Three Particle System
Suppose we have two particles of masses m1 and m2 already fixed in space at distance r12 from each other. Let us bring in a third particle of mass m3, from ∞ to some point P near the first two particles, so that m3 finally is at distance r13 from m1 and at distance r23 from m2.
Now, at any instant, there are two forces acting on m3, viz. the gravitational force F31 due to m1 and F32 due to m2. The total work done in moving m3 to point P is given by, Note that the two forces act independently of each other along respective radial directions. That is, for example, we have Note that the two forces act independently of each other along respective radial directions. That is, for example, we have where r and dr in the above integral refer to distances along the radial direction joining particles 1 and 3, at time t. Similarly, we get For conservative forces, the work done is interpreted as the negative change in potential energy. Hence, the increase in gravitational potential energy of the system by joining of third particle is (-W3). The total potential energy of three-particle system becomes,U = U12 + ( -W3 ) Thus, the total potential energy of the system is the sum of potential energies of each pair of particles taken independently.Remember that ( -W3 ) is not the potential energy 'of mass m3'; it is the sum of potential energies of masses (m1 and m3)and masses (m2 and m3).If m3 = 1 (unit mass), we define the gravitational field at point P due to masses m1 and m2 as the net force acting on unit mass at P. where we are now writing r1 and r2 as the position vectors of point P relative to masses m1 and m2. [That is, in fact, r1 ≡ r31and r2 = r32].Gravitational potential at point P due to masses m1 and m2 gives the change in potential energy of the system when a unit mass is added to the system at point P. That is, potential ØP at P is the value of ( -W3 ) from m3 = 1 (unit mass). where r1 and r2 denote distances of P from m1 and m2.