Question

4. A satellite orbits the earth with a velocity of 5.8 km/s. The period of the satellite in the orbit is 20.2 hours. How high above the earth is the satellite? [Radius of the earth, R = 6400 km, Mass of the earth, M = 5.97×10²⁴ kg, π = 3.14, Universal Gravitational Constant, G = 6.67×10⁻¹¹ m³s⁻²kg⁻¹]

Accepted Answer

For circular orbital motion, v = 2π(R+h)/T.
5.8 = [2×3.14×(6400+h)] / (20.2×3600)
421776 / (2×3.142) = 6400+h
67161.78 = 6400 + h
h = 67161.78 − 6400 ≈ 60,761.78 km ≈ 60,762 km above the earth's surface.

WAEC Physics 2025 Theory — Question 4

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