Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 8 - Exercises and Problems - Page 148: 53

Answer

The proof is below.

Work Step by Step

We know that the gravitational potential energy is: $\frac{GMm}{r}$. We set this equal to kinetic energy, since the radius of the earth is essentially negligible compared to infinite distance: $\frac{GMm}{r}=\frac{1}{2}mv^2$. $v = \sqrt{\frac{2GM}{r}}$ This is more than the escape speed, which is $\sqrt{\frac{2GM}{r}}$.
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