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: 34

Answer

The object's speed as it hits the ground is $\sqrt{\frac{2~G~M~h}{R^2+R~h}}$

Work Step by Step

We can use conservation of energy to find the object's speed as it hits the ground. $K_f+U_f = K_0+U_0$ $K_f = 0+U_0-U_f$ $\frac{1}{2}mv^2 = -\frac{G~M~m}{R+h}-(-\frac{G~M~m}{R})$ $\frac{1}{2}mv^2 = \frac{G~M~m}{R}-\frac{G~M~m}{R+h}$ $\frac{1}{2}v^2 = \frac{(G~M)(R+h)}{(R)(R+h)}-\frac{G~M~R}{(R)(R+h)}$ $v^2 = \frac{2~G~M~h}{R^2+R~h}$ $v = \sqrt{\frac{2~G~M~h}{R^2+R~h}}$ The object's speed as it hits the ground is $\sqrt{\frac{2~G~M~h}{R^2+R~h}}$
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