Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 247: 48


a) $KE_R=2.57\times10^{29}J$ b) $KE_R=2.68\times10^{33}J$

Work Step by Step

Rotational kinetic energy: $$KE_R=\frac{1}{2}I\omega^2$$ a) The earth can be considered a solid sphere with the axis through its center, so $I=\frac{2}{5}MR^2$ We have Earth's mass $M=5.97\times10^{24}kg$ and radius $R=6.38\times10^6m$ The Earth turns an angle of $\theta=2\pi rad$ (1 revolution) in $t=24h=8.64\times10^4s$, so $$\omega=\frac{\theta}{t}=7.27\times10^{-5}rad/s$$ Therefore, $$KE_R=\frac{1}{2}\Big(\frac{2}{5}MR^2\omega^2\Big)=2.57\times10^{29}J$$ b) The earth is the only particle rotating around the Sun in this exercise, so $I=M_ER_{orbit}^2$ The radius of orbit $R_{orbit}=1.5\times10^{11}m$. Therefore, $I=1.34\times10^{47}kg.m^2$ The Earth finishes an angle of $\theta=2\pi rad$ (1 revolution) in $t=365.25\text{ days}=3.16\times10^7s$, so $$\omega=\frac{\theta}{t}=2\times10^{-7}rad/s$$ Therefore, $$KE_R=\frac{1}{2}I\omega^2=2.68\times10^{33}J$$
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