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(C) The gravitational potential energy $U$ of a system of three particles is given by:$U = -\frac{G m_1 m_2}{r_{12}} - \frac{G m_2 m_3}{r_{23}} - \fra

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$1.00 	imes 10^{-11} \, J$

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(C) The gravitational potential energy $U$ of a system of three particles is given by:$U = -\frac{G m_1 m_2}{r_{12}} - \frac{G m_2 m_3}{r_{23}} - \frac{G m_1 m_3}{r_{13}}$Given $m_1 = m_2 = m_3 = 100 \, g = 0.1 \, kg$ and $r_{12} = r_{23} = r_{13} = 20 \, cm = 0.2 \, m$.Since all masses and distances are equal,$U = -3 	imes \frac{G m^2}{r}$.$U = -3 	imes \frac{6.67 	imes 10^{-11} 	imes (0.1)^2}{0.2}$$U = -3 	imes \frac{6.67 	imes 10^{-11} 	imes 0.01}{0.2} = -3 	imes 3.335 	imes 10^{-12} = -1.00 	imes 10^{-11} \, J$.The work required to move the particles to infinity is $W = U_{\infty} - U_{initial} = 0 - (-1.00 	imes 10^{-11} \, J) = 1.00 	imes 10^{-11} \, J$.

Particles of mass $100 \, g$ each are placed at the vertices of an equilateral triangle of side $20 \, cm$. How much work must be done to increase the separation between them to infinity?

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