Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Exercises and Problems - Page 90: 20

Answer

$1.98\times10^{20}\ N$

Work Step by Step

To find the tension in the cable, we simply must find the gravitational force pulling on the cable: $F=\frac{GM_1M_2}{R^2}$ We know that the moon orbits the earth, so we consult the appendix for data regarding the mass of the moon and the earth to obtain: $F=\frac{(6.67\times10^{-11})(.0735\times10^{24})\times(5.97\times10^{24})}{(3.84\times10^8)^2}=1.98\times10^{20}\ N$
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