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: 85

Answer

a) $r=7.61\times10^6m$ b) $F_G=3.79\times10^4N$ c) $alt=1.22\times10^6m$

Work Step by Step

a) $r=^3\sqrt{\frac{T^2Gm}{4\pi^2}}=^3\sqrt{\frac{(6600s)^2(6.67\times10^{-11}\frac{Nm^2}{kg^2})(5.98\times10^{24}kg)}{4\pi^2}}$ $=7.61\times10^6m$ b) $F_G=G\frac{m_sm_E}{r^2}=(6.67\times10^{-11}\frac{Nm^2}{kg^2})\frac{(5500kg)(5.98\times10^{24})}{7.61\times10^6m}=3.79\times10^4N$ c) $alt=r-r_E=7.61\times10^6m-6.38\times10^6m$ $=1.22\times10^6m$
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