A planet is nudged slightly outward from its circular orbit around a star in a 1/r² gravitational field. Which correctly describes what happens?
AThe planet spirals outward — any outward perturbation increases potential energy, making the orbit unstable
BThe planet's orbit becomes slightly elliptical, oscillating around the original circular radius without drifting away
CThe planet escapes because the gravitational force decreases at larger r, removing the restoring force
DThe orbit becomes elliptical and slowly precesses — the major axis rotates over many orbits
In a 1/r² field, the effective potential has a minimum at r₀ — the circular orbit sits at the bottom of a bowl. A small outward nudge increases U_eff, producing a restoring force that pushes r back. The orbit becomes slightly elliptical, rocking around r₀ but not departing from it. Option C is the common misconception: gravity weakening at larger r doesn't eliminate the restoring force — it's the *shape* of U_eff (a minimum) that matters, not the absolute strength of gravity. Option D is interesting: precession does occur for other force laws, but for 1/r² specifically, ω_r = ω_orbit so orbits close on themselves exactly — no precession.
Question 2 Multiple Choice
Why are gravitationally bound orbits in Newtonian gravity closed ellipses rather than precessing rosettes?
AAll stable orbits under any force law are closed — stability implies closure
BBecause the unperturbed orbit is circular, perturbations always return to the same path without rotation
CBecause for the 1/r² force specifically, the radial oscillation frequency equals the orbital frequency — ω_r = ω_orbit — so the orbit traces the same path every revolution
DBecause Kepler's laws guarantee all bound orbits are ellipses regardless of the force law
The equality ω_r = ω_orbit is a special, non-generic property of the 1/r² force. When these frequencies match, the radial oscillation (r expanding and contracting) completes one full cycle in exactly the same time as one full orbit, so the orbit traces the same closed path every revolution. For other force laws, ω_r ≠ ω_orbit, and bound orbits precess — the approximate ellipse slowly rotates, tracing a rosette. Option A is false: a stable orbit can precess (stability and closure are distinct properties). Option D is wrong: Kepler's laws are specific to 1/r² and are not universal.
Question 3 True / False
A circular orbit is stable if the effective potential has a minimum at the orbital radius, meaning small perturbations produce bounded oscillations rather than runaway departures.
TTrue
FFalse
Answer: True
This is the central stability criterion. At the circular orbit radius r₀, U_eff has a minimum: dU_eff/dr = 0 (the circular orbit condition) and d²U_eff/dr² > 0 (stability). A small displacement from r₀ moves the orbit into a region of higher U_eff, which generates a restoring force — exactly like a ball in a bowl. The positive second derivative is the mathematical signature of stability; a maximum (negative second derivative) at r₀ would make the orbit unstable to any perturbation.
Question 4 True / False
Most stable circular orbits precess — the fact that Earth's orbit appears to repeat annually is due to the very slow rate of precession.
TTrue
FFalse
Answer: False
Precession is not a universal feature of stable orbits. For the 1/r² force specifically, ω_r = ω_orbit, so the orbit repeats exactly after one revolution — zero precession. Earth's orbit genuinely closes annually (to high accuracy) rather than slowly rotating. Precession occurs for other force laws and for real orbits when perturbations from other planets are present (Mercury precesses noticeably), and general relativity adds an additional correction. But in a pure two-body 1/r² problem, stable orbits are exactly closed ellipses.
Question 5 Short Answer
What does the effective potential U_eff tell us about whether a circular orbit is stable, and how does the shape of U_eff differ between force laws that allow stable circular orbits and those that don't?
Think about your answer, then reveal below.
Model answer: U_eff combines the gravitational potential with the centrifugal term L²/(2μr²). A circular orbit corresponds to a minimum of U_eff. If that minimum exists and is bowl-shaped (d²U_eff/dr² > 0 at r₀), perturbations produce bounded radial oscillation — the orbit is stable. If U_eff has no local minimum (only a maximum or a monotone region), any perturbation causes runaway. The 1/r² force produces a true minimum; some force laws (F ∝ 1/rⁿ with n > 2) produce no local minimum, so stable circular orbits are impossible.
The effective potential framework reduces 2D orbital mechanics to a 1D stability question: is there a bowl-shaped potential minimum at the orbital radius? This geometric reasoning explains why planetary orbit stability is specific to the 1/r² force law — it depends on the particular mathematical form of Newtonian gravity, not on 'gravity' generically. General relativity slightly modifies U_eff, producing the small precession of Mercury's perihelion even in the two-body case — a measurable deviation from the perfectly-closed Newtonian ellipse.