Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 373: 18


The mass of the sun is $2.01\times 10^{30}~kg$.

Work Step by Step

We can convert the earth's orbital period to units of seconds. $T = (1.0~year)[\frac{(365~days)(24~hours/day)(3600~s/hr)}{1~year}]$ $T = 3.154\times 10^7~s$ We then use the earth's orbital period $T$ and the earth's orbital radius $R$ to find the mass of the sun $M_s$. $T^2 = \frac{4\pi^2~R^3}{G~M_s}$ $M_s = \frac{4\pi^2~R^3}{G~T^2}$ $M_s = \frac{(4\pi^2)(1.50\times 10^{11}~m)^3}{(6.67\times 10^{-11}~m^3/kg~s^2)(3.154\times 10^7~s)^2}$ $M_s = 2.01\times 10^{30}~kg$ The mass of the sun is $2.01\times 10^{30}~kg$.
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