A satellite orbiting the earth in a circular orbit 7000 km from Earth\'s center

Question

A satellite orbiting the earth in a circular orbit 7000 km from Earth's center experiences a centripetal acceleration of 8.20 m/s2. (This is about 600 km above the surface of the earth, but you would not use this number for any calculations here.)
(A) Find the velocity of the satellite: (m/s)
(B) Find the angular velocity &omega of the satellite: (rad/s)
Suppose the satellite's mass is 100 kg. Determine the force of gravity on it, in N. (You ought to be able to do this in your head. If not, you are doing it the hard way.) (N)
(D) Now, suppose we consider the same satellite, but we have moved it to an orbit of radius 21,000 km. What would be the centripetal acceleration on this satellite? (Just use the 1/r2 law.) (m/s/s)

Explanation / Answer

A) Centripetal acceleration = v^2 / R v = v a . R = v 8.2 . 7 . 10^6 = 7.58 . 10^3 m/s b) w = v/r = 7576 / 7 . 10^6 = 1.08 . 10^–3 rad/s c) F = m . a = 100 . 8.2 = 820 N d) 3 times the distance means one ninth the force and so 1/9 of the acceleration a = 8.2 / 9 = 0.91 m/s/s

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