A satellite moves in a circular orbit around Earth at a speed of 4988 m/s.Its altitude: 9.74x10^6. Determine the period of the satellites orbit in hrs

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Since the satellite is orbiting 9.74 x 10^6 meters above the earth's surface, it is actually 9.74 x 10^6 + 6.38 x 10^6 meters from the center of mass of the earth. (6.38 x 10^6 is the approximate radius of the earth.)

The circumference of the orbital path is then...

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Since the satellite is orbiting 9.74 x 10^6 meters above the earth's surface, it is actually 9.74 x 10^6 + 6.38 x 10^6 meters from the center of mass of the earth. (6.38 x 10^6 is the approximate radius of the earth.)

The circumference of the orbital path is then 2*pi*1.612x10^7 meter.

Distance/velocity = time.   1.612x10^7/4.988 x 10^3 m/s = 2.03x10^4 seconds.  2.03x10^4 seconds = 5.64 hours as the time for one orbit which is the period.

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