Essential University Physics: Volume 1 (4th Edition)

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

Chapter 8 - Exercises and Problems - Page 147: 26


$h=1.7\times 10^6m$

Work Step by Step

We know that 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: $\frac{1}{2}v^2=Gm_E(\frac{1}{r_1}-\frac{1}{r_2})$ We plug in the known values to obtain: $\frac{1}{2}(5100)^2=6.67\times 10^{-11}(5.97\times 10^{24})(\frac{1}{6.37\times 10^6}-\frac{1}{6.37\times 10^6+h})$ This simplifies to: $h=1.7\times 10^6m$
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