Questions: Reduced Mass Problem

Question 1

Why is the Earth-Sun two-body problem well-approximated by treating the Earth as orbiting a fixed Sun?

• A. Because the gravitational force on the Sun is much smaller than the force on the Earth
• B. Because the reduced mass μ ≈ m_Earth when m_Sun ≫ m_Earth, meaning the relative motion behaves as if the lighter body orbits a fixed center
• C. Because the Sun's orbital velocity is exactly zero in the solar system's rest frame
• D. Because the Earth's orbital period is short enough that the Sun's motion is negligible

Answer: (B)

When one mass dominates (m₁ ≫ m₂), the reduced mass μ = m₁m₂/(m₁+m₂) ≈ m₂. The equation of motion for the relative coordinate r becomes μr̈ = F(r), which with μ ≈ m₂ is essentially the equation for m₂ orbiting a fixed mass. The Sun does wobble slightly (the actual CM is near the Sun's surface), but the correction is one part in 300,000. The reduced-mass framework makes this approximation and its error quantitatively precise.

Question 2

Two stars of equal mass m are in mutual gravitational orbit. What is their reduced mass, and what does this tell you about their orbital geometry?

• A. μ = 2m; both stars orbit with the full combined mass
• B. μ = m/2; both stars orbit their common center of mass at equal distances
• C. μ = m; the stars are indistinguishable, so the reduced mass equals the individual mass
• D. μ = m/4; symmetry halves the effective mass twice

Answer: (B)

For two equal masses m, μ = m·m/(m+m) = m/2. The equation μr̈ = F(r) describes a particle of mass m/2 orbiting under the mutual force. In the CM frame, each star orbits the center at distance r/2 (where r is their separation), moving with equal and opposite velocities. The reduced mass m/2 reflects the fact that both bodies are in motion — neither is fixed — and the effective inertia resisting relative acceleration is reduced accordingly.

Question 3

In the reduced-mass formulation, the relative coordinate r obeys Newton's second law with the total mass M = m₁ + m₂ as the effective inertial mass.

• T. True
• F. False

Answer: False

False. The relative coordinate r obeys μr̈ = F(r), where μ = m₁m₂/(m₁+m₂) is the reduced mass — always less than the smaller of the two masses. The total mass M = m₁+m₂ describes the center-of-mass motion: MR̈ = F_external = 0 for an isolated system. The total mass and reduced mass play different roles: M governs the trivial CM drift, and μ governs the non-trivial relative orbital motion.

Question 4

The reduced-mass technique applies mainly to gravitational two-body problems, not to other types of central-force interactions like spring forces.

• T. True
• F. False

Answer: False

False. The reduced-mass transformation is purely kinematic — it relies only on the separation of CM and relative coordinates, not on the form of the force. As long as the force depends only on the relative separation r = r₁ − r₂, the relative motion equation takes the form μr̈ = F(r) regardless of whether F is gravitational, a spring, electrostatic, or any other central force. The same framework applies in atomic physics (hydrogen atom), molecular vibrations (diatomic molecules), and classical orbital mechanics.

Question 5

Explain why the two-body problem can always be reduced to an equivalent one-body problem. What is the key mathematical step, and what physical insight does it capture?

Model answer: Define the CM coordinate R = (m₁r₁ + m₂r₂)/(m₁+m₂) and the relative coordinate r = r₁ − r₂. The CM moves uniformly (no net external force), reducing to trivial uniform motion. Substituting into Newton's second law for r gives μr̈ = F(r), where μ = m₁m₂/(m₁+m₂). This is formally a one-body problem: a particle of mass μ orbiting under the mutual force. The physical insight is that only the relative motion is dynamically interesting; the CM drift is subtracted out.

The power of this reduction is that all results from single-particle mechanics — conservation of energy and angular momentum, Kepler's laws, the effective potential — apply directly to the equivalent one-body problem. You solve one equation for r, then recover each body's actual trajectory from r and R. This is the standard approach in celestial mechanics and quantum mechanics alike.