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: 29

Answer

The total gravitational potential energy of the three masses is $-1.96\times 10^{-7}~J$

Work Step by Step

We first find the distance $r_1$; $r_1 = \sqrt{(5.0~cm)^2+(20.0~cm)^2}$ $r_1 = 20.6~cm$ Note that $r_3$ is also 20.6 cm. We then find the gravitational potential energy of $m_1$ and $m_2$: $U_{1,2} = -\frac{G~m_1~m_2}{r_1}$ $U_{1,2} = -\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(20.0~kg)(10.0~kg)}{(0.206~m)}$ $U_{1,2} = -6.48\times 10^{-8}~J$ Next, we find the gravitational potential energy of $m_2$ and $m_3$: $U_{2,3} = -\frac{G~m_2~m_3}{r_2}$ $U_{2,3} = -\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(10.0~kg)(10.0~kg)}{(0.10~m)}$ $U_{2,3} = -6.67\times 10^{-8}~J$ We then find the gravitational potential energy of $m_1$ and $m_3$. $U_{1,3} = -\frac{G~m_1~m_3}{r_3}$ $U_{1,3} = -\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(20.0~kg)(10.0~kg)}{(0.206~m)}$ $U_{1,3} = -6.48\times 10^{-8}~J$ To find the total gravitational potential energy, we add the gravitational potential energy of each pair of masses. $U_{total} = U_{1,2}+U_{2,3}+U_{1,3}$ $U_{total} = (-6.48\times 10^{-8}~J)+(-6.67\times 10^{-8}~J)+(-6.48\times 10^{-8}~J)$ $U_{total} = -1.96\times 10^{-7}~J$ The total gravitational potential energy of the three masses is $-1.96\times 10^{-7}~J$.
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