Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 11 - Impulse and Momentum - Exercises and Problems: 45

Answer

(a) The recoil speed of the earth is $6.7\times 10^{-8}~m/s$. (b) The recoil speed is $2.23\times 10^{-10}~\%$ the earth's speed around the sun.

Work Step by Step

(a) We can use conservation of momentum to find the earth's recoil speed $v_f$. Let $m_e$ be the mass of the earth. Let $m_a$ be the mass of the asteroid. $p_f = p_0$ $(m_e+m_a)~v_f = m_a~v_0$ $v_f = \frac{m_a~v_0}{m_e+m_a}$ $v_f = \frac{(1.0\times 10^{13}~kg)(4.0\times 10^4~m/s)}{5.97\times 10^{24}~kg}$ $v_f = 6.7\times 10^{-8}~m/s$ The recoil speed of the earth is $6.7\times 10^{-8}~m/s$. (b) We can find the earth's speed as it revolves around the sun. $v = \frac{distance}{time}$ $v = \frac{2\pi ~r}{1~year}$ $v = \frac{(2\pi)(1.5\times 10^{11}~m)}{(1~year)(365~days/year)(24~hr/day)(3600~s/day)}$ $v = 3.0\times 10^4~m/s$ We can find the recoil speed as a percentage of the earth's speed as it revolves around the sun. $\frac{6.7\times 10^{-8}~m/s}{3.0\times 10^4~m/s} \times 100\%$ $= 2.23\times 10^{-10}~\%$ The recoil speed is $2.23\times 10^{-10}~\%$ of the earth's speed around the sun.
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