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 - Search and Learn - Page 137: 6

Answer

$2.79g$

Work Step by Step

$m=1.9\times10^{27}kg$ $r_{J}=7.1\times10^7m$ $T=35700s$ $g_J=G\frac{m_J}{r_J^2}=(6.67\times10^{-11}\frac{Nm^2}{kg^2})\frac{1.9\times10^{27}kg}{(7.1\times10^7m)^2}=25.1\frac{m}{s^2}$ $v=\frac{2\pi r}{T}=\frac{2\pi(7.1\times10^7m)}{35700s}=12500\frac{m}{s}$ $F_R=\frac{v^2}{r}=\frac{(12500\frac{m}{s})^2}{7.1\times10^7m}=2.20\frac{m}{s^2}$ $a_G=25.1\frac{m}{s^2}+2.20\frac{m}{s^2}=27.3\frac{m}{s^2}$ $\frac{27.3\frac{m}{s^2}}{9.8\frac{m}{s^2}}=2.79g$
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