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 - Page 354: 19

Answer

The escape speed is 4.34 km/s

Work Step by Step

We can use the equation for escape speed to find the escape speed at a height of 36,000 km above the earth's surface. Note that at this height, the distance $R$ from the center of the earth is 42,380 km. $v_{esc} = \sqrt{\frac{2~G~M}{R}}$ $v_{esc} = \sqrt{\frac{(2)(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{4.238\times 10^7~m}}$ $v_{esc} = 4.34\times 10^3 ~m/s$ $v_{esc} = 4.34~km/s$ The escape speed is 4.34 km/s.
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