Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems: 38

Answer

Since the total energy in the system is positive, the person will be able to escape from the asteroid.

Work Step by Step

We first find the initial speed of the jump on the earth. $v^2 = v_0^2+2gy$ $v_0^2= 0 - 2gy$ $v_0 = \sqrt{-(2)(-9.80~m/s^2)(0.50~m)}$ $v_0 = 3.13~m/s$ We then find an expression for the initial kinetic energy when jumping on the earth. Let $M_p$ be the person's mass. $K_0 = \frac{1}{2}M_p~v_0^2$ Note that on the asteroid, the initial kinetic energy will be the same. We can find the total energy in the system. Let $M_a$ be the asteroid's mass. $E = K_0+U_0$ $E = \frac{1}{2}M_p~v_0^2-\frac{G~M_a~M_p}{R}$ $E = \frac{1}{2}M_p~(3.13~m/s)^2-\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.0\times 10^{14}~kg)~M_p}{2000~m}$ $E = (4.90~m^2/s^2)~M_p-(3.34~m^2/s^2)~M_p$ $E = (1.56~M_p)~J$ Since the total energy in the system is positive, the person will be able to escape from the asteroid.
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