Question

A satellite is in circular orbit around the earth at an altitude of 2.80 x 10^6 m. Find (a) the period of the orbit, (b) the speed of the satellite, and (c) the acceleration of the satellite.

Answer 1

A)

Distance, d = 6371000 + 2800000 = 9.17 x 10^6 m

From Kepler’s law,

T^2 = [(4 pi^2/(6.67 x 10^-11 x 5.98 x 10^24)](9.17 x 10^6)^3

T = 8737.58 sec = 2.43 hrs

B)

V^2 = (6.67 x 10^-11 x 5.98 x 10^24)/(9.17 x 10^6)

V = 6595.21 m/s = 6.60 km/s

C)

Acceleration, a = v^2/r = 6595.21^2/9.17 x 10^6 = 4.74 m/^2

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