## Variations In Acceleration Due To Gravity

Acceleration due to gravity is not a constant. It changes due to the following reasons:

- The earth is an oblate spheroid; hence, the value of R is different at different points on the earth’s surface. Thus, the values of g at different places are different.
- The value of g differs at different heights from the earth’s surface.
- The value of g differs at different depths below the earth’s surface.
- Due to the diurnal motion of the earth, the value of g is different at different points on the surface of the earth.

**Effect Of The Earth’s Oblate Spheroidal Shape: **The earth is not a perfectly spherical body. It is oblate spheroidal in shape. The Arctic and the Antarctic regions are slightly depressed and the equatorial region is a little bulging. Hence, the equatorial radius is greater than the polar radius by about 21 km.

We know that, g ∝ \(\frac{1}{r^2}\), i.e., the value of g is inversely proportional to the square of the distance of any point on the earth’s surface from the centre of the earth. Now, as the polar region is closer to the centre, the value of g is greater than that in the equatorial region.

**Read and Learn More: Class 11 Physics Notes**

**Variation With Altitude:** Let the value of the acceleration due to gravity on the earth’s surface be g; the value of the acceleration due to gravity at a point P situated at height h from the earth’s surface be g’ and the distance of the earth’s surface from the centre of the earth O, i.e., the radius of the earth be R.

Hence, the distance of P from O = R+h. As the magnitude of the acceleration due to gravity is inversely proportional to the square of the distance of the point from the centre of the earth,

⇒ \(\frac{g^{\prime}}{g}=\frac{R^2}{(R+h)^2}\)

= \(\frac{1}{\left(1+\frac{h}{R}\right)^2}=\left(1+\frac{h}{R}\right)^{-2}=\left(1-\frac{2 h}{R}\right)\)

[higher terms in the above binomial expansion are neglected as the value of h is usually too small in comparison to that R]

∴ \(g^{\prime}=g\left(1-\frac{2 h}{R}\right)\)….(2)

This is the relationship between the value of acceleration due to gravity at a height h from the earth’s surface and that on the earth’s surface. Equation (2) shows that g’ decreases with the increase in h, which means that the acceleration due to gravity decreases with height from the earth’s surface.

Now, g – g’ = decrease in the acceleration due to gravity at a height h from the surface of the earth

= \(\frac{2 h g}{R}\)….(3)

Note that equation (2) is applicable when h<<R, but when h is comparable to R, equation (1) has to be applied.

**Variation With Depth:** Let the radius of the earth be R and a mass m be kept at a point P situated at a depth h from the surface of the earth.

The point O is the centre of the earth. A sphere is imagined with its centre at O and of radius (R-h). The surface of this imaginary sphere divides the Earth into two parts:

- An inner spherical core of radius (R-h) and
- An outer spherical shell of thickness h. It can be proved that the body of mass m, being placed just outside the inner core, experiences a gravitational pull due to this inner core only. Being inside the spherical shell, the mass m does not experience any gravitational pull due to the shell. Hence, to determine the gravitational pull, only the inner spherical core is to be taken into consideration.

The average density of the earth, \(\rho=\frac{\text { mass of the earth }}{\text { volume of the earth }}=\frac{M}{\frac{4}{3} \pi R^3}=\frac{3 M}{4 \pi R^3}\)

Volume of the inner spherical core =\( \frac{4}{3} \pi(R-h)^3\)

Hence, mass of the inner spherical core, \(M^{\prime}=\frac{4}{3} \pi(R-h)^3 \rho=\frac{4}{3} \pi(R-h)^3 \cdot \frac{3 M}{4 \pi R^3}=\frac{M(R-h)^3}{R^3}\)

Hence, the gravitational force acting on the mass m

F = \(\frac{G M^{\prime} m}{(R-h)^2}=\frac{G M m}{R^3}(R-h)\)

So, if the acceleration due to gravity at P, i.e., at a depth h be g’, then \(g^{\prime}=\frac{F}{m}=\frac{G M}{R^3}(R-h)=\frac{G M}{R^2} \frac{(R-h)}{R}=g\left(1-\frac{h}{R}\right)\)…(4)

In this case, g = \(\frac{G M}{R^2}\) = acceleration due to gravity on the surface of the earth.

From equation (4), we conclude that,

- The acceleration due to gravity from the centre of the earth to its surface is directly proportional to the distance (R-h) of the place from the centre of the earth. Thus the value of the acceleration due to gravity below the earth’s surface (g’) is less than that on the surface of the earth, and as depth h increases, the value of g’ decreases.
- At the centre of the earth, g’ = 0, because h = R. Hence, the value of the acceleration due to gravity at the centre of the earth is zero. In fact, when an object is kept at the centre of the earth, it is subjected to gravitational forces acting uniformly from all directions and the resultant of these forces is zero. Hence, the acceleration due to gravity is also zero.
- g-g’ = decrease in the acceleration due to gravity at a depth h from the earth’s surface = \(\frac{h g}{R}\)…(5)

It is seen from equations (2) and (4) that the value of the acceleration due to gravity decreases with height above the surface of the earth, as well as with depth below the surface of the earth. Hence, the acceleration due to gravity is maximum on the surface of the earth.

Comparing equations (3) and (5), it can be said that the acceleration due to gravity at a height h from the surface of the earth is less than that at a depth h, i.e., g decreases with height faster rate than with depth.

The graph shows the variation of g with an increase in distance from the centre of the earth. From the centre of the earth to the earth’s surface, the value of g is directly proportional to the distance and is represented by a straight line starting from the origin. After this, the value of g decreases with distance along the curved line PQ.

**Effect Of The Earth’s Diurnal Motion:** For its diurnal motion, the Earth rotates about its NS axis once in 24 hours. Therefore, except the objects at the poles, all objects on the surface of the earth rotate in a circular path. The centres of these circular paths lie on the earth’s NS axis.

Taking the surface of the earth as a reference frame, it can be stated that during circular motion, a centrifugal force acts on an object situated on the earth’s surface and partially balances the gravitational force on the body acting towards the centre of the earth.

This causes an apparent decrease in the weight of the body. As a result, the effective value of acceleration due to gravity is less compared to its actual value. The actual value means that of the acceleration due to gravity in the absence of the diurnal motion of the earth.

Let the radius of the earth be R, and the latitude of a point A on the surface of the earth be θ. A mass m is kept at A. The distance of the mass from the axis of rotation NS is r = Rcosθ. As the earth rotates with angular velocity ω about its own axis NS, in a circular path of radius r, the centrifugal force on the mass along the radius r acting away from the centre of the earth is mω²r.

Also, if g is the actual value of the acceleration due to gravity at A, the force of attraction directed towards the centre of the earth on the body = mg. At the same time, the component of the centrifugal force, directed away from the centre of the earth at that point = mω²rcosθ = mω²Rcos²θ. Hence, the resultant force on the mass m towards the centre of the earth,

F = \(m g-m \omega^2 R \cos ^2 \theta \quad \text { or, } F=m g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)

This is the apparent weight or the apparent force of gravity on the mass m at A. Hence, the apparent value of the acceleration due to gravity at A is

g’ = \(\frac{F}{m}=g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)…(6)

The above equation shows that the higher the value of cosθ, the lesser the value of the acceleration due to gravity will be at that place.

**Effect Of The Earth’s Diurnal Motion Special cases:**

1. At the poles θ = 90°; cosθ= 0

Hence, g’ = g. Therefore, the acceleration due to gravity does not decrease in the polar regions, owing to the diurnal motion of the earth.

2. At the equator θ = 0°; cosθ = 1

Hence, g’ \(=g\left(1-\frac{\omega^2 R}{g}\right) .\). Thus, due to the diurnal motion of the earth, the acceleration due to gravity is the lowest in the equatorial region. Putting R = 6.4 x 10^{6 }m, g = 9.8m · s^{-2} and \(\omega=\frac{2 \pi}{24 \times 3600} \mathrm{rad} \cdot \mathrm{s}^{-1}\), we get \(\frac{\omega^2 R}{g}=3.454 \times 10^{-3}\) and the value of the acceleration due to gravity in the equatorial region, g’ = g(1-3.454 x 10^{-3}) = 0.996546g.

Thus, due to the diurnal motion of the earth, the acceleration due to gravity in the equatorial region is 99.6546% of that in the polar region. So, the value decreases only by 0.3454%.

So, owing to the earth’s oblate spheroidal shape and diurnal motion, the acceleration due to gravity is the highest at the poles and the lowest at the equator.

## Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

## Variations In Acceleration Due To Gravity Numerical Examples

**Example 1. What will be the value of the acceleration due to gravity at a point 3 km above the earth’s surface? The diameter of the earth = 12800 km and g = 980 cm · s ^{-2} on the surface of the earth.**

**Solution:**

Radius of the earth R = \(\frac{12800}{2}\) = 6400 km.

As the height of the point (h = 3 km) is negligibly small in comparison to R, the acceleration due to gravity at that height,

g’ = \(g\left(1-\frac{2 h}{R}\right)=980\left(1-\frac{2 \times 3}{6400}\right)=980 \times \frac{6394}{6400}\) = 979.08 cm · s^{-2}.

**Example 2. An object of mass m is raised up to a small height h above the earth’s surface. If the acceleration due to gravity on the earth’s surface is g, prove that the weight of the object will decrease by \(\frac{2mgh}{R}\) with respect to that on the earth’s surface. R = radius of the earth.**

**Solution:**

Acceleration due to gravity at a small height h above the earth’s surface is given by

g’ = \(g\left(1-\frac{2 h}{R}\right)\)

Hence, weight at that height, \(m g^{\prime}=m g\left(1-\frac{2 h}{R}\right)=m g-\frac{2 m g h}{R}\)

or, \(m g-m g^{\prime}=\frac{2 m g h}{R}\)

So, the decrease in weight with respect to that on the earth’s surface = \(\frac{2mgh}{R}\)

**Example 3. At what depth below the surface of the earth, will the acceleration due to gravity decrease by 1% with respect to that on the earth’s surface? The earth can be taken as a uniform sphere of radius 6400 km.**

**Solution:**

If the acceleration due to gravity on the earth’s surface is g, then at a depth h below the earth’s surface,

g’ = \(g\left(1-\frac{h}{R}\right)=g-\frac{g h}{R}\)

∴ Decrease in value, g – g’ = \(\frac{g h}{R}\)

As per the question, \(\frac{g h}{R}\) = 0.01 g

or, h = 0.01 x R = 0.01 x 6400 = 64 km..

**Example 4. The acceleration due to gravity at a height h above the surface of the earth is the same as its value at a t depth d below the earth’s surface. Find the relationship between d and h.**

**Solution:**

Acceleration due to gravity at a height h, \(g_1=g\left(1-\frac{2 h}{R}\right)\)

Also, the value of acceleration due to gravity at a depth d, \(g_2=g\left(1-\frac{d}{R}\right)\)

Given \(g_1=g_2 or, g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right) or, d=2 h\).

**Example 5. If the present radius of the earth is doubled keeping the mass unchanged, how will the weight of a body on the surface of the earth change?**

**Solution:**

Weight of a body of mass m on the earth’s surface, W = mg = \(\frac{G M m}{R^2}\), where M and R are the mass and the radius of the earth.

When mass remains constant and radius doubles, then weight \(W_1=\frac{G M m}{(2 R)^2}=\frac{G M m}{4 R^2}\)

∴ \(\frac{W_1}{W}=\frac{G M m}{4 R^2} \cdot \frac{R^2}{G M m}=\frac{1}{4}\)

Hence, the new weight would be \(\frac{1}{4}\) th of the present weight.

**Example 6. What is the difference in the values of acceleration due to gravity at the polar region and at the equator due to the diurnal motion of the Earth? The radius of the earth = 6400 km**

**Solution:**

Angular velocity of diurnal motion of the earth \(\omega=\frac{2 \pi}{24 \times 60 \times 60} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Radius of the earth = 6400 km = 6.4 x 10^{8} cm

Acceleration due to gravity at latitude θ, g’ = \(g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)

= \(g-\omega^2 R \cos ^2 \theta\)

In the polar regions θ = 90°, i.e., cosθ = 0.

Therefore, the acceleration due to gravity is g_{1} = g.

At the equator θ = 0° and cosθ = 1; thus the acceleration due to gravity is g_{2} = g – ω²R.

So, the difference in the values of acceleration due to gravity in these two regions

⇒ \(g_1-g_2=\omega^2 R=\frac{(2 \pi)^2 \times 6.4 \times 10^8}{(24 \times 60 \times 60)^2}=3.38 \mathrm{~cm} \cdot \mathrm{s}^{-2} \text {. }\)

**Example 7. For what value of the angular velocity of the earth, the acceleration due to gravity in the equatorial region would have been zero? The average density of the earth’s material = 5.5 g · cm ^{-3} and G = 6.67 x 10^{-8} CGS unit. Find the ratio of this calculated value and the present value of the angular velocity of the earth.**

**Solution:**

The acceleration due to gravity at the equator g’ = g – ω²R

For g’ to be zero, the value of ω_{0} should be such that \(\)

As the average density of the earth is ρ and its radius is R, the mass of the earth M = \(\frac{4}{3} \pi R^3\)

Also, g = \(\frac{G M}{R^2}=\frac{G}{R^2} \cdot \frac{4}{3} \cdot \pi R^3 \rho=\frac{4 \pi G R \rho}{3}\)

Hence, the required angular velocity, \(\omega_0=\sqrt{\frac{g}{R}}=\sqrt{\frac{4 \pi G \rho}{3}}\)

∴ \(\omega_0=\sqrt{\frac{4 \times \pi \times 6.67 \times 10^{-8} \times 5.5}{3}}\)

= \(1.24 \times 10^{-3} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Present angular velocity of the earth \(\omega =\frac{2 \pi}{24 \times 60 \times 60}=7.27 \times 10^{-5} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

∴ \(\frac{\omega_0}{\omega}=\frac{1.24 \times 10^{-3}}{7.27 \times 10^{-5}}=17 \text { (approx) }\)

Hence, if the angular velocity of the diurnal motion of the earth had been 17 times its present value, the acceleration due to gravity at the equator would be zero.

**Alternative Method:** From known values of ω, R and g, \(\frac{\omega^2 R}{\boldsymbol{g}} \approx \frac{1}{289} \text { or, } \omega=\sqrt{\frac{g}{289 R}} \text {. }\)

If the acceleration due to gravity is zero at the equator, then the required angular velocity, \(\omega_0=\sqrt{\frac{g}{R}}\)

∴ \(\frac{\omega_0}{\omega}=\frac{\sqrt{g / R}}{\sqrt{g /(289 R)}}=\sqrt{289}=17 .\)

**Example 8. If the diurnal motion of the earth stops for some reason, what would be the percentage change in the weight of a body at the equator? The radius of the earth = 6400 km and g = 9.8 m · s ^{-2}.**

**Solution:**

Due to the diurnal motion of the earth, the apparent weight of a body at the equator W’ = W-mω²R

When the earth stops rotating, ω = 0 and hence, W’ = W. In this case, the increase in weight of the object,

ΔW=W-W’ = W-(W- mω²R) = mω²R

∴ Percentage increase in weight

= \(\frac{\Delta W}{W} \times 100=\frac{m \omega^2 R}{m g} \times 100=\left(\frac{2 \pi}{T}\right)^2 \times \frac{R}{g} \times 100\)

= \(\frac{4 \pi^2}{(24 \times 60 \times 60)^2} \times \frac{6400 \times 10^3}{9.8} \times 100=0.3454 \% .\)

**Example 9. Find the percentage decrease in weight of a body, when taken 16 km below the surface of the earth. Take the radius of the earth as 6400 km.**

**Solution:**

Here, R = 6400 km, x = 16 km.

The value of acceleration due to gravity at a depth x, \(g^{\prime}=g\left(1-\frac{x}{R}\right)=g\left(1-\frac{16}{6400}\right)\)

∴ \(g-g^{\prime}=\frac{1}{400} g=2.5 \times 10^{-3} \mathrm{~g}\)

If m is the mass of the body, then mg and mg’ are the weight of the body on the surface of the earth and at a depth of 16 km below the surface of the earth respectively. Then, the percentage decrease in the weight of the body

= \(\frac{m g-m g^{\prime}}{m g} \times 100\)

= \(\frac{g-g^{\prime}}{g} \times 100=\frac{2.5 \times 10^{-3}}{g} g \times 100=0.25 \%\)