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: 7

Answer

$7.8\times10^{-4o}$

Work Step by Step

a) $\theta=\arctan(\frac{F_M}{mg})$ $F_M=G\frac{m_1m_2}{r^2}=6.67\times10^{-11}\frac{Nm^2}{kg^2}\frac{mm_M}{(D_M)^2}$ $mg=G\frac{m_1m_2}{r^2}=6.67\times10^{-11}\frac{Nm^2}{kg^2}\frac{mm_E}{(r_E)^2}$ $\theta=\arctan\Big(\frac{mm_M}{(D_M)^2}\times\frac{(r_E)^2}{mm_E}\Big)=\arctan\Big(\frac{m_M(r_E)^2}{m_E(D_M)^2}\Big)$ b) $V=\frac{\pi r^2h}{3}=\frac{\pi(2000m)^2(4000m)}{3}=1.7\times10^{10}m^3$ $m=d\times V=(3000\frac{kg}{m^3})\times(1.7\times10^{10}m^3)$ $=5\times10^{13}kg$ c) $\arctan\Big(\frac{m_M(r_E)^2}{m_E(D_M)^2}\Big)=\arctan\Big(\frac{(5\times10^{13}kg)(6.38\times10^6m)^2}{(5.98\times10^{24}kg)(5\times10^3m)^2}\Big)=7.8\times10^{-4o}$
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