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 10 - Exercises and Problems - Page 192: 31

Answer

12,000 years

Work Step by Step

We treat the earth as a solid sphere. We know that there are 86,400 seconds in a day. Thus, the earth is rotating and $ 7.272205\times10^{-5} $ radians per second. If the day were a second less, the earth would rotate at $7.272289\times10^{-5}$ radians per second. Thus, we find: $\Delta K = \frac{1}{2}(\frac{2}{5}M_ER_E^2)(( 7.272121\times10^{-5})^2-(7.272289\times10^{-5})^2)=2.97\times10^{24}$ Dividing this by $16\times10^{12}W$ and then converting to seconds, we find: $\Delta t \approx 12,000\ years$
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