Central Forces and Kepler's Laws

Introduction

Central forces are directed toward or away from a fixed point and depend only on distance from that point. Gravity is the prime example, leading to Kepler's empirical laws of planetary motion. Understanding central force dynamics is fundamental to celestial mechanics and atomic physics.

Properties of Central Forces

Central force: F = f(r)\(\hat{r}\) (directed along position vector). Examples: Gravity f(r) = -GMm/r² (attractive), Coulomb f(r) = kq₁q₂/r² (can be attractive or repulsive), harmonic f(r) = -kr (spring). Central forces are conservative: F = -dU/dr \(\hat{r}\) where U depends only on r. They exert zero torque about the center: τ = r × F = 0.

Angular Momentum Conservation in Central Forces

Zero torque means angular momentum L is conserved. Consequences: (1) Motion is confined to a plane perpendicular to L; (2) Areal velocity dA/dt = L/(2m) is constant; (3) The force center lies in the orbital plane. This planar motion reduces the problem to two dimensions, significantly simplifying analysis.

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Introduction

Central forces are directed toward or away from a fixed point and depend only on distance from that point. Gravity is the prime example, leading to Kepler's empirical laws of planetary motion. Understanding central force dynamics is fundamental to celestial mechanics and atomic physics.

Properties of Central Forces

Central force: F = f(r)\(\hat{r}\) (directed along position vector). Examples: Gravity f(r) = -GMm/r² (attractive), Coulomb f(r) = kq₁q₂/r² (can be attractive or repulsive), harmonic f(r) = -kr (spring). Central forces are conservative: F = -dU/dr \(\hat{r}\) where U depends only on r. They exert zero torque about the center: τ = r × F = 0.

Angular Momentum Conservation in Central Forces

Zero torque means angular momentum L is conserved. Consequences: (1) Motion is confined to a plane perpendicular to L; (2) Areal velocity dA/dt = L/(2m) is constant; (3) The force center lies in the orbital plane. This planar motion reduces the problem to two dimensions, significantly simplifying analysis.

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Kepler's First Law

Planets move in elliptical orbits with the Sun at one focus. An ellipse has semi-major axis a, semi-minor axis b, eccentricity e = √(1-b²/a²). Distance from focus: r = a(1-e²)/(1+ecosθ). Circular orbits are special case (e = 0). Parabolic (e = 1) and hyperbolic (e > 1) orbits are unbound trajectories possible for ejected objects.

Kepler's Second Law

A line connecting planet to Sun sweeps out equal areas in equal times. This is direct consequence of angular momentum conservation: dA/dt = L/(2m) = constant. Planets move faster when closer to Sun (perihelion), slower when farther (aphelion). This law applies to any central force motion, not just gravity.

Kepler's Third Law

The square of orbital period is proportional to cube of semi-major axis: T² ∝ a³. For circular orbits: T² = (4π²/GM)a³. Ratio T²/a³ is same for all planets orbiting same central mass. Allows calculation of central mass from orbital data. Modified form for elliptical orbits uses semi-major axis a instead of radius.

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Solved Example: Satellite Orbital Period

Calculate orbital period of satellite at altitude 400 km above Earth. Earth mass M = 5.98×10^24 kg, radius R_E = 6.37×10^6 m. Solution: Orbital radius r = R_E + 400,000 = 6.77×10^6 m. Using Kepler's Third Law for circular orbit: T² = (4π²/GM)r³. T² = (4π² / (6.67×10⻹¹ × 5.98×10^24)) × (6.77×10^6)³ = (39.48 / 3.98×10^14) × 3.10×10^20 = 9.92×10⻹⁴ × 3.10×10^20 = 3.08×10^7 s². T = √(3.08×10^7) = 5550 s ≈ 92.5 minutes. This is typical Low Earth Orbit period. International Space Station at similar altitude has period about 90 minutes. Geostationary orbit at r ≈ 42,000 km has T = 24 hours. Verify with v = 2πr/T = 2π × 6.77×10^6 / 5550 = 7660 m/s, matches orbital velocity formula v = √(GM/r) = √(6.67×10⻹¹ × 5.98×10^24 / 6.77×10^6) = 7660 m/s ✓.

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