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: 28

Answer

Please see the work below.

Work Step by Step

We know that the initial energy of the block is given as $E_i=U+K=U+0=U=\frac{-GM_Em}{R_E}$ We plug in the known values to obtain: $E_i=\frac{-(6.67\times 10^{-11})(5.97\times 10^{24})(1)}{6.37\times 10^6}=-6.2512\times 10^7J$ The final energy is given as $E_f=\frac{-GM_Em}{2r}$ We plug in the known values to obtain: $E_f=\frac{-(6.67\times 10^{-11})(5.97\times 10^{24})(1)}{(2)(4.22\times 10^7)}=-4.718\times 10^6J$ Now we can find the required energy as $\Delta E=(-4.718\times 10^6)-(-6.2512\times 10^7)=5.8\times 10^7J$
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