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: 123


$v_{Ac}=5.27\times 10^3m/s$, $\Delta v=684m/s$

Work Step by Step

We can determine the required speed at $A$ and the change in the speed as follows: $v_{Ac}=\sqrt{\frac{GM_e}{h_{\circ}+r_e}}$ We plug in the known values to obtain: $v_{Ac}=\sqrt{\frac{66.73\times 10^{-12}\times 5.976\times 10^{24}}{8\times 10^6+6378\times 10^3}}$ This simplifies to: $v_{Ac}=5266.43m/s=5.27\times 10^3m/s$ Now we determine the change in speed as $v_p=\sqrt{\frac{2GM_er_a}{r_p(r_p+r_a)}}$ We plug in the known values to obtain: $v_p=\sqrt{\frac{2\times 66.73\times 10^{-12}\times 5.976\times 10^{24}\times 25.378\times 10^6}{14.378\times 10^6(14.378\times 10^6+25.378\times 10^6)}}$ $\implies v_p=5950.58m/s$ We know that $v_p=v_{Ae}$ $\implies v_{Ae}=5950.58m/s$ Now $\Delta v=v_{Ae}-v_{Ac}$ We plug in the known values to obtain: $\Delta v=5950.58-5266.43$ $\implies \Delta v=684m/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.