Question

If a satellite circulate around the earth at a height of 5,113.68 km above the earth's surface


If a satellite circulate around the earth at a height of 5,113.68 km above the earth's surface, given the earth radius is 3958.8 miles and mass is 5.98 x1024 kg, use G=6.67x 10-11 Nm2/kg2, find the period of this satellite in unit hours?

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Answer #1

using Kepler's 3rd law

T^2 = 4 pi^2 R^3 / ( GM)

T^2 = 4* 3.14^2 * (5113.68* 10^3 + 3958.8* 1.6* 10^3)^3 / ( 6.67* 10^-11* 5.98* 10^24)

T = 3.383 hours

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