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 135: 61

Answer

$1.8\times10^8$ years

Work Step by Step

For an object in circular motion around mass M, there are two expression for the velocity; $v=\sqrt{\frac{GM}{r}}$ and $v=\frac{2\pi r}{T}$. Equate the two expressions and solve for $T$. $$v=\sqrt{\frac{GM}{r}}=\frac{2\pi r}{T}$$ $$T=s\pi \sqrt{\frac{r^3}{GM}}=2\pi \sqrt{\frac{(3\times10^4ly\frac{(3\times10^8m/s)(3.16\times10^7s)}{1ly})^3}{(6.67\times10^{-11}Nm^2/kg^2)(4\times10^{41}kg)}}=5.8\times10^{15}s$$ $$=1.8\times10^8yr$$
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