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 147: 27

Answer

$v=4300\frac{m}{s}$

Work Step by Step

According to law of conservation of energy $\frac{1}{2}mv^2=Gm_Em(\frac{1}{r_1}-\frac{1}{r_2})$ This simplifies to: $v=\sqrt{2Gm_E(\frac{1}{r_1}-\frac{1}{r_2})}$ We plug in the known values to obtain: $v=\sqrt{(2)(6.67\times 10^{-11})(5.97\times 10^{24})(\frac{1}{6.37\times 10^6}-\frac{1}{6.37\times 10^6+1100\times 10^3})}$ $v=4300\frac{m}{s}$
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