Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 8 - Exercises and Problems: 32

Answer

Please see the work below.

Work Step by Step

We know that the escape speed is given as $v=\sqrt{\frac{2GM}{r}}$ This simplifies to: $r=\frac{2GM}{v^2}$ We plug in the known values to obtain: $r=\frac{(2)(6.67\times 10^{-11})(5.97\times 10^{24})}{(30000)^2}$ $r=8.8\times 10^5m$
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