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 Me /Re2 (1) where
- m=mass of the body
- Me = mass of the Earth
- Re= 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 mMe/Re2)
- (G mMe/Re2) is same as g(acceleration due to gravity)
- Therefore, the expression forAcceleration due to gravity.
- g=G Me/Re2
Problem: Calculate the acceleration due to gravity on the (a) earth’s surface (b) at a height of 1.5×105 m from the earth surface .Given: Re=6.4×106m; Me = 5.98×1024 kg?
(a) g= GMe/Re2
= 6.67×10-11×5.98×1024/(6.4 x106)2
(b) h= 1.5×105m (given)
g (h)= GMe/(Re+h)2
= 6.67×10-11×5.98×1024/ (6.4 x106+ 1.5×105)
Problem:- Calculate the mass of the moon if the free fall acceleration near its surface is known to be 1.62m/s2.Radius of the moon is 1738km?
Rm = 1738km =1738 x 103 m
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 = Me / (4/3π Re3) equation(1)
- Me = mass of the earth
- Volume of sphere= 4/3π Re3
- Re = radius of the earth.
- As entire mass is concentrated at the centre of the earth.
- Therefore density can be written as
ρ = Ms / (4/3π Rs3) equation (2)
- Comparing equation (1) and (2)
- Me/Ms = Re3/Rs3 where Rs = (Re-d) 3
- d= distance of the body form the centre to the surface of the earth.
- Me/Ms = Re3/(Re-d)3
- Ms = Me(Re-d)3/Re3 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 (Re–d) contributes to g.
- F= G m Ms/(Re-d)2
- g=F/m where g= acceleration due to gravity at point d below the surface of the earth.
- g= GMs/(Re-d)2
- Putting the value of Ms from equation (3)
= GMe (Re-d)3/Re3(Re-d)2
- We know g=GMe/Re2 equation (4)
- g(d) = GMe/ Re2 (1-d/Re)
- From equation (4)
- g (d) = g(1-d/Re)
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 mMe/(Re+h)2 where
- m = mass of the object, Re = radius of the earth
- g(h) = F/m = GMe/(Re+h)2 = GMe/[Re2(1+h/Re)2]
- h << Re (as radius of the earth is very large)
- By calculating we will get,
g(h) = g(1-2h/Re)
- Conclusion: –The value of acceleration due to gravity varies on the surface, above the surface and below the surface of the earth.
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?
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 Re
= Radius of the Earth
Acceleration due to gravity at depth g (d) is given by the relation:
= (1- Re/2x Re) g= ½ g
Weight of the body depth d,
=m 1/2g =1/2mg=1/2W
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 = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth
For h= Re /2
g’ = g/(1+ Re/2x Re)2 = g(1+1/2)2 = 4/9g
Weight of a body of mass m at height h is given as: