Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 199: 9


The tension in the cable would be $1.98\times 10^{20}~N$.

Work Step by Step

The tension in the cable would be equal to the gravitational force between the earth and the moon. Therefore; $T = \frac{G~M_e~M_m}{r^2}$ $T = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.97\times 10^{24}~kg)(7.35\times 10^{22}~kg)}{(3.84\times 10^{8}~m)^2}$ $T = 1.98\times 10^{20}~N$ The tension in the cable would therefore be $1.98\times 10^{20}~N$.
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