devarshi-dt-logo

Question:

A communication satellite of 500 kg revolves around the earth in a circular orbit of radius 4.0 × 10⁷ m in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is

≈0.13 × 10¹⁴ kg m²s⁻¹

≈0.58 × 10¹⁴ kg m²s⁻¹

≈2.58 × 10¹⁴ kg m²s⁻¹

≈1.3 × 10¹⁴ kg m²s⁻¹

Solution:

As the satellite is moving in equatorial plane with orbital radius 4 × 10⁷ m.
∴Satellite is geostationary satellite.
Hence, the time taken by satellite to complete its one revolution, T = 24 h = 86400 s
Velocity of satellite, v = 2πr/T
Angular momentum, L = mvr
L = m(2πr/T)r = 2πmr²/T
∴L = 2 × 3.14 × 500/86400 × (4 × 10⁷)² = 0.58 × 10¹⁴ kg m² s⁻¹.