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 354: 29


The mass of the planet is $1.01\times 10^{26}~kg$

Work Step by Step

We can find the distance $d$ that the moon travels when it completes one orbit. $d = v~t$ $d = (7500~m/s)(28~hr)(3600~s/hr)$ $d = 7.56\times 10^8~m$ We can find the orbital radius $R$. $2\pi R = d$ $R =\frac{d}{2\pi}$ $R = \frac{7.56\times 10^8~m}{2\pi}$ $R = 1.20\times 10^8~m$ We can use the equation for orbital speed to find the mass of the planet. $v = \sqrt{\frac{G~M}{R}}$ $v^2 = \frac{G~M}{R}$ $M = \frac{v^2~R}{G}$ $M = \frac{(7500~m/s)^2(1.20\times 10^8~m)}{6.67\times 10^{-11}~m^3/kg~s^2}$ $M = 1.01\times 10^{26}~kg$ The mass of the planet is $1.01\times 10^{26}~kg$
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