Questions: Orbital Stability and Perturbation Analysis

Question 1

A physicist proposes a new force law F = −k/r⁴ (corresponding to n = −4). Would circular orbits under this force be stable?

• A. Yes, because the force is attractive and any attractive central force allows stable circular orbits
• B. No, because n = −4 < −3 violates the stability condition, so the effective potential has a maximum rather than a minimum at the circular orbit radius
• C. Yes, because angular momentum always creates a minimum in the effective potential regardless of the force law
• D. It depends on the mass of the orbiting particle — heavier particles have more stable orbits

Answer: (B)

The stability condition for circular orbits under a power-law force F = −kr^n is n > −3. For n = −4, this condition fails: the effective potential V_eff does not have a minimum at the circular orbit radius — it has a maximum. Any small radial perturbation therefore grows rather than producing oscillation back toward the orbit. Stable circular orbits require a restoring force for radial displacements, which exists only when V_eff is concave up (minimum) at r₀. The specific power of gravity (n = −2) satisfies n > −3, which is part of why stable planetary orbits exist.

Question 2

A particle in a stable circular orbit is given a small outward radial push. According to effective potential analysis, what happens?

• A. The particle escapes to infinity because any perturbation breaks orbital balance
• B. The particle spirals inward and eventually crashes into the central body
• C. The particle oscillates radially around the equilibrium orbit radius while continuing to orbit, tracing a rosette-like path
• D. The orbit remains perfectly circular because conservation of angular momentum prevents any radial motion

Answer: (C)

A stable circular orbit sits at a minimum of V_eff — a potential well. A small outward push displaces the particle from the minimum, and the restoring force (from the concavity of V_eff) pulls it back. The particle then overshoots inward, comes back, and oscillates radially — epicyclic motion. It continues orbiting azimuthally while oscillating radially, so the path traces a rosette rather than a closed ellipse in general. This is exactly the same physics as a mass in a potential well oscillating about its equilibrium position.

Question 3

Under Newtonian gravity, the epicyclic (radial oscillation) frequency equals the orbital frequency, which is why orbits under pure gravity close exactly as ellipses.

• T. True
• F. False

Answer: True

For the gravitational 1/r² force, one can show that the epicyclic frequency κ equals the orbital frequency Ω. This means the radial oscillation completes one full cycle in exactly the same time as one full azimuthal orbit. The particle returns to its starting radial position after each orbit, so the path closes — it repeats identically. Closed elliptic orbits are a special property of the 1/r² force law, as Bertrand's theorem formalizes. For most other force laws, κ ≠ Ω and orbits precess, tracing open rosettes.

Question 4

A circular orbit is stable whenever the effective potential V_eff has a maximum at the orbit radius r₀.

• T. True
• F. False

Answer: False

A maximum of V_eff at r₀ means the potential is concave down there — any small displacement experiences a force that pushes the particle further away from r₀, not back toward it. This is an unstable equilibrium. Stability requires a minimum (concave up, d²V_eff/dr² > 0), where displacements produce a restoring force. The analogy is direct: a ball on top of a hill (maximum) rolls away when nudged; a ball in a bowl (minimum) oscillates back to the bottom. The same logic applies to orbits via the effective potential.

Question 5

Using the effective potential framework, explain why orbital stability is equivalent to asking whether d²V_eff/dr² > 0 at the circular orbit radius.

Model answer: A circular orbit sits at the radius r₀ where dV_eff/dr = 0 — a stationary point of the effective potential. Whether this equilibrium is stable depends on the sign of the second derivative. If d²V_eff/dr² > 0, the effective potential is concave up at r₀ (a minimum): a small radial displacement produces a restoring force proportional to the displacement, and the orbit is stable with oscillation frequency ω² = (1/m)(d²V_eff/dr²). If d²V_eff/dr² < 0, the potential is concave down (a maximum): displacements produce a force that amplifies the deviation and the orbit is unstable.

This is a direct application of the one-dimensional stability criterion: expand the potential to second order around equilibrium. The sign of the quadratic term determines stability. The effective potential reduces the three-dimensional orbit problem to a one-dimensional radial problem, so the same mathematical test applies. For gravity, d²V_eff/dr² > 0 is always satisfied at the minimum, guaranteeing stability. For force laws with n < −3, this condition fails and no stable circular orbits exist — a profound consequence of the specific exponent in the force law.