__What is called Gravitational Potential?__

*Gravitational potential at a point in the gravitational field of a body is defined as the amount of work done in bringing a body of unit mass from infinity to that point.*

Therefore, gravitational potential at a point in the gravitational field is given by –

V = \left ( \frac {\text {Work done}}{\text {Mass}} \right )

Or, \quad V = \left ( \frac { W }{ m } \right )

*The gravitational potential is a scalar quantity.*

__Earth’s Gravitational Potential__

From definition, work done in moving a particle through a distance is given by –

W = ( \text {Force} \ \times \ \text {Distance moved by body} )

Therefore, work done in bringing a body of mass ( m ) to a point at distance ( r ) * *from the centre of the earth will be –

W = - \left ( \frac { G M m }{ r^2 } \right ) \ \times \ r

Or, \quad W = - \left ( \frac { G M m }{ r } \right )

- Hence gravitational potential due to the earth at that point will be –

V = \left ( \frac { W }{ m } \right ) = - \left ( \frac { G M m }{ r } \right ) \ \div \ m

- Therefore, \quad V = - \left ( \frac { G M }{ r } \right )

At the surface of the earth \quad ( r = R )

- Therefore, gravitational potential at the earth’s surface will be –

V_{ surface } = - \left ( \frac { G M }{ R } \right )

__Gravitational Field__

*Gravitational field of a body is a space surrounding that body within which its gravitational force of attraction can be felt by another body.*

- Gravitational field is a natural phenomenon.
- All material bodies produce its own gravitational field around itself in concentric spherical fashion.

__Intensity of Gravitational Field__

*Intensity of gravitational field at a point is defined as the force experienced by a body of unit mass placed at that point.*

Therefore, \quad \text {Gravitational field intensity} = \left ( \frac {\text {Gravitational force}}{\text {Mass}} \right )

- Gravitational field intensity is a vector quantity and denoted by ( \vec { E } ) .
- It always acts towards the source body producing that gravitational field.

Consider about a source body of heavy *mass* ( M ) as shown in figure. The body will produce its gravitational field around it.

A test body of mass ( m ) is brought at point P within the gravitational field of the source body. Let, the point P is at a distance ( r ) from centre O of the source body.

- From
*Newton’s laws of gravitation*, the test body will experience a gravitational pull of magnitude –

\vec { F } = \left ( \frac { G M m }{ r^2 } \right )

- Therefore, gravitational field intensity at point P will be –

\vec { E } = \left ( \frac { \vec { F }}{ m } \right ) ……… (i)

Or, \quad E = \left ( \frac {F}{m} \right ) = \left ( \frac { G M }{ r^2 } \right ) ……….. (ii)

At infinity, gravitational field intensity becomes zero.

*Thus gravitational field intensity decreases as distance ( r ) increases and becomes zero at infinity.*

- The direction of ( \vec {E} ) is same as that of ( \vec {F} )

If the test mass is free to move, it will move towards mass ( m ) with an *acceleration* ( a ) under the influence of force ( F ) .

Therefore, \quad a = \left ( \frac { F }{ m } \right ) …….. (iii)

From equations (ii) and (iii), we get –

a = E

*Thus the intensity of gravitational field at any point is equal to the free acceleration produced in the test mass when placed at that point.*

__Intensity of Earth’s Gravitational Field__

*The gravitational field intensity of earth at any point near the earth’s surface is equal to the acceleration produced in freely falling body **at that point.*

Consider about a body placed at a distance of ( r ) * *from the centre of the earth O * *as shown in figure.

- The gravitational field intensity at point P will be –

\vec { E } = \left ( \frac { \vec { F }}{ m } \right ) = \left ( \frac { G M }{ r^2 } \right )

For a point on the surface of the earth ( r = R ) .

- Therefore, \quad E_{surface} = \left ( \frac { G M }{ R^2 } \right ) = g

__Gravitational Potential Energy__

*Gravitational potential energy of a body is the energy possessed by that body due to its position in the gravitational field of another body.*

- Potential energy is the amount of
*work done*in bringing a body from infinity to a given point in the gravitational field.

Consider about a sphere of mass ( M ) and radius ( R ) *.* A smaller body of mass ( m ) is located at point P such that ( OP = r ) where ( r > R ) as shown in figure.

Let, at an instant the smaller body is at A * *such that ( OA = x )

- The
*gravitational force*on the body at will be –

F = \left ( \frac { G M m }{ x^2 } \right )

Now let, the body is moved through a small distance ( dx ) .

- The small
*work done*in moving the body through this distance will be –

d W = F dx = \left ( \frac { G M m }{ x^2 } \right ) dx

- Then total work done in bringing the body from infinity i.e. ( x = \infty ) to the point P i.e. at ( x = r ) will be –

W = \int { d W } = \int\limits_{ \infty }^{ r } \left ( \frac { G M m }{ x^2 } \right ) dx

Or, \quad W = \left ( G M m \right ) \int\limits_{ \infty }^{ r } x^{ -2 } dx

= ( G M m ) \left [ - \frac { 1 }{ x } \right ]_{ \infty }^{ r }

= - ( G M m ) \left [ \frac { 1 }{ r } - \frac { 1 }{ \infty } \right ]

Therefore, \quad W = - \left ( \frac { G M m }{ r } \right )

- This work done is stored as the gravitational potential energy ( U ) in the body.

Therefore, \quad U = - \left ( \frac {GMm}{r} \right ) ……. (1)

**TO BE NOTED –**

- In above equation, negative sign indicates that work is being done on the body by the gravitational force.
- As the distance ( r ) increases, the gravitational potential energy increases.
- At ( x = \infty ) the maximum ( PE ) becomes zero.

### Change in Potential Energy

Rewriting the equation for gravitational potential energy, we get –

U = - \left ( \frac { G M m }{ r } \right ) = \left ( - \frac { G M }{ r } \right ) \times m

But \quad \left ( - \frac { G M }{ r } \right ) = V ( Gravitational potential ).

Therefore, \quad U = ( V ) \times ( m )

- Hence, \text {Gravitational potential energy} = \text {Gravitational potential} \ \times \ \text {Mass}

If a body of mass ( m ) is moved from a point at distance ( r_1 ) to a point at distance ( r_2 ) then the change in (PE ) of body will be –

dU = \int\limits_{ r_1 }^{ r_2 } \left ( \frac {GM m}{ x^2 } \right ) dx

= ( G M m ) \left [ - \frac { 1 }{ x } \right ]_{ r_1 }^{ r_2 }

= ( G M m ) \left [ \frac { 1 }{ r_1 } - \frac { 1 }{ r_2 } \right ]

- Therefore, if ( r_1 > r_2 ) then ( \Delta U ) will be negative.

So *when a body is brought closer to earth’s surface, its gravitational potential energy decreases.*

*Therefore, if ( r_1 = R ) and [ r_2 = ( R + h ) ] \quad Then \quad \Delta U = mgh *

See numerical problems based on this article.