Question

explain why the value of g changes from poles to equator on the surface of the earth. state how its value changed on moving:

1) towards the centre of the earth

2)away from the earth?

Answer 1

Acceleration due to gravity of our earth varies at poles and equator due to the following reasons:

1. Rotation of the earth:

The centrifugal on any object on the earth depends on the distance of the object from the axis of rotation of the earth.

g^/ = g - ω²R

Any object at the equator is at the maximum distance from the axis. Whereas, an object at the pole will be near the axis. So, acceleration due to gravity is less at the equator than at the poles.

2. Non sphericity of the earth:

The radius in the equatorial plane is about 21 Km larger than the radius along the poles. The value of g is accordingly larger at poles and less at the equator.

Variation of g with height:

Consider the body which is at height h from the surface of earth whose radius is considered to be R.

Let the mass of earth be M.

So, acceleration due to gravity acted on a body at surface of earth = GM/R²

=> g = GM/R² ............ ... ... ... (i)

Again,

Acceleration due to gravity acted on a body at the height h above the surface of earth,

g^/ = GM/(R+h)²

=> g^/ = GM/(R+h)² ... ... ... ... ... (ii)

Now, dividing (i) by (ii), we get,

g/g^/ = (GM/R²)/(GM/(R + h)²)

=> g/g^/ = (GM × (R+h)²)/(GM × R²)

=> g/g^/ = (R+h)²/R²

=> g/g^/ = (R/R + h/R)²

=> g/g^/ = (1+h/R)²

=> g^/ = g(1+h/R)^-2

Now, expanding binomially, we get,

g^/ = g(1- 2h/R)

Variation of g with depth:

Consider the body at the depth of x from the surface of earth whose mass is considered to be M and radius to be R.

But at this situation body will experience acceleration due to gravity only by that mass in which it is standing. Let that mass is considered to be M^/. Similarly radius will be (R-x).

Now, acceleration due to gravity at the surface of earth, g = GM/R²

=> g = GM/R² ... ... ... ... ... ... ... ... (i)

Similarly, acceleration due to gravity below the depth of x from surface of earth = GM^//(R-x)²

=> g^/ = GM^//(R-x)² ... ... ... ... ... ... (ii)

Dividing (ii) by (i), we get,

g^//g = (GM^//(R-x)²)/(GM/R²)

=> g^//g = M^/R²/[M(R-x)²]... ... ... ... (iii)

Now, Density of earth, ρ = M/V

=> ρ = M/[(4/3)Ï€ R³]

=> M = ρ[(4/3)Ï€R³] ... ... ... ... (iv)

Density of earth below x metre from the surface of earth, ρ = M^//V^/

=> ρ = M^//[(4/3)Ï€(R - x)³]

=> M^/ = ρ[(4/3)Ï€(R - x)³] ... ... ... (v)

Putting (iv) and (v) in (iii), we get,

g^//g = [{ρ(4/3)Ï€(R - x)³} × R²]/[{ρ(4/3)Ï€R³}(R-x)²]

=> g^/ = g(1- x/R)