Motion of Satellites and Orbital Mechanics

Introduction

Satellite orbits follow Kepler's laws and Newton's law of gravitation. Circular orbits require specific velocity; energy determines orbit type (bound or unbound). Understanding orbital mechanics is essential for spacecraft design, satellite deployment, and mission planning for exploration of our solar system and beyond.

Circular Orbits

For circular orbit at radius r around mass M: Centripetal force equals gravity: mv²/r = GMm/r². Orbital velocity: v = √(GM/r). Independent of satellite mass m. Period: T = 2πr/v = 2π√(r³/GM). Total energy: E = -GMm/(2r) (negative, bound orbit). Kinetic energy equals magnitude of total energy; potential energy is twice total energy (with negative sign).

Orbital Velocity and Period Relations

Key relations: v ∝ 1/√r (higher orbits have slower speeds); T ∝ r^(3/2) (Kepler's third law); v²T = 2πGM (constant for given central mass). Geostationary orbit: T = 24 hours, requires r ≈ 42,164 km from Earth's center (≈ 35,786 km altitude). Low Earth orbit (r ≈ R_E): v ≈ 7.9 km/s, T ≈ 90 minutes.

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Introduction

Satellite orbits follow Kepler's laws and Newton's law of gravitation. Circular orbits require specific velocity; energy determines orbit type (bound or unbound). Understanding orbital mechanics is essential for spacecraft design, satellite deployment, and mission planning for exploration of our solar system and beyond.

Circular Orbits

For circular orbit at radius r around mass M: Centripetal force equals gravity: mv²/r = GMm/r². Orbital velocity: v = √(GM/r). Independent of satellite mass m. Period: T = 2πr/v = 2π√(r³/GM). Total energy: E = -GMm/(2r) (negative, bound orbit). Kinetic energy equals magnitude of total energy; potential energy is twice total energy (with negative sign).

Orbital Velocity and Period Relations

Key relations: v ∝ 1/√r (higher orbits have slower speeds); T ∝ r^(3/2) (Kepler's third law); v²T = 2πGM (constant for given central mass). Geostationary orbit: T = 24 hours, requires r ≈ 42,164 km from Earth's center (≈ 35,786 km altitude). Low Earth orbit (r ≈ R_E): v ≈ 7.9 km/s, T ≈ 90 minutes.

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Elliptical Orbits

General bound orbits are ellipses with semi-major axis a. Orbital parameters: r_min = a(1-e) at periapsis, r_max = a(1+e) at apoapsis where e is eccentricity. Vis-viva equation gives speed at any point: v² = GM(2/r - 1/a). Total energy: E = -GMm/(2a) depends only on semi-major axis, not eccentricity. Angular momentum determines eccentricity: larger L gives more circular orbit.

Energy and Angular Momentum in Orbits

Specific orbital energy (energy per unit mass): ε = v²/2 - μ/r = -μ/(2a) where μ = GM. Specific angular momentum: h = r × v = rv_perpendicular = √(μa(1-e²)). For given a, h is maximum for circular orbit (e = 0) and zero for degenerate line orbit (e = 1). Orbital elements a and e completely specify orbit size and shape.

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Practical Considerations

Orbital maneuvers: Changing speed at a point alters orbit. Hohmann transfer: most fuel-efficient between circular orbits (elliptical transfer tangent to both). Orbital decay: atmospheric drag causes satellites to spiral inward, losing energy and angular momentum. Geostationary satellites: match Earth's rotation, fixed position overhead. Lagrange points: positions where combined gravitational and centrifugal forces give stable equilibrium.

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Solved Example: Geostationary Orbit

Find altitude, orbital velocity, and energy of geostationary satellite (period T = 24 hours = 86400 s). Earth: M_E = 5.98×10^24 kg, R_E = 6.37×10^6 m. Solution: Using Kepler's Third Law: r³ = GMT²/(4π²) = 6.67×10⻹¹ × 5.98×10^24 × (86400)² / (4π²) = 7.54×10^22 m³. r = (7.54×10^22)^(1/3) = 4.22×10^7 m = 42,200 km from center. Altitude = 42,200 - 6,370 = 35,830 km ≈ 35,786 km (standard GEO altitude). Orbital velocity: v = 2πr/T = 2π × 4.22×10^7 / 86400 = 3070 m/s ≈ 3.07 km/s. Compare to LEO: v_LEO ≈ 7.8 km/s. GEO satellites move slower despite being farther because v ∝ 1/√r, but angular velocity ω = 2π/T matches Earth's rotation. Energy per kg: ε = -GM/(2r) = -3.98×10^14/(2×4.22×10^7) = -4.72×10^6 J/kg = -4.72 MJ/kg. Less negative than LEO (-31 MJ/kg), so GEO is higher energy orbit (less bound).

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