Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 8 - Section 8.4 - Gravitational Energy - Example - Page 136: 8.5


$6.9\ Mm$

Work Step by Step

We know that kinetic energy plus potential energy is constant. Using this fact and solving for r, we find: $r = (\frac{1}{R_E}-\frac{v_0^2}{2GM_E})^{-1}=(\frac{1}{6.37 \times 10^6}-\frac{3100^2}{2(6.67\times10^{-11}\times5.97 \times10^{24})})^{-1} = 6.9 \ Mm$
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