Effective potential and the orbit equation for a central force

D-006 central-force-orbit-equation Home PU-201 Threads force · energy Depends on D-001, D-002 verified

Statement

Motion under any central force reduces to one-dimensional motion in an effective potential, and for an inverse-square force the resulting orbit is a conic section.

Why it matters

This is the single move that turns the two-body gravitational problem into something solvable by hand. It delivers all three of Kepler's laws as corollaries rather than separate discoveries, and it fixes the exact boundary — the inverse-square law — beyond which orbits stop closing. It is the classical baseline that relativity (PU-305) and the quantum atom (PU-101) are measured against.

Assumptions

Derivation

Result

Limiting cases

• The ℓ²/2μr² term is repulsive and dominates at small r — this is the centrifugal barrier, and it is why a body with any angular momentum at all cannot reach the origin.
• The minimum of V_eff is the circular orbit. Its curvature gives the frequency of small radial oscillations about it; when that frequency equals the orbital frequency, the orbit closes. Bertrand's theorem shows this happens only for the inverse-square and the linear force laws.
• Kepler's third law follows immediately: T² = 4π²μa³/k.

Breaks when

Failure modes

• Calling the centrifugal term a force. It is a term in an effective potential that appeared from eliminating θ̇; there is no outward force in the inertial frame.
• Using m instead of μ when the masses are comparable. Fine for Earth–Sun, badly wrong for a binary star.
• Assuming a small change in the exponent of the force law produces a small change in the orbit. It produces precession, which accumulates without bound.

Worked number

Integrate a force law F ∝ 1/r^2+ε and measure the precession rate as a function of ε: at ε = 0 the orbit closes exactly, and any nonzero ε drives a perihelion that advances without bound. The forward link PU-305 recovers Mercury's 43″/century from general relativity. Run the check →

Discussion

The engine of D-006 is not a clever integral but a bookkeeping of symmetry. A central force has no preferred direction in the plane and no explicit clock, so two quantities are handed to us for free: angular momentum ℓ (rotational symmetry, step 2) and energy E (time-translation symmetry, step 3). Each conserved quantity removes one degree of freedom. What began as two coupled second-order equations for r(t) and θ(t) collapses to a single first-order statement, E = ½μṛ² + V_eff(r), that a first-year student could read off a graph. The angular motion has not vanished; it has been repackaged as the ℓ²/2μr² centrifugal term, a wall that any body with ℓ ≠ 0 cannot climb through — which is why nothing carrying angular momentum ever reaches the force centre.

The reason the answer is a conic section and not some transcendental curve is visible in step 6. The Binet equation is a harmonic oscillator in the angle θ driven by −(μ/ℓ²u²)F(1/u). Only for the inverse-square law does the awkward 1/u² prefactor cancel the u² hidden in F = −ku², leaving a constant drive. A constant-forced oscillator has the solution "constant plus one cosine", u = (1 + e cos θ)μk/ℓ², and that single cosine is the ellipse. The eccentricity rides entirely on the energy, e = √(1 + 2Eℓ²/μk²): bound orbits (E < 0) close into ellipses, E = 0 is the marginal parabola, and E > 0 opens into a hyperbola. Kepler's first law is therefore a corollary of the algebra, not an independent empirical law.

The deeper statement is that the inverse-square orbit closes on itself — the perihelion never drifts — and this is overdetermination. Bertrand's theorem singles out only two force laws (the inverse square and the linear Hooke spring) for which every bound orbit closes; for these the radial oscillation frequency and the orbital frequency are locked in a rational ratio. Behind the Kepler case sits an extra conserved vector, the Laplace–Runge–Lenz vector A = p × L − μkr̂, which points along the major axis and pins the ellipse in place. Its conservation reflects a hidden SO(4) dynamical symmetry of the bound Kepler problem, larger than the obvious SO(3) of rotations; the same symmetry, promoted to quantum operators, is what makes the hydrogen energy levels depend only on the principal quantum number and not on ℓ. Break the inverse-square law by any amount and A is no longer conserved — the axis rotates and the orbit precesses.

The same six lines describe far more than planets. Flip the sign of k and demand E > 0 and the identical orbit equation gives the hyperbolic trajectory of Rutherford scattering; the impact-parameter–to–deflection relation and the famous dσ/dΩ ∝ sin⁻⁴(θ/2) cross-section fall straight out of it. The parabolic E = 0 case is the marginal-escape trajectory used for interplanetary transfers. And the forward link to PU-305 is exactly the failure mode named above: general relativity adds a small effective −1/r³ term to the potential, the axis-fixing vector A ceases to be conserved, and Mercury's perihelion advances the observed 43″ per century.

Common misconceptions.

• The centrifugal term is not a real force. It is a piece of V_eff that appeared purely from eliminating θ̇ using ℓ = const; in the inertial plane there is no outward push, only the inward F(r).
• Eccentricity is fixed by the energy alone, not by the angular momentum. Two orbits of the same E but different ℓ have the same e and the same major axis; only their size across (the semi-latus rectum ℓ²/μk) differs.
• Using the full mass m instead of the reduced mass μ. Harmless for Earth–Sun where μ ≈ m, but for a comparable-mass binary it corrupts both the period and the eccentricity.
• Assuming a tiny change in the force exponent gives a tiny change in the orbit. It gives precession, an angular error that grows without bound orbit after orbit — a qualitative change, not a small one.

Worked examples

Example 1 — Reconstructing a comet's orbit from one perihelion measurement. A comet is seen at perihelion at distance r_p = 8.0×10¹⁰ m from the Sun moving at v_p = 5.0×10⁴ m s⁻¹. Take GM_⊙ = 1.327×10²⁰ m³ s⁻². Because the comet mass is negligible, μ ≈ m, so we work per unit mass: ε ≡ E/μ, h ≡ ℓ/μ, and k/μ = GM. Find e, the semi-major axis, the aphelion distance and the period.

Example 2 — The circular orbit at the bottom of V_eff, and why it closes. A satellite in a circular orbit sits exactly at the minimum of the effective potential. For a low Earth orbit at r_c = 7.00×10⁶ m (GM_⊕ = 3.986×10¹⁴ m³ s⁻²), find the orbital speed and period, then show the frequency of small radial oscillations equals the orbital frequency — the statement that the inverse-square orbit closes.

Problems

1. (Classification.) A probe around the Sun has specific energy ε = −1.5×10⁸ J kg⁻¹ and specific angular momentum h = 6.0×10¹⁵ m² s⁻¹ (GM_⊙ = 1.327×10²⁰). Find the eccentricity, classify the orbit, and give the semi-major axis.
   Solution

   Use e = √(1 + 2εh²/(GM)²), the per-mass form of the result box. Compute 2εh² = 2(−1.5×10⁸)(3.6×10³¹) = −1.08×10⁴⁰; (GM)² = (1.327×10²⁰)² = 1.761×10⁴⁰. Ratio = −0.613, so e = √(1−0.613) = √0.387 = 0.62. Since ε < 0 (equivalently e < 1) the orbit is a bound ellipse. Semi-major axis a = −GM/2ε = (1.327×10²⁰)/(3.0×10⁸) = 4.42×10¹¹ m (≈ 2.95 AU).

2. (Apsides.) From the orbit equation r(θ) = (ℓ²/μk)/(1 + e cos θ), show that the perihelion and aphelion distances are r_± = p/(1 ∓ e) with p = ℓ²/μk, and hence that a = p/(1−e²) and a = −k/2E.
   Solution

   r is extremal where cos θ = ±1. At θ = 0, r_min = p/(1+e) (perihelion); at θ = π, r_max = p/(1−e) (aphelion). The major axis is 2a = r_min + r_max = p[1/(1+e) + 1/(1−e)] = p·2/(1−e²), so a = p/(1−e²). Now substitute e² = 1 + 2Eℓ²/μk² so that 1−e² = −2Eℓ²/μk². Then a = (ℓ²/μk)/(−2Eℓ²/μk²) = −k/2E. The size of the ellipse depends only on the energy; the shape (through e) mixes energy and angular momentum.

3. (Kepler III for a binary — use μ.) Two stars of masses M₁ = 2M_⊙ and M₂ = 3M_⊙ orbit their common centre of mass; the semi-major axis of the relative orbit is a = 3.0×10¹¹ m. Find the period. (M_⊙ = 1.989×10³⁰ kg, G = 6.674×10⁻¹¹.)
   Solution

   The result gives T² = 4π²μa³/k. Here k = GM₁M₂ and μ = M₁M₂/(M₁+M₂), so μ/k = 1/[G(M₁+M₂)] and T² = 4π²a³/[G(M₁+M₂)]. The individual masses cancel into their sum — the correct handling of the reduced mass. With M₁+M₂ = 5M_⊙ = 9.945×10³⁰ kg: G(M₁+M₂) = 6.637×10²⁰; a³ = 2.7×10³⁴. So T² = 39.48×2.7×10³⁴/6.637×10²⁰ = 1.606×10¹⁵, giving T = 4.0×10⁷ s ≈ 1.27 yr. (Had we mistakenly used m = M₂ instead of μ, the period would be badly wrong.)

4. (Inverse problem via Binet.) A particle is observed to move on a logarithmic spiral r(θ) = r₀e^kθ. What central force law produces this orbit?
   Solution

   Let u = 1/r = (1/r₀)e^−kθ. Then du/dθ = −ku and d²u/dθ² = k²u. Insert into the Binet equation (step 5): d²u/dθ² + u = (k² + 1)u = −(μ/ℓ²u²)F(1/u). Solving, F = −(ℓ²/μ)(1+k²)u³ = −(ℓ²(1+k²)/μ)·1/r³. The spiral is produced by an attractive inverse-cube force. Note this orbit does not close — the inverse cube is not one of Bertrand's laws — consistent with the spiral falling in to (or out from) the centre.

5. (Precession from a perturbed potential.) Add an inverse-cube perturbation, F(r) = −k/r² − λ/r³ with λ > 0 small. Solve the Binet equation, show the orbit is a precessing ellipse, and find the apsidal angle and the perihelion advance per revolution. Relate this to the forward link to PU-305.
   Solution

   With F(1/u) = −ku² − λu³, the Binet equation becomes d²u/dθ² + u = −(μ/ℓ²u²)(−ku²−λu³) = μk/ℓ² + (μλ/ℓ²)u. Collect the u terms: d²u/dθ² + Ω²u = μk/ℓ², with Ω² ≡ 1 − μλ/ℓ². This is still a constant-forced oscillator, but with shifted frequency Ω < 1. Its solution is u = μk/(ℓ²Ω²) + A cos(Ωθ) — an ellipse-like curve whose radius returns to perihelion each time Ωθ increases by 2π. The angle between successive perihelia is therefore Δθ = 2π/Ω, exceeding 2π because Ω < 1. The advance per orbit is Δφ = 2π(1/Ω − 1). For small λ, Ω ≈ 1 − μλ/2ℓ², so 1/Ω ≈ 1 + μλ/2ℓ² and Δφ ≈ πμλ/ℓ² per revolution (prograde). This is the mechanism behind the forward link: general relativity contributes an effective −1/r³ term to the potential, which by exactly this calculation makes Mercury's perihelion advance — 43″ per century after Newtonian planetary perturbations are subtracted. A change in the force law that looks small produces an angular drift that accumulates without bound.