An artificial satellite maintains a circular orbit with “gravity as the centripetal force.” Learn to accurately calculate the relationships among orbital speed, period, and altitude.
Basic Knowledge
An artificial satellite of mass m moving in a circular orbit at altitude h (r = R + h from Earth's center):
gravity = centripetal force: GMm/r² = mv²/r
→ orbital speed: v = √(GM/r)
→ orbital period: T = 2πr/v = 2π√(r³/GM) (consistent with the third law)
Note: Orbital speed and period are independent of the satellite's mass m.
Geostationary satellite: A satellite located at an orbital altitude whose period matches Earth's rotation period (T = 24h), remaining directly above the same point on Earth.
Geostationary orbital radius r₀: T² = 4π²r₀³/(GM) → r₀ = [GMT²/(4π²)]^(1/3) ≈ 42,000 km (from Earth's center)
Orbital speed: v = √(GM/r)
Orbital period: T = 2π√(r³/GM)
Geostationary orbit condition: T = 24 h (Earth's rotation period)
Larger r → smaller v (higher orbits are slower)
Larger r → longer T (higher orbits have longer periods)
Further Exploration (Background and Significance)
The orbital speed v = √(GM/r) becomes smaller as r increases. Higher orbits are slower. This may seem counterintuitive, but gravity is weaker at higher orbits, so a smaller speed is sufficient to maintain a circular orbit.
The speed in a circular orbit just above Earth's surface (r = R) is called the “first cosmic velocity,” v1 = √(gR) ≈ 7.9 km/s. If an object is launched horizontally at this speed, it becomes a satellite orbiting Earth.
Geostationary satellites (≈ 35,800 km above the equator) are used as communication and weather satellites. Because they synchronize with Earth's rotation, ground antennas can remain fixed and always point toward the same satellite.
Kepler's third law, T²/r³ = 4π²/(GM), follows directly from the artificial-satellite equations, and satellite motion is a special case of Kepler's laws (with Earth as the central body).
- Gravity = centripetal force → v = √(GM/r)
- Orbital speed is independent of satellite mass
- Higher orbit → slower speed and longer period
- First cosmic velocity v1 = √(gR) ≈ 7.9 km/s
- Geostationary satellite: T = 24 h, above the equator
- T = 2π√(r³/GM) (consistent with Kepler's third law)
- r is the distance from Earth's center
Points of Caution (Common Confusions)
① r is the distance from Earth's center. When altitude h is given, convert it using r = R + h. ② Orbital speed does not depend on satellite mass m (m cancels). ③ Geostationary satellites can orbit only above the equator. Satellites in polar orbits cannot remain stationary. ④ T = 24 h is Earth's rotation period. Do not confuse it with the Moon's orbital period (27.3 days).
Practice
- Given Earth's radius R = 6.4×10⁶ m and g = 9.8 m/s², find the circular-orbit speed just above Earth's surface (the first cosmic velocity).
- Express the period T of an artificial satellite with orbital radius r in terms of G, M, and r.
- Express the orbital radius of a geostationary satellite in terms of G, M, and T (=24h).