Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 13 - Gravitation - Problems - Page 383: 68b

Answer

$v_0 = 7.68\times 10^3~m/s$

Work Step by Step

We can find the speed: $K = \frac{1}{2}mv_0^2 = \frac{GMm}{2r}$ $v_0^2 = \frac{GM}{r}$ $v_0 = \sqrt{\frac{GM}{r}}$ $v_0 = \sqrt{\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)}{6.37\times 10^6~m+4.0\times 10^5~m}}$ $v_0 = 7.68\times 10^3~m/s$
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