Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.6 - Conservation of Energy - Problems - Page 228: 85



Work Step by Step

We can determine the required speed of the satellite as follows: We know that $v=\frac{-GM_em}{r}$ We plug in the known values to obtain: $v=\frac{-66.73\times 10^{-12}\times 5.976(10^{24})\times 60}{20\times 10^6}$ $\implies v=-1.196\times 10^9J$ Similarly $v_2=\frac{-66.73\times 10^{-12}\times 5.97\times 10^{24}\times 60}{80\times 10^6}=-0.299\times 10^9J$N We know that $T_1=\frac{1}{2}mv_A^2$ $\implies T_1=\frac{1}{2}\times 60\times (\frac{40\times 10^6}{3600})^2=3.7\times 10^9J$ According to the conservation of energy equation $T_1+v_1=T_2+v_2$ We plug in the known values to obtain: $3.7(10)^9-1.196(10)^9=30v_B^2-0.299(10)^9$ This simplifies to: $v_B=9666.1m/s=\frac{9666.1\times 3600}{10^6}=34.8Mm/h$
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