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: 56

Answer

(a) $\rho = \frac{3\pi}{GT^2}$ (b) $\rho = 5400 ~kg/m^3$

Work Step by Step

(a) The volume of a sphere is $V = \frac{4}{3}\pi r^3$ $T^2 = \frac{4\pi^2 r^3}{Gm}$ $\frac{m}{4\pi r^3} = \frac{\pi}{GT^2}$ $\frac{m}{\frac{4}{3}\pi r^3} = \frac{3\pi}{GT^2}$ $\frac{m}{V} = \frac{3\pi}{GT^2}$ This works for an orbit that is very near the surface of a planet because in that case, the orbital radius r is approximately the same as the radius of the planet. (b) $\rho = \frac{3\pi}{GT^2} = \frac{3\pi}{(6.67 \times 10^{-11}~N\cdot m^2/kg^2)(85*60 ~s)^2}$ $\rho = 5400 ~kg/m^3$
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