF'/F]^{-1/2}$ completely, then impose radius-independence and integrate to the power law (4). The strategy: a circular orbit sits at the minimum of the effective potential, so a slightly perturbed orbit executes simple harmonic radial motion at a frequency fixed by the well's curvature; the apsidal angle is $\pi$ times the ratio of angular to radial frequency. Watch one bookkeeping point — the circular-orbit condition is used twice to eliminate $L$ .

Step 1. Set up and locate the circular orbit. With $V'(r)=-F(r)$ , the radial problem is $m\ddot r=-V_{\mathrm{eff}}'(r)$ with $V_{\mathrm{eff}}'(r)=-F(r)-\dfrac{L^2}{mr^3}$ ; the circular radius satisfies $V_{\mathrm{eff}}'(r_0)=0$ , i.e. $\dfrac{L^2}{mr_0^3}=-F(r_0)>0$ . Why: conservation of $L$ buries the angular motion in the centrifugal term; a circular orbit is radial equilibrium, requiring attraction at $r_0$ .

Step 2. Compute the well's curvature. $V_{\mathrm{eff}}''(r_0)=-F'(r_0)+\dfrac{3L^2}{mr_0^4}=-F'(r_0)-\dfrac{3F(r_0)}{r_0}$ , using Step 1. Why: the curvature is the spring constant felt by a radial perturbation $\eta$ : $m\ddot\eta=-V_{\mathrm{eff}}''(r_0)\,\eta$ .

Step 3. Form the frequency ratio. $\omega_r^2=V_{\mathrm{eff}}''(r_0)/m$ and $\omega_\varphi^2=L^2/m^2r_0^4=-F(r_0)/mr_0$ , so $\dfrac{\omega_r^2}{\omega_\varphi^2}=\dfrac{-F'-3F/r_0}{-F/r_0}=3+\dfrac{r_0F'(r_0)}{F(r_0)}\equiv\beta^2$ . Why: every dimensional quantity cancels; the ratio is the force law's local logarithmic slope plus three.

Step 4. Convert to the apsidal angle. Perihelion to aphelion is half a radial period $\pi/\omega_r$ , during which $\varphi$ advances $\Delta\varphi=\omega_\varphi\cdot\pi/\omega_r=\pi/\beta$ — equation (3). Why: the apsidal angle is how far the angle clock turns during half a tick of the radius clock.

Step 5. Demand radius-independence and integrate. Closure at every $r_0$ with one rational $\beta$ requires $rF'/F=\beta^2-3$ for all $r$ : a separable ODE, $\ln|F|=(\beta^2-3)\ln r+\text{const}$ , hence $F=-c\,r^{\beta^2-3}$ — equation (4). Why: if $\beta$ varied with $r_0$ it would sweep through irrational values between any two radii; a constant logarithmic derivative is precisely the definition of a power law.

Go deeper · Derivation B · 2 minThe finite-eccentricity expansion selects two force laws

Derivation BThe finite-eccentricity expansion selects two force laws

The goal is the selection rule (6), $\beta^2(\beta^2-1)(\beta^2-4)=0$ , which eliminates every power law except Kepler and Hooke. This is the technical half of Bertrand's proof, run honestly as a structured sketch: the full bookkeeping lives in Goldstein's Appendix A, but every conceptual move is here. The strategy is a Poincaré–Lindstedt expansion of the Binet equation about the circular solution: nonlinear terms shift the fundamental oscillation's frequency at second order in amplitude; closure demands that shift vanish identically, which becomes an algebraic equation in $\beta$ .

Step 1. Specialize Binet to the power law. With $F=-c\,r^{\beta^2-3}$ , equation (2) reads $u''+u=J(u)$ with $J(u)=\dfrac{mc}{L^2}\,u^{1-\beta^2}$ . Why: only for a power-law force is $J$ a pure power of $u$ , making the Taylor coefficients below simple functions of $\beta$ alone.

Step 2. Expand about the circular fixed point. Set $u=u_0+x$ with $u_0=J(u_0)$ : $x''+\beta^2x=\tfrac12J''(u_0)\,x^2+\tfrac16J'''(u_0)\,x^3+\cdots$ , using $J'(u_0)=1-\beta^2$ . Why: the linear term reproduces the near-circular result — apsidal angle $\pi/\beta$ ; the quadratic and cubic terms are the anharmonicity that can spoil it.

Step 3. Feed the ansatz through. Write $x=a\cos\beta\varphi+a^2(b_0+b_2\cos2\beta\varphi)+\cdots$ and collect orders. At order $a^2$ the quadratic term generates a constant offset $b_0$ and a second harmonic $b_2$ , both proportional to $J''(u_0)$ . Why: a squared cosine contains DC and double-frequency pieces; neither shifts the fundamental's frequency yet, but both feed back at the next order.

Step 4. Impose no frequency shift at order $a^3$ . The third-order terms proportional to $\cos\beta\varphi$ — sourced by $J''\times(b_0,b_2)$ cross-terms and by $J'''$ directly — would resonantly shift the fundamental unless their net coefficient vanishes. Inserting the power-law values of $J''$ and $J'''$ in terms of $\beta$ , the amplitude-dependence of $\Delta\varphi$ cancels only when $\beta^2(\beta^2-1)(\beta^2-4)=0$ . Why: the standard Lindstedt secular-term condition; the algebra collapses because $J$ is a pure power.

Step 5. Read off the survivors. $\beta^2=0$ is the $n=-3$ stability boundary; $\beta^2=1$ is Kepler, $n=-2$ ; $\beta^2=4$ is Hooke, $n=+1$ . Every other rational $\beta$ fails at finite eccentricity. Why: closure for all amplitudes is an infinite family of conditions, and the first nontrivial one is already this restrictive; higher orders check out for the survivors because their orbits are exactly solvable.

Both survivors check out exactly. For $n=-2$ Binet's equation is linear with constant drive, $u''+u=mk/L^2$ , solved by the focal conic $u=(mk/L^2)(1+e\cos\varphi)$ — apsides diametrically opposite, $\Delta\varphi=\pi$ at every eccentricity (Kepler's first law; the Kepler problem article runs it in full). For $n=+1$ the substitution $w=u^2$ linearizes Binet to $w''+4w=\text{const}$ : the radius oscillates twice per revolution, $\Delta\varphi=\pi/2$ at every amplitude — the centred ellipse. The factor-of-two echo, four apses for a centred ellipse against two for a focal one, is the geometric face of $\beta=2$ versus $\beta=1$ .

Limiting cases

Case	Result	Meaning
Kepler, $n=-2$	$\Delta\varphi=\pi$ at all $e$	Focal ellipse; closes after one revolution; LRL vector conserved.
Hooke, $n=+1$	$\Delta\varphi=\pi/2$ at all $e$	Centred ellipse; closes after half a revolution; Fradkin tensor conserved.
Generic power law	$\Delta\varphi=\pi/\sqrt{3+n}$ near-circular only	Drifts with eccentricity: rosette; fails the amplitude test.
Logarithmic force $F\propto-1/r$	$\Delta\varphi=\pi/\sqrt2$	Irrational multiple of $\pi$ : never closes even near the circle.
$n\to-3$	$\beta\to0$ , $\Delta\varphi\to\infty$	Stability boundary: circular orbits marginal, orbits spiral.

Common trapA force law with some closed orbits passes Bertrand's test

It does not — the most common misreading of the theorem. Every attractive central potential has closed circular orbits, and many have special energies where one orbit closes as a many-petalled flower. Bertrand demands closure for ALL bound initial conditions simultaneously, with the same rational apsidal angle everywhere — an infinitely stronger condition. Two related traps: $\Delta\varphi=\pi/\sqrt{3+n}$ is a near-circular result, not exact at finite eccentricity (the whole point of the amplitude test); and watch the convention — $n$ here is the force exponent ( $n=-2$ Kepler), while the potential-exponent version reads $\pi/\sqrt{n_V+2}$ ( $n_V=-1$ Kepler).

Worked examples

1 · A puck on a spring table (confidence builder)

A $0.5\ \mathrm{kg}$ puck on a frictionless table is tied to a central spring, $k=20\ \mathrm{N/m}$ (Hooke force $F=-kr$ ), on a slightly non-circular orbit. Find the two frequencies, the apsidal angle, and the orbit's shape.

Step 1. For $F=-kr$ the circular condition $m\omega_\varphi^2 r_0=kr_0$ holds at every radius, so $\omega_\varphi=\sqrt{k/m}=\sqrt{20/0.5}=\boxed{6.32\ \mathrm{rad/s}}$ — independent of $r_0$ .

Step 2. The force exponent is $n=+1$ , so $\beta^2=3+n=4$ and $\omega_r=2\omega_\varphi=\boxed{12.65\ \mathrm{rad/s}}$ : the radius breathes twice per revolution.

Step 3. Apsidal angle: $\Delta\varphi=\pi/\beta=\boxed{90°}$ — perihelion to aphelion is a quarter turn, four apses per revolution.

Step 4. Period check: a revolution takes $2\pi/\omega_\varphi=0.99\ \mathrm{s}$ , a radial cycle $0.50\ \mathrm{s}$ ; after one revolution the puck has completed exactly two radial breaths and retraces its path: a closed ellipse centred on the post — Bertrand's second survivor, on a tabletop.

2 · How fast does a slightly-wrong force law precess? (exam level)

Suppose gravity followed $F=-c/r^{2.02}$ — force exponent $n=-2.02$ , a 1% tilt. For a near-circular planet, find the apsidal angle, the perihelion drift per orbit, and how many orbits until the perihelion laps the full circle.

Step 1. $\beta^2=3+n=3-2.02=0.98$ , so $\beta=\sqrt{0.98}=0.98995$ . The frequency ratio is no longer exactly 1: the radial clock now ticks slightly slower than the angular one.

Step 2. Apsidal angle: $\Delta\varphi=180°/0.98995=\boxed{181.83°}$ — each half radial cycle overshoots half a turn by $1.83°$ .

Step 3. Perihelion drift per orbit: $\Delta\varpi=2\Delta\varphi-360°=2\times181.83°-360°=\boxed{3.65°}$ per orbit, prograde (the apsides advance with the motion, as for any force falling faster than inverse-square).

Step 4. Full apsidal revolution after $360°/3.65°=\boxed{98.5\ \mathrm{orbits}}$ . A 1% exponent error makes the orbit visibly non-closing within a hundred revolutions — which is why Newton, via his revolving-orbits theorem, could already argue from the slowness of observed apsidal motion that the exponent must be extremely close to 2.

Origins · 1 minJoseph Bertrand and how this was discovered

Joseph Bertrand

1822–1900 · Paris

Bertrand entered the École Polytechnique at eleven as an auditor and held a doctorate by seventeen. As professor at the Collège de France and permanent secretary of the Académie des Sciences he shaped French mathematics for half a century — his prime conjecture, probability paradox, and critiques of economic equilibrium all carry his name. The closed-orbit theorem appeared in 1873 as a three-page Comptes Rendus note, so compressed that filling in its steps became a standard graduate exercise for the next century. Characteristically, Bertrand posed it as a question about what observation could prove: if all orbits close, the force law is already determined — an early, crisp example of an inverse problem in physics.

History

1687

— Newton's Principia includes the theorem of revolving orbits, tying apsidal motion to the force exponent.

1873

— Bertrand proves only the inverse-square and linear laws close all bounded orbits.

1915

— Einstein computes Mercury's stray 43 arcseconds per century: the measured failure of closure fixed by general relativity.

1965

— Fradkin exhibits the conserved tensor behind the oscillator's closed orbits, completing the symmetry picture.

2020

— The GRAVITY interferometer measures the Schwarzschild precession of the star S2 around the galactic-centre black hole.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics (3rd ed.), §3.6 and Appendix A · Landau & Lifshitz, Mechanics (Vol. 1), §14 and Problem after §14 · Morin, Introduction to Classical Mechanics, Ch. 7 · Taylor, Classical Mechanics, Ch. 8 · J. Bertrand, Comptes Rendus 77, 849 (1873) · Arnold, Mathematical Methods of Classical Mechanics, §8 The key checkpoints of the article are: (1) the near-circular apsidal angle $\Delta\varphi=\pi[3+rF'/F]^{-1/2}$ derived from $V_{\mathrm{eff}}''$ ; (2) the power-law identity $\beta^2=n+2$ giving $\Delta\varphi=\pi/\sqrt{n+2}$ with values $\pi$ ( $n=-1$ ) and $\pi/2$ ( $n=2$ ) in the potential-exponent convention ( $n_V=\beta^2-2$ ; the article text uses the force exponent $n=\beta^2-3$ ); (3) the cubic selection condition $\beta^2(\beta^2-1)(\beta^2-4)=0$ from the finite-eccentricity expansion; and (4) the explicit Binet solutions $u=\tfrac{mk}{L^2}(1+e\cos\varphi)$ and the $w=u^2$ harmonic reduction. Mercury's $43''$ /century is the standard GR value. Worked-example numbers: $\omega_\varphi=6.32$ , $\omega_r=12.65$ rad/s for $m=0.5$ kg, $k=20$ N/m; for $n=-2.02$ : $\beta=0.98995$ , $\Delta\varphi=181.83°$ , drift $3.65°$ /orbit, full turn in 98.5 orbits.

Builds on: The Effective Potential · The Orbit Equation and Binet's Equation · Angular Momentum Conservation in Central Fields · Classification of Orbits by Energy Leads to: The Kepler Problem · Kepler's First Law: Elliptical Orbits · Conic-Section Orbits

Is it just a coincidence that the two most important forces in physics are the two Bertrand forces?

Partly yes, partly no. Gravity and electrostatics are inverse-square for a deep reason — flux conservation in three dimensions (Gauss's law) — and a linear restoring force arises universally as the leading Taylor term near stable equilibrium. So both laws were guaranteed starring roles regardless of Bertrand. What the theorem adds is that these two, and only these two, produce closed orbits — why solar-system dynamics is so clean and the harmonic oscillator became the exactly-solvable workhorse. The two problems physics solves beautifully are the two with maximal hidden symmetry.

The apsidal angle formula gives a rational value for lots of exponents, like n = 6 giving one third of pi. Why do those fail?

Because $\Delta\varphi=\pi/3$ for $n=6$ holds only in the strict near-circular limit. Feed in a finite eccentricity and the anharmonic terms of the Binet equation shift the radial frequency by an amount proportional to $e^2$ — equation (5) — so the apsidal angle drifts away from $\pi/3$ as the orbit grows. An orbit that closes at one amplitude precesses at the next; the family as a whole never closes. The amplitude test of Derivation B is exactly the demand that this drift vanish, and its cubic (6) has no root at $\beta=3$ . Only $\beta=1$ and $\beta=2$ hold the apsidal angle rigid at every eccentricity.

Where does the observed direction of Mercury's precession come from — why prograde?

A force falling faster than inverse-square ( $n<-2$ , so $\beta<1$ ) makes the radial oscillation slower relative to the revolution: $\Delta\varphi=\pi/\beta>\pi$ , and the apsides advance with the motion — prograde. General relativity's correction adds extra attraction at small radius, effectively steepening the law, which is why Mercury's perihelion advances rather than regresses. The worked example's $n=-2.02$ toy shows the same sign: $\Delta\varphi=181.83°>180°$ . A force falling more slowly than inverse-square gives $\Delta\varphi<\pi$ and regression instead.

Does Bertrand's theorem apply to the actual solar system with all its planets?

Strictly, no — and that is its practical power. The theorem concerns one body in an exact central $1/r^2$ field; the real Sun is slightly oblate, the planets tug on each other, and general relativity corrects the potential, so real orbits precess. Mercury's perihelion moves about $532''$ per century from planetary perturbations plus $43''$ from GR. The theorem is why these numbers are measurable: against the rigid backdrop of would-be-closed Kepler ellipses, any drift is signal, not noise. Astronomers use Bertrand closure as the null hypothesis every deviation is tested against.

Why does demanding closure at every radius force a power law, rather than just some smooth force?

Because "the apsidal angle is the same at every circular radius" is a differential statement. Equation (3) says $\Delta\varphi$ depends on $r_0$ only through $rF'/F$ — the logarithmic slope of the force. Holding $\Delta\varphi$ constant across all radii means holding $rF'/F$ constant, and the only functions with constant logarithmic slope are pure powers, $F\propto r^{\beta^2-3}$ . Any other smooth force has a slope that varies with radius, so its apsidal angle sweeps through a continuum of values — necessarily hitting irrational multiples of $\pi$ — and closure fails somewhere in the family.

What actually happens physically at the n = −3 boundary where the formula blows up?

At $n=-3$ the attractive force scales as $1/r^3$ — exactly the same power as the centrifugal term $L^2/mr^3$ . The two no longer cross at an isolated radius: either the centrifugal term wins everywhere (escape) or the attraction wins everywhere (spiral to the centre), depending on the ratio of $L^2$ to the force constant. There is no potential well, hence no stable circular orbit, and $\beta^2\to0$ signals it: the radial oscillation frequency goes to zero, so a displaced orbit never gets pulled back. For $n<-3$ circular orbits exist algebraically but sit on a potential maximum — the stability side of the story is the circular-orbit stability article.