Question

A spy satellite is in circular orbit around Earth. It makes one revolution in 7.20 h. Mass of Earth is 5.974 × 10²⁴ kg, radius of Earth is 6371 km and Gravitational constant G is = 6.674 × 10⁻¹¹ N·m²/kg².

How high above Earth’s surface is the satellite?

Answer 1

Time,

T^2 = 4 pi^2 r^3/GM

(7.20 x 3600)^2 = (4 pi^2 r^3)/(6.67 x 10^-11 x 5.97 x 10^24)

r^3 = 6.78 x 10^21

r = 1.89 x 10^7 m = 1.89 x 10^4 km

H = (1.89 x 10^4) - 6371 = 12552.6 km

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