Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 355: 56


The radius of the orbit is $6.71\times 10^7~m$

Work Step by Step

The period of the satellite's orbit is 48 hours. We can convert the period $T$ to units of seconds. $T = (48~hrs)(3600~s/hr)$ $T = 172,800~s$ We can use the orbital period and the mass of the earth $M_e$ to find the orbital radius $R$. $T^2 = \frac{4\pi^2~R^3}{G~M_e}$ $R^3 = \frac{G~M_e~T^2}{4\pi^2}$ $R = (\frac{G~M_e~T^2}{4\pi^2})^{1/3}$ $R = (\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)(172,800~s)^2}{4\pi^2})^{1/3}$ $R = 6.71\times 10^7~m$ The radius of the orbit is $6.71\times 10^7~m$
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