"An Earth satellite moves in a circular orbit 730 km above the Earth's surface. The period of the motion is 99.2 min. What is the magnitude of the centripetal acceleration of the satellite?" OK so to solve this I used the formula centripetal acceleration = v^2/r To calculate velocity (v) I did the following: v = distance travelled (d) / time taken (t) d = 2piR = 2 x pi x 730,000 = 4586725.274m t = 99.2 x 60 = 5952s v = d/t = 4586725.274/5952 = 770.619ms^-1 Then we already have r which is 730,000m so v^2/r is: 770.619^2/730,000 = 0.813ms^-2 But this is answer is wrong...HOW?!