Question

derive an expression for the variation of acceleration due to gravity with height h from the surface of the earth.

Answer 1

Force of gravity acting on a body of mass m on the earth surface = F = GMm/R^2 ____________ (1)

Where R is the radius of the earth (considering it a homogenous sphere)
and M here is the total mass of earth concentrated at its centre. G is the gravitational constant.

Now, from Newton’s 2nd Law of Motion,
F = mg.___________________(2)

Just to recapitulate, as a body falls downwards it is continuously acted upon by a force of gravity. The body thus possesses an acceleration, called Acceleration Due to gravity(g).

From equation 1 and 2 we can write

mg = (GMm) / R^2

=> g = GM / R^2 _______________ (3)

Another expression of g here:

If mean density of earth is p then mass of the earth = M = volume X density = (4/3) Pi R^3 p
(Pi = 22/7)

g = G.( (4/3) Pi R^3 p) / R^2

g = (4/3) Pi R p G ________________ (4)

Answer 2

Acceleration due to gravity – on the earth’s surface

Definition: Acceleration due to gravity of an object is its rate of change of velocity due to the sole effect of the earth’s gravitational pull or gravity on that object that directs towards the centre of the earth.

Force of gravity acting on a body of mass m on the earth surface = F = GMm/R^2 ____________ (1)

Where R is the radius of the earth (considering it a homogenous sphere)
and M here is the total mass of earth concentrated at its centre. G is the gravitational constant.

Now, from Newton’s 2nd Law of Motion,
F = mg.___________________(2)


Just to recapitulate, as a body falls downwards it is continuously acted upon by a force of gravity. The body thus possesses an acceleration, called Acceleration Due to gravity(g).

From equation 1 and 2 we can write

mg = (GMm) / R^2

=> g = GM / R^2 _______________ (3)

Another expression of g here:

If mean density of earth is p then mass of the earth = M = volume X density = (4/3) Pi R^3 p
(Pi = 22/7)

g = G.( (4/3) Pi R^3 p) / R^2

g = (4/3) Pi R p G ________________ (4)

Variation of Acceleration due to gravity with height 

At a height of h from the surface of the earth, the gravitational force on an object of mass m is

F = GMm/(R+h)^2

Here (R + h) is the distance between the object and the centre of earth.

Say at that height h, the gravitational acceleration is g1.

So we can write, mg1 = GMm / (R+h)^2

=> g1 = GM/(R+h)^2 _________________ (5)

Now we know on the surface of earth, it is
g =  GM / R^2

Taking ratio of these 2,

g1/g = R^2 /(R+h)^2

= 1/(1 + h/R)^2 = (1 + h/R)^(-2) = (1 – 2h/R)

=> g1/g = (1 – 2h/R)

=> g1 = g (1 – 2h/R) ___________________ (6)

So as altitude h increases, the value of acceleration due to gravity falls.