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(D) The work done by the gravitational force $(W_g)$ is equal to the negative change in the gravitational potential energy of the particle $D$ due to

Vedclass · Accepted Answer

$-\frac{G m^2}{L} \left( \frac{2 \sqrt{2} + 1}{\sqrt{2}} \right)$

Vedclass · Answer

(D) The work done by the gravitational force $(W_g)$ is equal to the negative change in the gravitational potential energy of the particle $D$ due to the other particles $(A, B, C)$.$W_g = -\Delta U = -(U_{	ext{final}} - U_{	ext{initial}}) = U_{	ext{initial}} - U_{	ext{final}}$.When the particle $D$ is at infinity,the potential energy $U_{	ext{final}} = 0$.Therefore,$W_g = U_{	ext{initial}}$.The initial gravitational potential energy of particle $D$ due to particles $A, B,$ and $C$ is:$U_{	ext{initial}} = -\frac{G m_A m_D}{r_{AD}} - \frac{G m_B m_D}{r_{BD}} - \frac{G m_C m_D}{r_{CD}}$Given $m_A = m_B = m_C = m_D = m$,$r_{AD} = L$,$r_{CD} = L$,and $r_{BD} = \sqrt{2} L$ (diagonal of the square).$U_{	ext{initial}} = -\frac{G m^2}{L} - \frac{G m^2}{\sqrt{2} L} - \frac{G m^2}{L} = -\frac{G m^2}{L} \left( 2 + \frac{1}{\sqrt{2}} ight)$.$U_{	ext{initial}} = -\frac{G m^2}{L} \left( \frac{2 \sqrt{2} + 1}{\sqrt{2}} ight)$.Thus,the work done by the gravitational force is $-\frac{G m^2}{L} \left( \frac{2 \sqrt{2} + 1}{\sqrt{2}} ight)$.

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