A freely falling body experiences a force that of gravity. This force increases the velocity of the falling object at a constant rate. This constant rate of change of velocity of a freely falling object is known as acceleration due to gravity.

The universal law of gravitation states that two bodies are attracted to each other by a force which is-
i. directly proportional to the product of their masses, and
ii. inversely proportional to the square of the distance \(d\) between their centres
Let us consider the mass of the earth as \(M\) and that of an object is \(m\).
The gravitational force on the object of mass \(m\),
\(
\begin{aligned}
&F \propto M \times m \\
&F \propto \frac{1}{d^2}
\end{aligned}
\)
Combining them, \(F \propto \frac{M m}{d^2}\)
\(
F=\frac{G M m}{d^2}
\)
\(G\) is called the universal gravitational constant and is equal to \(6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2\).

Force is defined as the product of mass and acceleration.
Force \(=\) mass \(\times\) acceleration
\(
F=m a
\)
For gravity, it is- \(F=m g\)
Therefore, \(F=m g=\frac{G M m}{d^2}\)
\(
g=\frac{G M}{d^2}
\)

For an object falling towards the earth, the distance between it and the surface is negligible compared to the radius \(r\) of the earth. So, we can equate \(d\) to \(r\).
\(
g=\frac{G M}{r^2}
\)
Therefore, \(g\) depends on the-
i. mass of the earth, \(M\)
ii. the radius of the earth, \(r\)

The mass of the object does not affect \(g\). All objects fall at the same acceleration simultaneously from the same height, irrespective of their weights. This was demonstrated by the Italian astronomer and scientist Galileo Galilei from the Leaning Tower of Pisa.

If a leaf or piece of paper falls slowly, it is because of air resistance to such light objects. Air resistance makes a parachute slow down the fall. A feather and a block of iron fall to the ground simultaneously in a vacuum.

Value of Acceleration Due to Gravity

Universal gravitational constant, \(G=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\)
Mass of the earth, \(M=6 \times 10^{24} \mathrm{~kg}\)
The average radius of the earth, \(r=6.4 \times 10^6 \mathrm{~m}\)
\(
\therefore g=\frac{G M}{r^2}=\frac{6.67 \times 10^{11} \times 6 \times 10^{24}}{\left(6.4 \times 10^6\right)^2}=9.8 \mathrm{~m} / \mathrm{s}^{-2}
\)
A freely falling body increases its velocity by \(9.8 \mathrm{~m} / \mathrm{s}\) every second as it nears the surface.

Acceleration Due to Gravity in Terms of Density

The density of an object is the ratio of its mass to its volume.
Let us assume the earth to be a perfect sphere. So, its volume is:
\(
V=\frac{4}{3} \pi r^3
\)
Density \(\rho=\frac{M}{V}\)
\(
\begin{aligned}
&g=\frac{G M}{r^2}=\frac{G}{r^2} \rho V=\frac{G \rho}{r^2} \times \frac{4}{3} \pi r^3 \\
&g=\frac{4 \pi}{3} G \rho r
\end{aligned}
\)
Acceleration due to gravity is directly proportional to the density of the earth.

Factors Affecting Acceleration Due to Gravity

Three factors affect acceleration due to gravity. They are:

• The shape of the earth.
• Altitude or height.
• Depth.

Example 1: The mass of the earth is 10 times that of another planet. Radius of the earth is 2 times its radius. Compare the weight of an object on the planet with that on the earth.

Solution: Given,
\(
\begin{aligned}
&M=10 M_p \\
&r=2 r_p
\end{aligned}
\)

Dividing,
\(
\frac{g}{g_p}=\frac{M}{M_p} \times\left(\frac{r_p}{r}\right)^2=10 \times\left(\frac{1}{2}\right)^2=2.25
\)
Since weight \(W=m g\), and \(m\) remains the same everywhere, \(\frac{W}{W_p}=2.25\)
Weight on the planet is approximately half of that on earth.

Example 2: Two balls of equal mass 2 kilograms are separated by 1 metre from their centres. What is the force of attraction between them?

Solution: \(F=\frac{G M m}{d^2}\)
Universal gravitational constant, \(G=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\)
\(
\begin{aligned}
&F=\frac{6.67 \times 10^{11} \times 2 \times 2}{1^2} \\
&F=2.668 \times 10^{-10} \mathrm{~N}
\end{aligned}
\)