Three masses $200 \mathrm{~kg}, 300 \mathrm{~kg}$ and 400 kg are placed at the vertices of an equila

Question

Three masses $200 \mathrm{~kg}, 300 \mathrm{~kg}$ and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m . They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process $\_\_\_\_$ J.
(Gravitational constant $\mathrm{G}=6.7 	imes 10^{-11} \mathrm{~N} \mathrm{~m}^2 / \mathrm{kg}^2$ )

Professor PC Thomas Classes · Answer

The gravitational potential energy of the system of three masses separated by distance $a$ is $U = -G\frac{m_1 m_2 + m_2 m_3 + m_3 m_1}{a}$. Let $M_{prod} = (200)(300) + (300)(400) + (400)(200) = 60000 + 120000 + 80000 = 260000 	ext{ kg}^2$. The work done by an external agent is the change in potential energy: $W = U_f - U_i = -G \cdot M_{prod} (\frac{1}{a_f} - \frac{1}{a_i}) = G \cdot M_{prod} (\frac{1}{a_i} - \frac{1}{a_f})$. Substituting the given values: $W = 6.7 	imes 10^{-11} 	imes 260000 	imes (\frac{1}{20} - \frac{1}{25}) = 1.742 	imes 10^{-5} 	imes (0.05 - 0.04) = 1.742 	imes 10^{-7} 	ext{ J}$.