# Finding G.P.E. of a system of particles

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.

• 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. – sammy gerbil Nov 8 '18 at 21:57

This would be $$-\dfrac {Gm_1m_2}{r_{12}}- \dfrac {Gm_1m_3}{r_{13}} - \dfrac {Gm_2m_3}{r_{23}}$$
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 $$-\dfrac {Gm_1m_2}{r_{12}}- \dfrac {Gm_1m_3}{r_{13}} - \dfrac {Gm_2m_3}{r_{23}}-\dfrac {Gm\,m_1}{R_{1}}- \dfrac {Gm\,m_2}{R_{2}} - \dfrac {Gm\,m_3}{R_{3}}$$