Question

Derivation for acceleration due to gravity on its effects of height and depth... Explain plllzzz

Answer 1

1)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)

In the next 2 sections, we will discuss how this g varies as we go (a)higher from the surface and go (b) deeper from the surface.

2)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.

In the next section we will see how g varies with depth below the surface of the earth.

3)Variation of Acceleration due to Gravity with depth

Let’s say, a body of mass m is resting at point A , where A is at a depth of h from the earth’s surface.

Distance of point A from the centre of the earth = R – h,
where R is the radius of the earth.

Mass of inner sphere = (4/3). Pi. (R-h)^3. p

Here p is the density.

Now at point A, the gravitational force on the object of mass m is

F = G M m/ (R-h)^2 = G. [(4/3). Pi. (R-h)^3. p] m/(R-h)^2 = G. (4/3). Pi. (R-h). p. m

Again at point A, the acceleration due to gravity (say g2) = F/m = G. (4/3). Pi. (R-h). p _________________ (7)

Now we know at earth’s surface, acceleration due to gravity = g = (4/3) Pi R p G

Taking the ratio, again,

g2/g = [G. (4/3). Pi. (R-h). p ]/ [(4/3) Pi R p G] = (R-h) / R = 1 – h/R.

=> g2 = g (1 – h/R) _________________________ (8)

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

In the next paragraph we will compare these 2 equations to get a clearer picture.

Equations – formula- Comparison
g1 = g (1 – 2h/R) ___________________ (6) at a height h from the earth’s surface

g2 = g (1 – h/R) _________________________ (8) at a depth h from the earth’s surface

1) Now from eqn 6 and 8 we see that both g1 and g2 are less than g on the earth’s surface.
2) We also noticed that, g1 < g2

And that means:
1) value of acceleration due to gravity falls as we go higher or go deeper.
2) But it falls more when we go higher.