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: 22


The radius of the orbit is $2.93\times 10^9~m$

Work Step by Step

We can convert the orbital period $T$ to units of seconds as: $T = (1.0~day)(24~hr/day)(3600~s/hr)$ $T = 86,400~s$ We then use the orbital period and the mass of the sun $M_s$ to find the orbital radius $R$; $T^2 = \frac{4\pi^2~R^3}{G~M_s}$ $R^3 = \frac{G~M_s~T^2}{4\pi^2}$ $R = (\frac{G~M_s~T^2}{4\pi^2})^{1/3}$ $R = (\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg)(86,400~s)^2}{4\pi^2})^{1/3}$ $R = 2.93\times 10^9~m$ The radius of the orbit is $2.93\times 10^9~m$.
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