An Earth satellite moves in a circular orbit 519 km above Earth\'s surface with
ID: 1659724 • Letter: A
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