A satellite revolves at a height equal to Earth's radius R above the surface. Given g = π² m s⁻² at Earth's surface. Its time period is:
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Question

A satellite revolves at a height equal to Earth's radius R above the surface. Given g = π² m s⁻² at Earth's surface. Its time period is:

G1 Physics · Accepted Answer

Orbital radius is r = R + R = 2R For a circular orbit, T² = 4π²r³/(GM) At Earth's surface, g = GM/R² So, GM = gR² Therefore, T² = 4π²(2R)³/(gR²) = 32π²R/g Given g = π² m s⁻², T² = 32R Hence, T = √(32R) s Option A follows.