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: 26

Answer

(a) The orbital period of satellite 1 is 250 minutes. The orbital period of satellite 3 is 459 minutes. (b) The force on satellite 2 is 20,000 N The force on satellite 3 is 4440 N (c) $\frac{K_1}{K_3} = 1.5$

Work Step by Step

(a) We can write an expression for the orbital period $T_1$ of satellite 1. Let $M_p$ be the mass of the planet. $T_1^2 = \frac{4\pi^2~(2R)^3}{G~M_p}$ $T_1^2 = 8\times ~\frac{4\pi^2~R^3}{G~M_p}$ $T_1 = \sqrt{8\times ~\frac{4\pi^2~R^3}{G~M_p}}$ Since the orbital period $T_2$ of satellite 2 has the same orbital radius as satellite 1, the orbital period of satellite 2 is also 250 minutes. We can write an expression for the orbital period $T_3$ of satellite 3. $T_3^2 = \frac{4\pi^2~(3R)^3}{G~M_p}$ $T_3^2 = 27\times ~\frac{4\pi^2~R^3}{G~M_p}$ $T_3 = \sqrt{27\times ~\frac{4\pi^2~R^3}{G~M_p}}$ We can divide $T_3$ by $T_1$. $\frac{T_3}{T_1} = \sqrt{\frac{27}{8}}$ $T_3 = \sqrt{\frac{27}{8}}~T_1$ $T_3 = \sqrt{\frac{27}{8}}~(250~minutes)$ $T_3 = 459~minutes$ The orbital period of satellite 3 is 459 minutes. (b) We can write an expression for the force on satellite 1. $F_1 = \frac{G~M_p~m}{(2R)^2}$ $F_1 = \frac{1}{4}\times ~\frac{G~M_p~m}{R^2}$ We can write an expression for the force on satellite 2. $F_2 = \frac{G~M_p~(2m)}{(2R)^2}$ $F_2 = \frac{1}{2}\times ~\frac{G~M_p~m}{R^2}$ $F_2 = 2F_1$ $F_2 = (2)(10,000~N)$ $F_2 = 20,000~N$ The force on satellite 2 is 20,000 N We can write an expression for the force on satellite 3. $F_3 = \frac{G~M_p~m}{(3R)^2}$ $F_3 = \frac{1}{9}\times ~\frac{G~M_p~m}{R^2}$ We can divide $F_3$ by $F_1$. $\frac{F_3}{F_1} = \frac{1/9}{1/4}$ $\frac{F_3}{F_1} = \frac{4}{9}$ $F_3 = \frac{4}{9}~F_1$ $F_3 = \frac{4}{9}~(10,000~N)$ $F_3 = 4440~N$ The force on satellite 3 is 4440 N (c) The orbital speed of an object around a planet is $v = \sqrt{\frac{G~M_p}{R_0}}$, where $R_0$ is the orbital radius. We can use the speed $v$ to find an expression for the kinetic energy of an object of mass $M_0$ in orbit around a planet. $K = \frac{1}{2}M_0v^2$ $K = \frac{1}{2}M_0~(\frac{G~M_p}{R_0})$ $K = \frac{G~M_p~M_0}{2R_0}$ We can write an expression for the kinetic energy of satellite 1. $K_1 = \frac{G~M_p~m}{(2)(2R)}$ $K_1 = \frac{1}{4}\times \frac{G~M_p~m}{R}$ We can write an expression for the kinetic energy of satellite 3. $K_3 = \frac{G~M_p~m}{(2)(3R)}$ $K_3 = \frac{1}{6}\times \frac{G~M_p~m}{R}$ We can find the ratio of $\frac{K_1}{K_3}$. $\frac{K_1}{K_3} = \frac{1/4}{1/6}$ $\frac{K_1}{K_3} = \frac{6}{4}$ $\frac{K_1}{K_3} = 1.5$
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