Essential University Physics: Volume 1 (4th Edition)

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

Chapter 8 - Exercises and Problems - Page 147: 40


Please see the work below.

Work Step by Step

We know that $F=(\frac{GM}{r^2})m$ We plug in the known values to obtain: $F=(\frac{6.67\times 10^{-11})(4\times 10^6)}{(15)^2})m$ $F=1.1858\times 10^{-6}m$ Now we can find the decrease as $(\frac{F}{mg})\times 100=(\frac{m(1.1858\times 10^{-6})}{m(9.8)})\times 100=1.2\times 10^{-5}$ This gives the fractional change: .99999988
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