### 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 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} N m^{2}/kg^{2}

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##### 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 $\vec{F_1}$ , $\vec{F_2}$ are the forces exerted on particle 1 by particle 2 and on particle 2 by particle 1 respectively,

then $\vec{F_1} = – \vec{F_2} $ . Since the forces $\vec{F_1}$ and $\vec{F_2}$ 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 $\vec{F}$ exerted on it must be equal to the vector sum of the forces due to all the n particles.

$\vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} + ….+\vec{F_n} $

**Solved Example :** 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

$\vec{F} = \vec{F_A} + \vec{F_B} + \vec{F_C} $

(a) As shown in the figure, when P is at the mid-point of a side, $\vec{F_A}$ and $\vec{F_B}$ 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.

$ \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 $\vec{F_A}$ , $\vec{F_B}$ and $\vec{F_C}$ will be equal in magnitude and will be at 120° with each other.

Hence the resultant force on P at O is ,

$\vec{F} = \vec{F_A} + \vec{F_B} + \vec{F_C} = 0 $

__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{E_p} = \frac{\vec{F_p}}{m} $

, where F_{p}^{→} 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 M^{0}LT^{-2}

For earth, the gravitational field

$ \displaystyle \vec{E} = \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.

### Gravitational Field due to a Point mass at a distance r :

To find E_{P} , 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 E_p = \frac{F_p}{m} = \frac{G M }{r^2} $ ; and it is directed towards the mass M.

### Acceleration due to gravity :

Acceleration due to gravity : The acceleration of a particle caused by the gravitational force of the earth is known as the acceleration due to gravity. It is a vector quantity and is a measure of the intensity of the earth’s gravitational field.

The magnitude of the gravitational force F_{g} exerted by the earth on a particle of mass m , calculated by using Newton’s law of gravitation, is given by

$ \displaystyle F_g = \frac{G M m}{r^2} $ ……. (i)

If the particle is free to fall, its acceleration a is given by

$ \displaystyle a = \frac{F_g}{m} = \frac{GM}{r^2}$ ……… (ii)

This quantity has already been introduced as g (gravitational field strength or intensity.

$ \displaystyle a = g = \frac{GM}{r^2}$

where M represents the mass of earth and r is the distance of the observation point from the centre of the earth.

If the particle is very close to the earth’s surface (or it is placed on earth’s surface)

then

r = R + h ≈ R . ( h << R ) , R being the radius of the earth.

Using the measured values of

g ≅ 9.8 m/s^{2} and G = 6.672 × 10^{-11} MKS, R ≅ 6371 km, we get,

Mass of the earth, M = 5.97 × 10^{24}kg

No wonder, Cavendish’s first measurement of G was referred to as “weighing the earth”.