Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 12 - Gravity - Problems and Conceptual Exercises - Page 414: 102

Answer

(a) increased by $\sqrt 2$ (b) $15813m/s$ (c) same

Work Step by Step

(a) We know that $v_e=\sqrt{\frac{2GM_E}{R_E}}$ Given that the mass is doubled then $v_e=\sqrt{\frac{2G(2M_E)}{R_E}}$ $\implies v_e=\sqrt{\frac{4GM_E}{R_E}}$ Thus, the escape speed is increased by a factor of $\sqrt 2$. (b) We can calculate the required escape speed as $v_e=\sqrt{\frac{4GM_E}{R_E}}$ We plug in the known values to obtain: $v_e=\sqrt{\frac{4(6.67\times 10^{-11}N.m^2/Kg^2)(5.97\times 10^{24}Kg)}{6.37\times 10^6m}}$ $v_e=15813m/s$ (c) We know that $v_e=\sqrt{\frac{2GM_E}{R_E}}$. This equation shows that the escape speed depends on mass and not on the mass of the rocket. Thus, the escape speed remains the same.
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