(a) Explanation Given info: The circular orbit of the satellite is 150 mi above Earth. Explanation: The radius of the satellite orbit will be, r = R e + h (1) R e is the radius of earth h is the distance of the satellite from earth The centripetal force required for the circular motion of the satellite will be provided by the gravitational force. Hence, m v t 2 r = G M e m r 2 M e is the mass of earth m is the mass of the satellite G is the universal gravitational constant r is the distance between the centre of earth to the satellite v t is the tangential speed On re-arrangement for the tangential speed, v t = G M e r (2) Combining (1) and (2), the tangential speed of the satellite will be, v t = G M e R e + h Substitute 6.67 × 10 − 11 N ⋅ m 2 ⋅ kg -2 for G , 5.98 × 10 24 kg for M e , 6.38 × 10 6 m for R e , 150 mi for h to determine the orbital speed of the satellite, v t = ( 6 (b) To determine The time taken by the satellite to complete one complete revolution.

Bundle: College Physics, Volume 1, 11th + WebAssign Printed Access Card for Serway/Vuille's College Physics, 11th Edition, Single-Term

11th Edition

ISBN: 9781337741583

Author: Raymond A. Serway, Chris Vuille

Publisher: Cengage Learning

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Question

Chapter 7, Problem 53AP

(a)

To determine

The orbital speed of the satellite.

(b)

To determine

The time taken by the satellite to complete one complete revolution.

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Chapter 7 Solutions

Bundle: College Physics, Volume 1, 11th + WebAssign Printed Access Card for Serway/Vuille's College Physics, 11th Edition, Single-Term

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The orbital speed of the satellite.

Solution Summary: The author explains the orbital speed of the satellite and the centripetal force required for its circular motion.

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The orbital speed of the satellite.

The time taken by the satellite to complete one complete revolution.

Chapter 7 Solutions