Questions: Central Force Motion and Orbital Dynamics

Question 1

Why does a central force conserve angular momentum?

• A. Because the magnitude of the force depends only on the distance from the center
• B. Because the force is always perpendicular to the velocity of the particle
• C. Because the force is parallel to the position vector, making the torque r × F equal to zero
• D. Because the kinetic energy is constant throughout the motion

Answer: (C)

Angular momentum changes only when torque is applied: dL/dt = τ = r × F. For a central force, F = F(r)r̂ — it points along r̂, which is parallel to r. Since r and F are parallel, their cross product r × F = 0. Zero torque means L is conserved. Option A (magnitude depending only on r) is part of the definition of a central force but not itself the reason angular momentum is conserved — it is the direction of the force that matters.

Question 2

A particle orbiting under a 1/r² attractive force has total energy E = −3 J. The effective potential has a minimum of U_eff_min = −5 J. What type of orbit does the particle follow?

• A. Unbound (hyperbolic), because the total energy is negative
• B. Circular, because E equals the minimum of U_eff
• C. Bound, oscillating radially between a minimum and maximum radius
• D. The orbit type cannot be determined without knowing the angular momentum

Answer: (C)

E = −3 J > U_eff_min = −5 J means the particle has more energy than the minimum of the effective potential but still less than the escape threshold (E < 0 for this force). It is trapped in the effective potential well, oscillating between two turning points where E = U_eff(r). This is a bound, non-circular orbit (an ellipse for a 1/r² force). Option B would be a circular orbit, which requires E = U_eff_min exactly. Option A is wrong because E < 0 alone does not determine boundedness — the shape of U_eff matters.

Question 3

A particle with nonzero angular momentum moving under an attractive central force will eventually spiral into the origin (r = 0) if the attractive force is strong enough.

• T. True
• F. False

Answer: False

The centrifugal term L²/(2μr²) in the effective potential diverges to +∞ as r → 0. For any particle with L ≠ 0, this creates a repulsive barrier that prevents collapse to the origin, regardless of how strongly attractive U(r) is at small r. Only a particle with exactly zero angular momentum — falling straight toward the center with no transverse motion — can actually reach r = 0. Angular momentum is the 'guard' that keeps orbiting particles from collapsing.

Question 4

For a 1/r² attractive force, all bound orbits are closed ellipses. For most other central force laws, bound orbits precess rather than closing.

• T. True
• F. False

Answer: True

This is Bertrand's theorem: only two central force laws produce closed bound orbits for all energies — the 1/r² force (gravity, Coulomb) and the harmonic oscillator (F ∝ r). For any other force law, the radial oscillation period and the orbital period are generally incommensurable, causing the orbit to precess. The anomalous precession of Mercury's perihelion was a key confirmation of general relativity precisely because Newtonian 1/r² gravity predicts perfectly closed ellipses, and the observed excess precession had no Newtonian explanation.

Question 5

Explain physically what the centrifugal potential term L²/(2μr²) represents and why it appears in the effective potential.

Model answer: When the 2D orbit problem is reduced to a 1D radial equation, the particle's rotational kinetic energy gets absorbed into the effective potential. Angular momentum conservation gives φ̇ = L/(μr²), so the rotational kinetic energy (1/2)μr²φ̇² = L²/(2μr²) depends only on r. This term acts as a repulsive barrier: as r decreases, the particle must rotate faster to conserve angular momentum, and more of its energy is tied up in rotation rather than available for radial motion. The closer to the center, the stronger the effective repulsion.

The effective potential trick converts a 2D problem into a 1D one by using conservation of L to eliminate the angular degree of freedom. The price you pay is that the centrifugal term — representing the cost of rotating faster as you move inward — appears as an effective repulsion. The resulting 1D problem in U_eff(r) fully encodes the orbital behavior: bound vs unbound, circular vs elliptical, stable vs unstable, all readable from the shape of U_eff.