College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 229: 42


We need a launch speed of 2.38 km/s

Work Step by Step

We need to find the escape speed from the moon's surface. We can assume that kinetic energy and gravitational potential energy after a long time are both equal to zero. We can use conservation of energy to find the escape speed: $KE+U_g = 0$ $\frac{1}{2}mv^2-\frac{G~M_{moon}~m}{R_{moon}} = 0$ $\frac{1}{2}mv^2 = \frac{G~M_{moon}~m}{R_{moon}}$ $\frac{1}{2}v^2 = \frac{G~M_{moon}}{R_{moon}}$ $v = \sqrt{\frac{2G~M_{moon}}{R_{moon}}}$ $v = \sqrt{\frac{(2)(6.67\times 10^{-11}~m^3/kg~s^2)~(7.35\times 10^{22})}{1.737\times 10^6~m}}$ $v = 2380~m/s = 2.38~km/s$ We need a launch speed of 2.38 km/s.
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