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 - Problems - Page 134: 49

Answer

$7.05\times10^3s$

Work Step by Step

The speed of an object in a circular orbit of radius $r$ around mass $M$ is given by $v=\sqrt{\frac{GM}{r}}$ and also $v=\frac{2\pi r}{T}$, where $T$ is the period of orbiting object. Equate both expressions and solve for $T$. $$v=\sqrt{\frac{GM}{r}}=\frac{2\pi r}{T}$$ $$T=2\pi \sqrt{\frac{r^3}{GM}}=2\pi \sqrt{\frac{(1.74\times10^6m+9.5\times10^4m)^3}{(6.67\times10^{-11}Nm^2/kg^2)(7.35\times10^{22}m)}}$$ $$=7.05\times10^3s$$
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