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(B) The gravitational potential energy of a system of three particles of mass $M$ at the corners of an equilateral triangle of side $a$ is given by th

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$-\frac{3GM^2}{a}, \frac{3GM^2}{2a}$

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(B) The gravitational potential energy of a system of three particles of mass $M$ at the corners of an equilateral triangle of side $a$ is given by the sum of potential energies of all pairs: $U_1 = -\frac{GM^2}{a} - \frac{GM^2}{a} - \frac{GM^2}{a} = -\frac{3GM^2}{a}$.When the side length is changed to $2a$,the new potential energy is $U_2 = -\frac{3GM^2}{2a}$.The work done by an external agent is equal to the change in potential energy: $W = U_2 - U_1$.$W = -\frac{3GM^2}{2a} - (-\frac{3GM^2}{a}) = -\frac{3GM^2}{2a} + \frac{3GM^2}{a} = \frac{3GM^2}{2a}$.

If three particles each of mass $M$ are placed at the corners of an equilateral triangle of side $a$,the potential energy of the system and the work done if the side of the triangle is changed from $a$ to $2a$,are

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