__Expression for Acceleration due to gravity__

- Consider any object of mass ‘m’ at a point A on the surface of the earth.
- The force of gravity between the body and earth can be calculated as
- F =G m M
_{e }/R_{e}^{2}(1) where- m=mass of the body
- M
_{e}= mass of the Earth - R
_{e}= distance between the body and the earthis same as the radius of the earth

- Newton’s Second law states that
- F=ma (2)

Comparing the equations (1) and (2)

- F=m (G mM
_{e}/R_{e}^{2}) - (G mM
_{e}/R_{e}^{2}) is same as g(acceleration due to gravity) - Therefore, the expression forAcceleration due to gravity.
**g=G M**_{e}/R_{e}^{2}

** Problem**: Calculate the acceleration due to gravity on the (a) earth’s surface (b) at a height of 1.5×10

^{5 }m from the earth surface .Given: R

_{e}=6.4×10

^{6}m; M

_{e }= 5.98×10

^{24 }kg?

** Answer**: –

(a) g= GM_{e}/R_{e}^{2}

= 6.67×10-11×5.98×10^{24}/(6.4 x10^{6})^{2}

= 9.74m/s^{2}

(b) h= 1.5×10^{5}m (given)

g (h)= GM_{e}/(R_{e}+h)^{2}

= 6.67×10-11×5.98×10^{24}/ (6.4 x10^{6}+ 1.5×10^{5})

=9.30m/s^{2}

** Problem:- **Calculate the mass of the moon if the free fall acceleration near its surface is known to be 1.62m/s

^{2}.Radius of the moon is 1738km?

__Answer:__

g=1. 62m/s^{2}

R_{m} = 1738km =1738 x 10^{3} m

G=6.67×10^{-11}Nm^{2}/kg^{2}

g= GM_{m}/R_{m}^{2}

M_{m}=gR_{m}^{2}/G

=1.62x (1738×10^{3})^{2}/6.67×10^{-11}

=7.34×10^{22 }kg

**Acceleration due to gravity below the surface of earth**

- To calculate acceleration due to gravity below the surface of the earth (between the surface and centre of the earth).
- Density of the earth is constant throughout. Therefore,
- ρ = M
_{e }/ (4/3π R_{e}^{3}) equation(1) - where
- M
_{e }= mass of the earth - Volume of sphere= 4/3π R
_{e}^{3} - R
_{e}= radius of the earth.

- M

- As entire mass is concentrated at the centre of the earth.
- Therefore density can be written as

ρ = M_{s }/ (4/3π R_{s}^{3}) equation (2)

- Comparing equation (1) and (2)
- M
_{e}/M_{s}= R_{e}^{3}_{/}R_{s}^{3}where R_{s}= (R_{e}-d)^{3} - d= distance of the body form the centre to the surface of the earth.
- Therefore,
- M
_{e}/M_{s}= Re^{3}/(R_{e}-d)^{3} - M
_{s}= M_{e}(R_{e}-d)^{3}/R_{e}^{3}from equation(3) - To calculate Gravitational force (F) between earth and point mass m at a depth d below the surface of the earth.

Above figure shows the value of g at a depth d. In this case only the smaller sphere of radius (R_{e}–d) contributes to g.

- F= G m M
_{s}/(R_{e}-d)^{2} - g=F/m where g= acceleration due to gravity at point d below the surface of the earth.
- g= GM
_{s}/(R_{e}-d)^{2} - Putting the value of M
_{s}from equation (3)

= GM_{e }(R_{e}-d)^{3}/R_{e}^{3}(R_{e}-d)^{2}

=GM_{e }(R_{e}-d)/R_{e}^{3}

- We know g=GM
_{e}/R_{e}^{2}equation (4)

- g(d) = GM
_{e/}R_{e}^{2 }(1-d/R_{e}) - From equation (4)
**g (d) = g(1-d/R**_{e})

**Acceleration due to gravity above the surface of earth**

- To calculate the value of acceleration due to gravity of a point mass m at a height h above thesurface of the earth.

Above figure shows the value ofacceleration due to gravity g at a height h above the surface of the earth.

- Force of gravitation between the object and the earth will be
- F= G mM
_{e}/(R_{e}+h)^{2}where

- m = mass of the object, R
_{e}= radius of the earth - g(h) = F/m = GM
_{e}/(R_{e}+h)^{2}= GM_{e}/[R_{e}^{2}(1+h/R_{e})^{2}] - h << R
_{e}(as radius of the earth is very large) - By calculating we will get,

**g(h) = g(1-2h/R _{e})**

The value of acceleration due to gravity varies on the surface, above the surface and below the surface of the earth.__Conclusion: –__

** Problem:- ** ( Gravitation )

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

__Answer:__

Weight of a body of mass m at the Earth’s surface, W = mg = 250 N

Body of mass m is located at depth, d=1/2 R_{e}

Where,

= Radius of the Earth

Acceleration due to gravity at depth g (d) is given by the relation:

g’= (1-d/R_{e})g

= (1- R_{e}/2x R_{e}) g= ½ g

Weight of the body depth d,

W’=mg’

=m 1/2g =1/2mg=1/2W

=1/2 x250=125N

** Problem:- **( Gravitation )

A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

** Answer: **Weight of the body, W = 63 N

Acceleration due to gravity at height h from the Earth’s surface is given by the relation:

g’=g/(1+h/R_{e})^{2}

Where,

g = Acceleration due to gravity on the Earth’s surface

R_{e} = Radius of the Earth

For h= R_{e} /2

g’ = g/(1+ R_{e}/2x R_{e})^{2} = g(1+1/2)^{2} = 4/9g

Weight of a body of mass m at height h is given as:

W’=mg’

=mx4/9g=4/9mg

=4/9W

=4/9×63=28N