Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - General Problems - Page 228: 94

Answer

$L=6.24\times10^{34}\frac{kgm^2}{s}$

Work Step by Step

$L=I\omega=\frac{2\pi I}{T}$ $1 year=31536000s$ $L_t=\frac{2\pi I}{11.9 y\times31536000\frac{s}{y}}$ $I_J=\frac{2MR^2}{5}=\frac{2(190\times10^{25}kg)(6.99\times10^7m)^2}{5}=3.71\times10^{42}$ $L_J=\frac{2\pi I}{T}=\frac{2\pi 3.71\times10^{42}}{(11.9y)(31536000)\frac{s}{y}}=6.22\times10^{34}$ $I_S=\frac{2MR^2}{5}=\frac{2(56.8\times10^{25}kg)(5.82\times10^7m)^2}{5}=7.70\times10^{41}$ $L_J=\frac{2\pi I}{T}=\frac{2\pi 7.70\times10^{41}}{(29.5y)(31536000)\frac{s}{y}}=1.50\times10^{32}$ $I_U=\frac{2MR^2}{5}=\frac{2(8.68\times10^{25}kg)(2.53\times10^7m)^2}{5}=2.22\times10^{41}$ $L_U=\frac{2\pi I}{T}=\frac{2\pi 2.22\times10^{41}}{(84.0y)(31536000)\frac{s}{y}}=5.27\times10^{31}$ $I_N=\frac{2MR^2}{5}=\frac{2(10.2\times10^{25}kg)(2.46\times10^7m)^2}{5}=2.47\times10^{41}$ $L_U=\frac{2\pi I}{T}=\frac{2\pi 2.47\times10^{41}}{(165y)(31536000)\frac{s}{y}}=2.98\times10^{31}$ $L_t=6.22\times10^{34}+1.50\times10^{32}+5.27\times10^{31}+2.98\times10^{31}=6.24\times10^{34}\frac{kgm^2}{s}$
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