♦ Learn about :Newton’s Law of Gravitation ,Characteristics of the Gravitational force , Numerical problems ♦
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 m1m2 /r2
where m1 and m2 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. m2/kg2.
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 F1→ , F2→ are the forces exerted on particle 1 by particle 2 and on particle 2 by particle 1 respectively,
then F1→ = – F2→ Since the forces F1→ and F2→ 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→ = FA→ + FB→ + FC→
(a) As shown in the figure, when P is at the mid-point of a side, FA→ and FB-> 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 = FC
(b) At the centre of the triangle O, the forces FA→ , FB→ and FC→ will be equal in magnitude and will be at 120° with each other.
Hence the resultant force on P at O is F→ = FA→ + FB→ + FC→ = O