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
$ \displaystyle F \propto m_1 m_2 $
$ \displaystyle F \propto \frac{1}{r^2} $
on combining
$ \displaystyle F \propto \frac{m_1 m_2}{r^2} $
$ \displaystyle F = G \frac{m_1 m_2}{r^2} $
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 N 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 : 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.
$ \displaystyle F = F_C = G \frac{m . m}{(CP)^2} $
$ \displaystyle F = G \frac{m^2}{(a sin60)^2} $
$ \displaystyle F = 4 G \frac{m^2}{3 a^2} $ ; along PC
(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
Gravitational Field
The space around a material body, where it exerts a gravitational force on other bodies, is known as the gravitational field.
The gravitational force field is a vector field because a particle placed at any point P within the field experiences a force which depends on the coordinates of the point P.
Gravitational Field Strength (Intensity ) :
The intensity of the gravitational field at a point P is the gravitational force per unit mass exerted on a test particle placed at point P.
The strength of the gravitational field
$ \displaystyle \vec{g_p} = \frac{\vec{F_p}}{m} $
, where Fp→ is the net force acting on a test particle of mass m kept at the point P.
Its SI unit is N/kg and dimensions are M0LT-2
For earth, the gravitational field
$ \displaystyle \vec{g} = \frac{\vec{W}}{m} $
The above expression is equal to the acceleration due to gravity g→
⇒ Gravitational field strength at a point on the earth is equal to the acceleration due to gravity at that point.
To find g, due to a point mass M kept at a point O (the origin), we place a point test mass m at P, the observation point, and measure the force exerted by M on the test mass m
$ \displaystyle F_p = \frac{G M m }{r^2} $
$ \displaystyle g_p = \frac{F_p}{m} = \frac{G M }{r^2} $ ; and it is directed towards the mass M.