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: 58

Answer

$v=\sqrt{2GM(\frac{3}{R}-\frac{1}{r_1})}$

Work Step by Step

We use conservation of energy to find: $\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=GMm(\frac{1}{R}-\frac{1}{r_1})$ $\frac{1}{2}v^2-\frac{1}{2}v_0^2=GM(\frac{1}{R}-\frac{1}{r_1})$ $v=\sqrt{2GM(\frac{1}{R}-\frac{1}{r_1})+v_0^2}$ We now need to substitute a value for $v_0^2$. We know that escape speed equals: $\sqrt{\frac{2GM}{R}}$ Since the actual speed is $\sqrt{2}$ greater, we find: $v_0^2 =\frac{4GM}{R}$. Plugging this into the above equation gives: $v=\sqrt{2GM(\frac{3}{R}-\frac{1}{r_1})}$
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