Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.7 - Central-Force Motion and Space Mechanics - Problems - Page 172: 122


$r_{\circ}=11.1Mm$, $\Delta v=814m/s$

Work Step by Step

We can determine the $r_{\circ}$ and the change in velocity as follows: $r_{\circ}=\frac{2GM_M}{v^2_A}$ We plug in the known values to obtain: $r_{\circ}=\frac{2\times 66.73\times 10^{-12}\times 0.1074\times 5.976\times 10^{24}}{(10\times \frac{10^6}{3600})^2}$ This simplifies to: $r_{\circ}=11.1\times 10^6=11.1Mm$ Now $v_{Ac}=\sqrt{\frac{66.73\times 10^{-12}(0.1074)\times 5.976\times 10^{24}}{11.1\times 10^6}}$ $\implies v_{Ac}=1964.19m/s$ $\Delta v=v_{Ac}-v_A$ $\implies \Delta v=1964.19-(10\times\frac{10^6}{3600})$ This simplifies to: $\Delta v=814m/s$
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