Satellite CommunicationHardQuestion 20297
Question
For a geostationary satellite to remain fixed over the equator, its orbital velocity must match Earth's angular velocity ω_E = 7.27 × 10⁻⁵ rad/s. The required orbital radius (G = 6.67×10⁻¹¹, M_E = 6×10²⁴ kg) is approximately:
- A6400 km
- B42,241 kmCorrect
- C384,400 km
- D10,000 km
Correct answer
42,241 km
Explanation
For geostationary orbit: GM/r² = ω²r → r³ = GM/ω² = (6.67×10⁻¹¹ × 6×10²⁴)/(7.27×10⁻⁵)² = 4×10¹⁴/5.285×10⁻⁹ = 7.57×10²² → r = (7.57×10²²)^(1/3) ≈ 42,241 km from Earth's centre, i.e., altitude ≈ 35,786 km. Matches the known value.