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: 53

Answer

The energy required to move the earth to the new orbit is $1.76\times 10^{25}~J$

Work Step by Step

We can find the change in mechanical energy when the earth moves to an orbit with a radius that is 1.0 km larger. Let $M_e$ be the earth's mass and let $M_s$ be the sun's mass. Let $R$ be the earth's orbital radius. $\Delta E_{mech} = \frac{1}{2}\Delta U$ $\Delta E_{mech} = \frac{1}{2}(U_f-U_0)$ $\Delta E_{mech} = \frac{1}{2}[-\frac{G~M_e~M_s}{R+1.0~km}-(-\frac{G~M_e~M_s}{R})]$ $\Delta E_{mech} = \frac{1}{2}(\frac{G~M_e~M_s}{R}-\frac{G~M_e~M_s}{R+1.0~km})$ $\Delta E_{mech} = \frac{1}{2}(G~M_e~M_s)(\frac{1}{R}-\frac{1}{R+1.0~km})$ $\Delta E_{mech} = \frac{1}{2}(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)(1.99\times 10^{30}~kg)(\frac{1}{1.50\times 10^{11}~m}-\frac{1}{1.50000001\times 10^{11}~m})$ $\Delta E_{mech} = 1.76\times 10^{25}~J$ The energy required to move the earth to the new orbit is $1.76\times 10^{25}~J$
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