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: 60

Answer

a) $14,833 \ m/s$ b) $22,804 \ m/s$

Work Step by Step

We use the law of conservation of energy to obtain: $\frac{1}{2}mv^2-\frac{1}{2}mv_0^2 =GMm(\frac{1}{r_1}-\frac{1}{r_2})$ $\frac{1}{2}v^2-\frac{1}{2}v_0^2 =GM(\frac{1}{r_1}-\frac{1}{r_2})$ $v^2-v_0^2 =2GM(\frac{1}{r_1}-\frac{1}{r_2})$ $v^2 =2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_0^2$ $v =\sqrt{2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_0^2}$ Using this, we find that the two speeds of impact are: a) $14,833 \ m/s$ b) $22,804 \ m/s$
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