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we want to calculate the P.E(potential energy) of a system containing 3particles p1,p2,p3.the point of observation is P.so now we should add up the P.E at P due to p1,p2,p3 to get the net potential energy of the system,but why we take the P.E of particles due to each other into count instead of the previous method.I can't figure it out.

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  • $\begingroup$ You are confusing two different things : the potential at a point due to 3 masses, and the potential energy of a system of 3 masses. $\endgroup$ – sammy gerbil Nov 8 '18 at 21:57
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we want to calculate the P.E(potential energy) of a system containing 3particles p1,p2,p3.the point of observation is P

There is no point of observation when the potential energy of the system of three particles is evaluated.

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Assuming that the zero of potential energy is when all three particles is at infinity then to evaluate the potential energy of the system of three particles when they are in their final positions one must find the work done by an external force in bringing those three particles from infinity to their final positions.

This would be $$-\dfrac {Gm_1m_2}{r_{12}}- \dfrac {Gm_1m_3}{r_{13}} - \dfrac {Gm_2m_3}{r_{23}}$$

Now consider the three particles in their final position.

Each of the particles will contribute to the potential at a point $P$ which is $$-\dfrac {Gm_1}{R_{1}}- \dfrac {Gm_2}{R_{2}} - \dfrac {Gm_3}{R_{3}}$$ and if a particle of mass $m$ was placed at position $P$ then the potential energy of that one particle would be $$-\dfrac {Gm\,m_1}{R_{1}}- \dfrac {Gm\,m_2}{R_{2}} - \dfrac {Gm\,m_3}{R_{3}}$$ and the potential energy of the whole system of the four particles would be $$-\dfrac {Gm_1m_2}{r_{12}}- \dfrac {Gm_1m_3}{r_{13}} - \dfrac {Gm_2m_3}{r_{23}}-\dfrac {Gm\,m_1}{R_{1}}- \dfrac {Gm\,m_2}{R_{2}} - \dfrac {Gm\,m_3}{R_{3}}$$

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