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: 31

Answer

Please see the work below.

Work Step by Step

We know that (a) The escape speed is given as $v=\sqrt{\frac{2GM}{r}}$ We plug in the known values to obtain: $v=\sqrt{\frac{(2)(6.67\times 10^{-11})(0.107\times 10^{24})}{2.40\times 10^6}}$ $v=2400\frac{m}{s}$ (b) As we know that $v=\sqrt{\frac{2GM}{r}}$ We plug in the known values to obtain: $v=\sqrt{\frac{(2)(6.67\times 10^{-11})(1.99\times 10^{30})}{6\times 10^3}}$ $v=2.1\times 10^8\frac{m}{s}$
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