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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 ≡ r₃₁ and 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.