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: 17

Answer

The asteroid's orbital radius is $4.37\times 10^{11}~m$ The asteroid's orbital speed is 17.4 km/s

Work Step by Step

First, we convert the orbital period to units of seconds. $T = (5.0~years)[\frac{(365~days)(24~hours/day)(3600~s/hr)}{1~year}]$ $T = 1.577\times 10^8~s$ Then, we use the orbital period to find the orbital radius $R$. Let $M_s$ be the mass of the sun. $T^2 = \frac{4\pi^2~R^3}{G~M_s}$ $R^3 = \frac{T^2~G~M_s}{4\pi^2}$ $R = (\frac{T^2~G~M_s}{4\pi^2})^{1/3}$ $R = (\frac{(1.577\times 10^8~s)^2(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg}{4\pi^2})^{1/3}$ $R = 4.37\times 10^{11}~m$ The asteroid's orbital radius is $4.37\times 10^{11}~m$. We then use the orbital radius to find the orbital speed. $v = \sqrt{\frac{G~M_s}{R}}$ $v = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg)}{4.37\times 10^{11}~m}}$ $v = 1.74\times 10^4~m/s$ $v = 17.4~km/s$ The asteroid's orbital speed is 17.4 km/s.
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