An Earth satellite moves in a circular orbit 519 km above Earth\'s surface with

Question

An Earth satellite moves in a circular orbit 519 km above Earth's surface with a period of 94.81 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite? An Earth satellite moves in a circular orbit 519 km above Earth's surface with a period of 94.81 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite? An Earth satellite moves in a circular orbit 519 km above Earth's surface with a period of 94.81 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Explanation / Answer

A) speed = distance /time

= 2 pi r / T

= 2 pi *(6371000+519000) /(94.81*60)

= 7610 m/s

B) centripetal acceleration = v^2 /r

= 7610^2 /(6371000+519000)

= 8.405 m/s^2

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