Acceleration due to gravity is defined as the constant acceleration produced in a body when it falls freely under the effect of gravity alone. The earth attracts every other body lying near or on its surface towards its center. The origin of this force between the earth and a body is due to the gravitational interaction between their masses. The force of attraction by the earth on a body is the gravitational pull or gravity.
We know that when a force acts on a body, it produces acceleration. Similarly, a body under the effect of gravitational pull must accelerate. This acceleration under produced by gravitation is known as acceleration due to gravity.
The acceleration due to gravity does not depend upon, whether the body falling under gravity is a light one or a heavy one, i.e. the value of g is independent of the mass of body and only depends on the mass of Earth.
The value of acceleration due to gravity near the surface of Earth is 9.8 ms-2
Acceleration due to Gravity Formula
The formula of the acceleration due to gravity is given by
$ \displaystyle g=\dfrac{{GM}}{{{{R}^{2}}}}$
Suppose, Earth is a spherical body of mass M, radius R with center at O. Consider a body of mass m placed on the surface of Earth. If the density of the Earth is uniform, its mass can be supposed to be concentrated at its center O.
Let F be the force of attraction between the body of mass m and the Earth, then according to Newton’s law of universal gravitation
$\displaystyle F=\frac{{GMm}}{{{{R}^{2}}}}$
According to Newton’s second law of motion if $\displaystyle a$ is the acceleration produced on a body of mass m due to a force F then,
$\displaystyle F=ma$
if a = g then,
$\displaystyle F=mg$
i.e. $ \displaystyle mg=\dfrac{{GMm}}{{{{R}^{2}}}}$
or
$\displaystyle g=\dfrac{{GM}}{{{{R}^{2}}}}$
We can now see that the value of acceleration due to gravity is independent of mass, shape and size of body but only depends on the mass of radius of Earth.
Mass and Density of Earth
The mass of earth is 6.018 × 1024kg
Since, the acceleration due to gravity is given by
$\displaystyle g=\dfrac{{GM}}{{{{R}^{2}}}}$
$\displaystyle \therefore M=\dfrac{{g{{R}^{2}}}}{G}$
$ \displaystyle \because G=6.67\times {{10}^{{-11}}}\text{ N}{{\text{m}}^{\text{2}}}\text{k}{{\text{g}}^{{\text{-2}}}}$ and $\displaystyle g=9.8\text{ m}{{\text{s}}^{{\text{-2}}}}$ and $\displaystyle R=6.4\times {{10}^{5}}\text{ m}$
$\displaystyle \begin{array}{l}\therefore M=\dfrac{{9.8\times {{{\left( {6.4\times {{{10}}^{5}}} \right)}}^{2}}}}{{6.67\times {{{10}}^{{-11}}}}}\\\Rightarrow M=6.018\times {{10}^{{24}}}\text{ kg}\end{array}$
Effect of Altitude on Acceleration due to Gravity
The value of acceleration due to gravity changes with height (i.e. altitude), depth or shape of the earth and rotation of the earth about its own axis. Suppose, if the mass of the earth is M, radius R with center at O. The acceleration due to gravity at a point A on the surface of earth is g.
The acceleration due to gravity at a point B at an altitude of h above the earth surface is given by
$ \displaystyle {g}’=g\left( {1-\dfrac{{2h}}{R}} \right)$
The above value shows that as the height h increases the value of acceleration due to gravity $\displaystyle {{g}’}$ decreases. It is due to this reason that the value of acceleration due to gravity is less at mountains than in plains.
Effect of Depth on Acceleration due to Gravity
If we consider a B at a depth of d below the earth surface than the acceleration due to gravity at B is given by
$\displaystyle {g}’=g\left( {1-\dfrac{d}{R}} \right)$
The above relation shows that the value of acceleration due to gravity decreases with the increase in depth d.
Therefore, at the center of earth i.e. when d = R, the value of g’ = 0. It means that the acceleration due to gravity at the center of earth is zero. The weight of the body is zero at the center of the earth.
Effect of Shape of Earth on Acceleration Due to Gravity
The radius of the Earth is more at the equator than at the poles. Due to this, the value of $g$ is more at the poles than that at the equator.
Effect of Latitude and Rotational Motion on Acceleration Due to Gravity
The latitude at a point on the surface of the Earth is defined as the angle which the line joining that point to the center of the Earth makes with the equatorial plane. It is denoted by $\lambda$.
It follows that at the poles: $\lambda = 90^\circ$, and at the equator: $\lambda = 0^\circ$.
The Earth is not perfectly a sphere but is ellipsoidal in shape. It also possesses rotational motion about an axis through the poles. The value of acceleration due to gravity at a place (at a given latitude) is affected both due to the shape of the Earth and its rotational motion.
Effect of Rotational Motion of the Earth
The Earth has rotational motion of period 24 h about its own axis. As the Earth rotates, a body placed on its surface moves along a circular path and hence experiences centrifugal force. The magnitude of the centrifugal force varies with the latitude of the place.
At the poles, the body moves along a circular path of zero radius, and hence the centrifugal force on the body is also zero. On the other hand, at the equator, the body moves along a circular path of radius equal to that of the Earth.
Since it is the circular path of maximum radius, the centrifugal force $\left( m R \omega^2 \right)$ on the body at the equator is maximum due to earth’s rotational motion. Therefore, the apparent weight of the body varies with latitude due to variation in the magnitude of centrifugal force on the body. the acceleration due to gravity decreases due to rotational motion of the earth and this effect is maximum at equator and zero at poles.
