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 319: 42


$626 m/s$

Work Step by Step

The work needed to move the rocket a distance $r$ from the surface of the earth can be calculated as: $ Work \ done =G M_e \ m (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})$ The rocket will attain its maximum height when its kinetic energy reduces to $0$. That is, $\dfrac{1}{2} m V_0^2 =G M_e \ m (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})$ Now, the initial velocity will be $V_0=\sqrt {\dfrac{2G M_e \ m }{m} (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})}$ or, $V_0=\sqrt { 2G M_e(\dfrac{1}{R_e}-\dfrac{1}{r+R_e})}$ Plug in the given data: $V_0=\sqrt { 2 (6.67 \times 10^{-11}) (5.98 \times 10^{24}) (\dfrac{1}{6.37 \times 10^{6}}-\dfrac{1}{20000+6.37 \times 10^{6}})}$ $=626~m/s$
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