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: 24

Answer

The speed of the satellite is 1.44 km/s and the altitude above the surface of Mars is $1.71\times 10^7~m$.

Work Step by Step

We can convert the orbital period $T$ to units of seconds. $T = (24.8~hours)(3600~s/hr)$ $T = 89,280~s$ We then use the orbital period and the mass of Mars $M_m$ to find the orbital radius $R$: $T^2 = \frac{4\pi^2~R^3}{G~M_m}$ $R^3 = \frac{G~M_m~T^2}{4\pi^2}$ $R = (\frac{G~M_m~T^2}{4\pi^2})^{1/3}$ $R = (\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(6.42\times 10^{23}~kg)(89,280~s)^2}{4\pi^2})^{1/3}$ $R = 2.05\times 10^7~m$ We then use the radius and the orbital period to find the speed of the satellite; $v = \frac{distance}{time}$ $v = \frac{2\pi~R}{T}$ $v = \frac{(2\pi)(2.05\times 10^7~m)}{89,280~s}$ $v = 1443~m/s \approx 1.44~km/s$ The radius of the orbit is $2.05\times 10^5~m$. We can use the radius of Mars $R_m$ to find the satellite’s altitude $h$ above the surface of Mars. $R_m+h = R$ $h = R - R_m$ $h = (2.05\times 10^7~m) - (3.39\times 10^6~m)$ $h = 1.71\times 10^7~m$ The speed of the satellite is 1.44 km/s and the altitude above the surface of Mars is $1.71\times 10^7~m$.
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