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15 May, 02:48

An earth satellite moves in a circular orbit at a speed of 5000 m/s. part a what is its orbital period? express your answer using two significant figures

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  1. 15 May, 03:25
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    r = GM/v^2 r is the distance of the satellite from the center of Earth in meters G is newtons constant, 6.67428 x 10^-11 M is the mass of Earth, 5.9742 x 10^24 v is the orbital speed, which we know is 5,000 m/s r = 6.67428 x 10^-11 x 5.9742 x 10^24/5,000^2 r = 8,137,445 meters from the center of the Earth and : 8,137,445 - 6,378,100 = 1,749,345 meters from the surface of the Earth Now that we know the distance we can calculate the orbital period. Time = Distance/speed distance is the circumference of the orbit, 2Ď€ x 8,137,445 = 51,129,074 meters Time = (51,129,074) / 5,000 Time = 10,225 seconds = 2.84 hours so, orbital period is 2.84 hrs
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