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 149: 72

Answer

12,982 meters

Work Step by Step

We need to make the asteroid so that the velocity of the jumper is below the escape speed of the asteroid. Thus, we first find how fast a jumper that goes 3 meters up on earth is jumping using conservation of energy: $\frac{1}{2}mv^2 = mgh \\ v=\sqrt{2gh}=\sqrt{2\times9.81\times3}=7.672 \ m/s$ We know the following equation for escape speed: $v_{esc}=\sqrt{\frac{2GM}{r}}$ $v_{esc}=\sqrt{\frac{2G(\frac{4}{3}\pi r^3)(2500)}{r}}$ $ r = \sqrt{\frac{v_{esc}^2}{6666.67\pi G}}$ Plugging in the value for G (a constant) and v (found above) gives: $r=9491 m$ This means that: $\fbox{d=12,982 meters}$
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