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: 35

Answer

$5.644\space km/s$

Work Step by Step

Here we use the principle of conservation of mechanical energy. Final mechanical energy = Initial mechanical energy $K+U=K_{0}+U_{0}$ $-\frac{GM_{E}m}{r}=\frac{1}{2}mV_{0}^{2}-\frac{GM_{E}m}{R_{E}}$ Where m is the rocket's mass, r is the distance from Earth's center at the peak & Earth's radius $R_{E}$ is the distance at launch. Solving for r gives. $\frac{GM_{E}}{R_{E}}-\frac{1}{2}V_{0}^{2}=\frac{GM_{E}}{r}$ $2GM_{E}(\frac{1}{R_{E}}-\frac{1}{r})=V_{0}^{2} $; Let's plug known values into this equation. $V_{0}^{2}=2\times6.67\times10^{-11}Nm^{2}/ kg^{2}\times5.97\times10^{24}kg(\frac{1}{6.37\times10^{6}m}-\frac{1}{8.52\times10^{6}m})$ $V_{0}^{2}=79.64\times10^{7}(0.16-0.12)\space m^{2}/s^{2}$ $V_{0}=5644.1\space m/s= 5.644\space km/s$
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