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 13 - Newton's Theory of Gravity - Exercises and Problems: 51

Answer

The work done by the thrusters is $6.73\times 10^{8}~J$

Work Step by Step

We can find the change in mechanical energy when the lander moves to a higher orbit. Let $M_m$ be the moon's mass and let $M_l$ be the lander's mass. Let $R$ be the moon's radius. $\Delta E_{mech} = \frac{1}{2}\Delta U$ $\Delta E_{mech} = \frac{1}{2}(U_f-U_0)$ $\Delta E_{mech} = \frac{1}{2}[-\frac{G~M_m~M_l}{R+300~km}-(-\frac{G~M_m~M_l}{R+50~km})]$ $\Delta E_{mech} = \frac{1}{2}(\frac{G~M_m~M_l}{R+50~km}-\frac{G~M_m~M_l}{R+300~km})$ $\Delta E_{mech} = \frac{1}{2}(G~M_m~M_l)(\frac{1}{R+50~km}-\frac{1}{R+300~km})$ $\Delta E_{mech} = \frac{1}{2}(6.67\times 10^{-11}~m^3/kg~s^2)(7.35\times 10^{22}~kg)(4000~kg)(\frac{1}{1.787\times 10^6~m}-\frac{1}{2.037\times 10^6~m})$ $\Delta E_{mech} = 6.73\times 10^{8}~J$ The work done by the thrusters is equal to the change in mechanical energy. Therefore, the work done by the thrusters is $6.73\times 10^{8}~J$
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