*♦ Learn about :Newton’s Law of Gravitation ,Characteristics of the Gravitational force , Numerical problems ♦
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__Newton’s Law of Gravitation :__

Every particle in this universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.

The gravitational force acting between two particles

F ∝ product of masses

F ∝ 1/ (the separation between two particles)^{2}

Thus, F = G m_{1}m_{2 }/r^{2}

where m_{1} and m_{2} are the masses of the particles, r is the distance of separation between them and G is the Universal Gravitational Constant. The value of G was first-experimentally measured by **Cavendish in 1798** by using a torsion balance.

Magnitude of G = 6.67 × 10^{-11} Newton. m^{2}/kg^{2}.

__Characteristics of the Gravitational force:__

**(a) Attractive Force :** Gravitational force between two particles is always attractive and directed along the line joining the particles.

**(b) Independent of Medium:** It is independent of the nature of the medium surrounding the particles.

**(c) Universal :** It holds good for long distances like inter-planetary distances and also for short distances like inter-atomic distances.

**(d) Action – reaction:** Both the particles experience forces of equal magnitude in opposite directions. If F_{1}^{→} , F_{2}^{→} are the forces exerted on particle 1 by particle 2 and on particle 2 by particle 1 respectively,

then F_{1}^{→} = – F_{2}^{→} Since the forces F_{1}^{→} and F_{2}^{→} are exerted on different bodies, they are known as action-reaction pair.

**(e) Gravitation is conservative:** The work done by the gravitational force acting on a particle is independent of the path described by the particle. It depends upon the initial and final positions of the particle. Work done by gravity on a particle moving in a closed path is zero

**(f) Superposition principle:** If a particle is attracted by n particles, the net force exerted on it must be equal to the vector sum of the forces due to all the n particles.

**Illustration 1:** Three identical particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. Find the force exerted by this system on a particle P of mass m placed at the

(a) the mid point of a side

(b) centre of the triangle.

**Solution:** Using the superposition principle, the net gravitational force on P is

F^{→} = F_{A}^{→} + F_{B}^{→} + F_{C}^{→}

(a) As shown in the figure, when P is at the mid-point of a side, F_{A}^{→} and F_{B}^{->} will be equal in magnitude but opposite in direction. Hence they will cancel each other. So the net force on the particle P will be the force due to the particle placed at C only.

=> F = F_{C}

along PC

(b) At the centre of the triangle O, the forces F_{A}^{→} , F_{B}^{→} and F_{C}^{→} will be equal in magnitude and will be at 120° with each other.

Hence the resultant force on P at O is F^{→} = F_{A}^{→} + F_{B}^{→} + F_{C}^{→} = O