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: 33

Answer

(a) The satellite's orbital speed is 6.95 m/s (b) The escape speed from the asteroid is 12.3 m/s

Work Step by Step

(a) Let $M_a$ be the asteroid's mass. Let $R$ be the asteroid's radius. Note that the satellite's orbital radius is $R+5.0~km$. We can use the orbital radius to find the orbital speed. $v = \sqrt{\frac{G~M_a}{R+5.0~km}}$ $v = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.0\times 10^{16}~kg)}{1.38\times 10^4~m}}$ $v = 6.95~m/s$ The satellite's orbital speed is 6.95 m/s (b) We can find the escape speed from the asteroid. $v_{esc} = \sqrt{\frac{2~G~M_a}{R}}$ $v_{esc} = \sqrt{\frac{(2)(6.67\times 10^{-11}~m^3/kg~s^2)(1.0\times 10^{16}~kg)}{8.8\times 10^3~m}}$ $v_{esc} = 12.3~m/s$ The escape speed from the asteroid is 12.3 m/s
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