## Gravity

**Gravity Definition:** The force of attraction exerted by the earth on the objects on or near the surface of the earth is called gravity.

Because of gravity, a body, when released from a height, starts moving towards the earth’s surface. Clearly, gravity is a form of gravitation.

Let the mass of the earth be M and its radius be R. At a distance of r from the centre of the earth, let there be a body of mass m. Taking the earth as an isotropic, uniform and homogeneous sphere, its entire mass may be considered to be concentrated at its centre. According to the law of gravitation, the force of attraction on the body due to the earth is

F = \(\frac{G M m}{r^2}\)….(1)

From Newton’s second law of motion, it can be stated that the body of mass m, being acted upon by a force F, gets an acceleration a in the direction of F, i.e., towards the centre of the earth. Acceleration produced in a body due to gravity is called acceleration due to gravity. This is denoted by g.

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

** Acceleration Due To Gravity: **Acceleration produced in a freely falling body under the action of gravity is called acceleration due to gravity.

In the case of a mass m, using Newton’s second law of motion, we have, F = mg

Hence, comparing with equation (1) \(m g=\frac{G M m}{r^2} \text { or, } g=\frac{G M}{r^2}\)….(2)

Both G and M are constants, hence, \(g \propto \frac{1}{r^2}\)…(3)

In other words, the acceleration due to gravity at a point near the surface of the earth is inversely proportional to the square of the distance between the body and the centre of the earth.

It is easily understood from equation (2) that the value of the acceleration due to gravity at a place does not depend on the mass of the body. At any fixed place, the value of acceleration due to gravity remains the same for a light or a heavy body.

Thus points equidistant from the centre of the earth have the same value of g. For any point on the surface of the earth r = R. Hence, from equation (2), the value of the acceleration due to gravity on the earth’s surface is

g = \(\frac{G M}{R^2}\)…(4)

This is the relation between acceleration due to gravity g and the gravitational constant G.

If the average density of the earth is p, then M = \(\frac{4}{3}\)πR³ρ

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

or, ρ = \(\frac{3 g}{4 \pi R G}\)……(5)

This is the relation between the gravitational constant G and the average density of the earth.

Using equations (4) and (5), or by other experiments, the value of the acceleration due to gravity on the surface of the earth can be obtained. The average value of g on the surface of the earth is:

**CGS System:** g = 980.6 cm · s^{-2}

**SI:** 9.806 m · s^{-2} ≅ 9.8 m · s^{-2}

**Acceleration Of The Body And Acceleration Of The Earth:** Let us take the example of two bodies as shown. The earth of mass M pulls a body of mass m towards its centre with a force F. As per the law of gravitation, the mass m also pulls the earth towards it with the same force F. (This is an example of the action and reaction forces being equal and opposite.)

Hence, from Newton’s second law of motion,

acceleration of the body = \(\frac{\text { force on the body }}{\text { mass of the body }}=\frac{F}{m}\)

also acceleration of the earth = \(\frac{\text { force on the earth }}{\text { mass of the earth }}=\frac{F}{M}\)

Therefore, \(\frac{\text { acceleration of the body }}{\text { acceleration of the earth}} =\frac{F / m}{F / M}=\frac{M}{m}\)

As the masses of ordinary objects, situated on or near the surface of the earth, are negligible compared to the mass of the earth, \(\frac{M}{m}\) >>1.

Hence, \(\frac{\text { acceleration of the body }}{\text { acceleration of the earth }}\) >>1

∴ acceleration of a body >> acceleration of the earth

Hence, the acceleration of a body is many times greater than the acceleration of the earth. Practically, it is the body that moves towards the earth; the acceleration of the earth towards the body is so small that it cannot be felt.

## Newtonian Gravitation And Planetary Motion

## Gravity Numerical Examples

**Example 1. What will be the percentage, increase or decrease in the value of acceleration due to gravity on the surface of the earth, if the radius of the earth decreases by 1 % and the mass of the earth remains unchanged?**

**Solution:**

Acceleration due to gravity on the surface of the earth, for the present value of the earth’s radius R, is g = \(\frac{G M}{R^2}\).

Let R’ = radius of the earth when it decreases by 1%, i.e., R’ = 0.99R.

At that time, the value of the acceleration due to gravity on the earth’s surface is g’ = \(\frac{G M}{R^2}\)

∴ \(\frac{g^{\prime}}{g}=\frac{R^2}{R^{\prime 2}}=\frac{R^2}{(0.99 R)^2}=\frac{1}{(0.99)^2}=1.02\)

Hence, g’ = 1.02g = g + 0.02g

Hence, in this case, the value of acceleration due to gravity on the earth’s surface increases by 2%.

**Alternative Method:**

We know, g = \(\frac{G M}{R^2}\)

or, \(\frac{d g}{d R}=-G M \frac{2}{R^3}=-2 \frac{G M}{R^2} \cdot \frac{1}{R}=-\frac{2 g}{R}\)

∴ \(\frac{d g}{g}=-2 \frac{d R}{R}\)

Radius decreased by 1% implies that, \(\frac{d R}{R}\) x 100 = -1

∴ Percentage change in g = \(\frac{d g}{g}\) x 100 = -2\(\frac{d R}{R}\) x 100 = 2

Hence, the value of g increases by 2%.

**Example 2. The mass of the moon is \(\frac{1}{80}\)th of the mass of the earth and its radius is \(\frac{1}{4}\)th of that of the earth. Compare the accelerations due to gravity on the Earth’s and the Moon’s surfaces.**

**Solution:**

Let the mass of the earth be M and its radius R; the acceleration due to gravity on the earth’s surface is g = \(\frac{G M}{R^2}\).

Similarly, taking the mass and radius of the moon as m and r respectively, acceleration due to gravity on the moon’s surface is g’ = \(\frac{G M}{r^2}\).

∴ \(\frac{g^{\prime}}{g}=\frac{m}{M} \cdot \frac{R^2}{r^2}=\frac{m}{M} \cdot\left(\frac{R}{r}\right)^2=\frac{1}{80} \times\left(\frac{4}{1}\right)^2=\frac{1}{5} .\)

Hence, the value of the acceleration due to gravity on the moon’s surface is \(\frac{1}{5}\)th of that on the surface of the earth.

**Mass And Average Density Of The Earth**

**Mass:** Let R = radius of the earth, M = its mass and ρ = average density.

From equation (4), M = \(\frac{g R^2}{G}\)

In this case, g = 9.8 m · s^{-2}; R = 6400 km = 6.4 x 10^{6 }m and G = 6.7 x 10^{-11 }N · m² · kg^{-2}.

∴ M \(=\frac{9.8 \times\left(6.4 \times 10^6\right)^2}{6.7 \times 10^{-11}}=6.0 \times 10^{24} \mathrm{~kg}\)

This is the measure of the mass of the earth.

**Average Density:** From equation (5) get ρ = \(\frac{3 g}{4 \pi R G}\)

Substituting the values of g, R and G, we obtain ρ = 5500 kg · m^{-3} = 5.5 g · cm^{-3}.

The density of seawater and of soil on the upper crust of the earth is 1 g · cm^{-3} and 2.7 g · cm^{-3} respectively. Thus, the value of the average density of the earth (5.5 g · cm^{-3}) indicates that the earth is not a sphere of uniform mass density.

The density of materials in the inner layers is definitely high. In fact, the density at the central core is about 13 g · cm^{-3}. It is, therefore, understood that the inner layers of the earth’s crust consist of heavy materials like metals.