The acceleration of gravity is constant at a particular place but it varies from place to place. In this article, we shall study this variation in acceleration due to gravity.

Variation in Acceleration Due to Gravity due to Shape of the Earth:

The acceleration due to gravity on the surface of the earth is given by

We can see that the acceleration due to gravity at a place is inversely proportional to the square of the distance of the point from the centre of the earth. Now, the earth is not perfectly spherical. It is flattened at the poles and elongated on the equatorial region. The radius of the equatorial region is approximately 21 km more than that at the poles. Hence acceleration due to gravity is maximum at the poles and minimum at the equator. As we move from the equator to the poles the distance of the point on the surface of the earth from the centre of the earth decreases. Hence the acceleration due to gravity increases.

Variation in acceleration due to gravity Due to Latitude of the Place:

The latitude of a point is the angle Φ between the equatorial plane and the line joining that point to the centre of the earth. Latitude of the equator is 0° and that of poles is 90°.

Let us consider a body of mass ‘m’ at a point P with latitude ‘Φ’ as shown on the surface of the earth. Let ‘g_Φ’ be the acceleration due to gravity at point P.

OP = Radius of earth = R

O’P = distance of point P from the axis of the earth

Due to rotational motion of the earth about its axis, the body at P experiences a centrifugal force which is given by mrω². Let us resolve this centrifugal force into two rectangular components. Its component along the radius of the earth is mrω² cosΦ.

Now the body is acted upon by two forces its weight mg acting towards the centre of the earth and the component mrω² cosΦ acting radially outward. The difference between the two forces gives the weight of that body at that point.

mg_Φ = mg – mrω²cosΦ ………….. (1)

Now cos Φ = O’P / OP = r/R

∴ r  = R cos Φ

Substituting in equation (1)

mg_Φ = mg – m(R cos Φ)ω²cos Φ

∴   g_Φ = g – R ω²cos² Φ

This is an expression for acceleration due to gravity at a point P on the surface of the earth having latitude Φ.

At the equator Φ = 0°. Hence ‘g’ is minimum on the equator. For the poles Φ = 90°. Hence ‘g’ is maximum on the poles.

Numerical Problems:

Example – 01:

Find the weight of a body of mass 100 kg on the earth at a) equator b) pole c) latitude of 30°. R = 6400 km, g = 9.8 m/s²

Given: m = 100 kg, R = 6400 km = 6.4 × 10⁶ m,

To find: W_E =? W_P = ? W_Φ =?

Solution:

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

g_Φ = g – Rω²cos²Φ

The Weight of body at latitude Φ is given by

W_Φ = mg_Φ = mg – mRω²cos²Φ

At equator Φ = 0°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²0

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (1)²

W_E = 980 – 3.386 = 976.6 N

At poles Φ =90°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²90

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (0)²

W_E = 980 – 0 = 980 N

At latitude Φ = 30°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²30

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (0.866)²

W_E = 980 – 2.539 = 977.5 N

Ans: Weight of the body on the equator, on the pole and on latitude 30° are 976.6 N, 980 N and 977.5 N respectively.

Example – 02:

Find the difference in weight of a body of mass 100 kg on equator and pole. R = 6400 km, g = 9.8 m/s²

Given: m = 100 kg,  R = 6400  km = 6.4 × 10⁶ m,

To find: W_P – W_E =?

Solution:

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

g_Φ = g – Rω²cos²Φ

The Weight of body at latitude Φ is given by

W_Φ = mg_Φ = mg – mRω²cos²Φ

At equator Φ = 0°

W_E = mg – mRω²cos²0° = mg – mRω²(1)² = mg – mRω²    ………… (1)

At poles Φ = 90°

W_P = mg – mRω²cos²90° = mg – mRω²(0)² = mg    ………… (2)

Now, W_P – W_E = mg – (mg – mRω²) = mRω² 

W_P – W_E = mg – (mg – mRω²) = 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²   = 3.386 N

Ans: The required difference in Weight is 3.386 N

Variation in Acceleration Due to Gravity Due to Altitude:

Expanding binomially and neglecting terms of higher power of (h/R) we get

This is an expression for the acceleration due to gravity at small height ‘h’ from the surface of the earth. This expression shows acceleration due to gravity decreases as we move away from the surface of the earth.

Loss in Weight of a Body at Height h:

Numerical Problems:

Example – 03:

A mass of 5 kg is weighed on a balance at the top of a tower 20 m high. The mass is then suspended from the pan of the balance by a fine wire 20 m long and weighed. Find the change in the weight of a body in mgf assuming the radius of the earth as 6330 km.

Given: Mass of body = m = 5 kg, Radius of earth = R = 6330  km = 6.33 x 10⁶ m, height of tower =  h = 20 m.

To find: W – W_h =?

Solution:

Ans: The change in the weight of a body is 31.6 mgf

Example – 04:

At what height will a man’s weight become half his weight on the surface of the earth? Take the radius of the earth as R.

Given: W_h = 1/2W, Radius of earth = R

To find: h =?

Solution:

Now r = R + h

∴   h = r – R = 1.414 R – R = 0.414 R

Ans: At a height of 0.414R, the man’s weight become half his weight on the surface of the earth.

Example – 05:

A meteor is falling. How much gravitational acceleration it will experience when its height from the surface of the earth is equal to three times radius of the earth. Acceleration due to gravity on the surface of the earth is ‘g’.

Given: h = 3R.

To find: g_h =?

Solution:

Ans: The acceleration of the meteor is g/16.

Variation in Acceleration Due to Gravity Due to Depth:

The acceleration due to gravity on the surface of the earth is given by

Let ‘ρ’ be the density of the material of the earth.

Now, mass = volume x density

Substituting in the equation for g we get

Now, let the body be taken to the depth ‘d’ below the surface of the earth. Then acceleration due to gravity g_d at the depth ‘d’ below the surface of the earth is given by

Dividing equation (3) by (2) we get

This is an expression for the acceleration due to gravity at the depth ‘d’ below the surface of the earth. This expression shows acceleration due to gravity decreases as we move down into the earth. At the centre of the earth d = R, hence acceleration due to gravity at the centre of the earth is zero.

Relation between g_d and g_h:

We have

Thus the acceleration due to gravity at a small height ‘h’ from the surface of the earth is the same as the acceleration due to gravity at the depth ‘d = 2h’ below the surface of the earth. It means that the value of acceleration due to gravity at a small height from the surface of the earth decreases faster than the value of the acceleration due to gravity at the depth below the surface of the earth.

Variation of g with Altitude and Depth

Numerical Problems:

Example – 06:

Find percentage decrease in the weight of a body when taken 16 km below the surface of the earth. Take radius of the earth as 6400 km.

Given: depth = d = 16 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find: Percentage decrease in weight =?

Solution:

Ans: The percentage decrease in weight is 0.25.

Example – 07:

How much below the surface of the earth does acceleration due to gravity becomes 1 % of the value at the earth’s surface? Assume the radius of the earth as 6380 km.

Given: g_d = 1% g = 0.01 g, Radius of earth = R = 6380 km, g = 9.8 m/s².

To find: depth d =?

Solution:

Ans: At a depth of 6316 km below the surface of the earth the acceleration due to gravity becomes 1 % of the value at the earth’s surface.

Example – 08:

Compare the weight of 5 kg body 10 km above and 10 km below the surface of the earth. Given the radius of the earth = 6400 km.

Given: d = 10 km, d = 10 km, Radius of earth = R = 6400 km

To find:  W_h : W_d = ?

Solution:

Ans: The required ratio of weights is 0.998 : 1

Example – 09:

Compare the weight of body 0.5 km above and 1 km below the surface of the earth. Given the radius of the earth = 6400 km.

Solution:

Given: d = 1 km, d = 0.5 km, Radius of earth = R = 6400 km

To find:  W_h : W_d = ?

Ans: The required ratio of weights is 1:1

Example – 10:

The acceleration due to gravity at a height 1/20 th radius of the earth above the earth’s surface is 9 m/s². Find the value of acceleration due to gravity at an equal distance below the surface of the earth. Given the radius of the earth = 6400 km.

Given: h = 1/20 R, gh = 9 m/s², Radius of earth = R = 6400 km

To find:   g_d =? when d = 1/20 R

Solution:

Ans: The acceleration due to gravity at depth equal to R is 9.5 m/s²

Example – 11:

At what depth below the surface of the earth, a man’s weight becomes half his weight on the surface of the earth. Take the radius of the earth as R = 6400 km.

Given: W_d = 1/2 W, Radius of earth = R = 6400 km

To find:   d =?

Solution:

∴ R = 2R – 2d

∴ d = R i.e. d = R/2 = 6400/2 = 3200 km

Ans: At depth of 3200 km below the surface of the earth a man’s weight becomes half his weight on the surface of the earth.

Example – 12:

Find decrease in value of acceleration due to gravity at a point 1600 km below the earth’s surface. R = 6400 km, g = 9.8 m/s².

Given: d = 1600 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   decrease in value of acceleration due to gravity = g – g_d =?

Solution:

Ans: The decrease in acceleration due to gravity is 2.45 m/s²

Example – 13:

What is the decrease in the weight of a body of mass 500 kg when it is taken into a mine of depth 1000 m? R = 6400 km, g = 9.8 m/s².

Given: Mass of body = m = 500 kg, depth d = 1000 m = 1 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   decrease in weight = W – W_d =?

Solution:

Ans: The decrease in the weight of the body is 1 N.

Example – 14:

Find the acceleration due to gravity at a depth of 2000 km from the surface of the earth, assuming earth to be a homogeneous sphere. R = 6400 km, g = 9.8 m/s².

Given: d = 2000 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   acceleration due to gravity =  g_d =?

Solution:

Ans: The acceleration at depth of 2000 km below the surface of the earth is 6,738 m/s²

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