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 171: 119


$v_A=6.67\times 10^3m/s$,$v_B=2.77\times 10^3m/s$

Work Step by Step

We can determine the rocket's speed at $A$ and $B$ as follows: $v_A=\sqrt{\frac{2GMr_a}{r_p(r_p+r_a)}}$ We plug in the known values to obtain: $v_A=\sqrt{\frac{2\times 66.73(10^{-12})\times 0.6\times 5.97\times 10^{24}}{7.6\times 10^6(7.6\times 10^6+1.38\times 10^6)}}$ This simplifies to: $v_A=6670m/s=6.67\times 10^3m/s$ Similarly, $v_B=\frac{r_p v_A}{r_a}$ We plug in the known values to obtain: $v_B=\frac{7.6\times 10^3\times 6670}{13.8\times 10^3}$ This simplifies to: $v_B=2770m/s=2.77\times 10^3m/s$
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