Three masses $200\ \text{kg}, 300\ \text{kg}$ 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 $\text{G} = 6.7\times10^{-11}\ \text{N m}^2/\text{kg}^2$ )
Answer: (B) $1.74\times10^{-7}$
Potential energy of the system with side $a$:
$$U = -\frac{G(m_1m_2 + m_2m_3 + m_3m_1)}{a},\quad m_1m_2 + m_2m_3 + m_3m_1 = (6 + 12 + 8)\times10^4 = 2.6\times10^5\ \text{kg}^2$$
Work done $= U_f - U_i$:
$$W = G(2.6\times10^5)\left(\frac1{20} - \frac1{25}\right) = 6.7\times10^{-11}\times2.6\times10^5\times0.01 = 1.74\times10^{-7}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics