An Earth satellite moves in a circular orbit 665 km above the Earth\'s surface.

Question

An Earth satellite moves in a circular orbit 665 km above the Earth's surface. The period of the motion is 97.8 min. (a) What is the speed of the satellite? ...... m/s ? (b) What is the magnitude of the centripetal acceleration of the satellite? ..... m/s2 ? show steps please!

Explanation / Answer

a) radius of the earth = 6400E3 m ==> ---------------------------------------------------------------------------------------------------------------------------------------------------------------------- v = 2 pi R/t = 2*3.1416*(6400e3+665e3)/(97.8*60) = 7565 m/s ---------------------------------------------------------------------------------------------------------------------------------------------------------------------- b) a =v^2/R = (7565 *7565)/(6400e3+665e3) = 8.10 m/s2

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