Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.5 Work and Energy - Exercises - Page 318: 36

Answer

$1.99 \times 10^{10} \ J$

Work Step by Step

The work needed to increase the separation from a distance $r_1$ to a distance $r_2$ can be calculated as: $ Work \ done =\int_{r_1}^{r_2} \dfrac{G Mm}{r^2} \ dr \\=[\dfrac{-G Mm}{r^2}]_{r_1}^{r_2} \\=G Mm (\dfrac{1}{r_1}-\dfrac{1}{r_2})$ The satellite covers the distance from $r_1= r_e$to $r_2 =R_e +12,00,000$. Plug in the given data: $W= G Mm (\dfrac{1}{r_1}-\dfrac{1}{r_2}) \\=(6.67 \times 10^{-11} )(5.98 \times 10^{24}) (2000) $ By using a calculator, the required result is: $W \approx 1.99 \times 10^{10} \ J$
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