# Newton’s Law of Gravitation & Gravitational Field

### 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

$\displaystyle \vec{g_p} = \frac{\vec{F_g}}{m}$

, where Fg 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

$\displaystyle F_g = \frac{G M m }{r^2}$

$\displaystyle g_p = \frac{F_g}{m} = \frac{G M }{r^2}$ ; and it is directed towards the mass M.

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