Physics: Principles with Applications (7th Edition)

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

Chapter 5 - Circular Motion; Gravitation - General Problems - Page 136: 79

Answer

Inner Period: $2.01\times10^4s$ Outer Period: $7.14\times10^4s$

Work Step by Step

$a_R=\frac{v^2}{r}$; $v=\sqrt{ra_R}$ $F_G=G\frac{m_1m_2}{r^2}$; $a_R=G\frac{m}{r^2}$ $v=\sqrt{\frac{Gm}{r}}$ $T=\frac{2\pi r}{v}=2\pi r\sqrt{\frac{r}{Gm}}$ $T=2\pi (73,000,000m)\sqrt{\frac{73,000,000m}{\big(6.67\times10^{-11}\frac{Nm^2}{kg^2}\big)(5.7\times10^{26}kg)}}=2.01\times10^4s$ $T=2\pi (170,000,000m)\sqrt{\frac{170,000,000m}{\big(6.67\times10^{-11}\frac{Nm^2}{kg^2}\big)(5.7\times10^{26}kg)}}=7.14\times10^4s$ Saturn's period is 10 hours and 39 minutes, equivalent to $3.83\times10^4s$, which is in between the periods of the inner and outer chunks.
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