Question

Which of the following statements are true about acceleration due to gravity?
A) $g$ is zero at the centre of the earth.
B) $g$ decreases if the earth stops rotating about its axis.
C) $g$ decreases in moving away from the centre of the earth if $r < R$
D) $g$ decreases in moving away from the centre of the earth if $r > R$

Answer 1

Hint: We have to define the acceleration due to gravity and derive its formula using the force acting on the body due to gravity and the universal law of gravitation. Now, we have to take different positions concerning the radius of the earth and mark the distance taken as $r$. According to the taken positions of the object, find the acceleration due to gravity at that position and compare it with the acceleration due to gravity on the surface of the earth.

Complete answer:
When an object is falling freely under the influence of gravity, it attains some acceleration. This acceleration gained by the falling object due to gravitational force is known as acceleration due to gravity. It is denoted by $'g'$ and its standard value on the surface of the earth is $9.8m/{s^2}$.
Now, we will derive the formula of acceleration due to gravity. The force acting on a body due to gravity is given by:
$F = mg......(1)$
Where,
$F$ is the force.
$m$ is the mass of the object.
$g$ is the acceleration due to gravity.
But according to the universal law of gravitation,
$F = G\dfrac{{Mm}}{{{R^2}}}......(2)$
Where,
$F$ is the force.
$m$ is the mass of the object.
$M$ is the mass of the earth.
$R$ is the radius of the earth.
$G$ is the universal gravitational constant.
Equating equations (1) and (2) we get,
$mg = G\dfrac{{Mm}}{{{R^2}}}$
$ \Rightarrow g = \dfrac{{GM}}{{{R^2}}}......(3)$
Now, we have to find the acceleration due to gravity above the surface of the earth, below the surface of the earth, and at the center of the earth.
Let us assume that the earth has a uniform spherical surface and a uniform density. Then the density of the earth will be:
$\rho = \dfrac{M}{{\dfrac{4}{3}\pi {R^3}}}$
$ \Rightarrow M = \rho \times \dfrac{4}{3}\pi {R^3}......(4)$
Where,
$\rho $ is the density.
$M$ is the mass of the earth.
$R$ is the radius of the earth.
From equation (3) substitute the value of $M$ in equation (4) we get,
$\dfrac{{g{R^2}}}{G} = \rho \times \dfrac{4}{3}\pi {R^3}$
$ \Rightarrow g = \dfrac{4}{3}\left[ {\pi \rho RG} \right].....(5)$
Let the object is at a depth $'d'$ below the surface of the earth such that $R - d = r$ where $r < R$. The acceleration due to gravity at a depth is given by:
$ \Rightarrow {g_d} = \dfrac{4}{3}\pi \rho \left( {R - d} \right)G.....(6)$
From equations (5) and (6), we get,
${g_d} = \dfrac{{g(R - d)}}{R}$
$ \Rightarrow {g_d} = \dfrac{{gr}}{R}$
$ \Rightarrow {g_d} \propto r$
So, the value of ${g_d}$ increases with increase in $r$.
If the object is at the center of the earth then the depth, $d = R$. Therefore equation (6) becomes,
${g_d} = \dfrac{4}{3}\pi \rho \left( {R - R} \right)G$
$ \Rightarrow {g_d} = 0$
So, the value of ${g_d}$ at the center is zero.
Let the object be at a height $'h'$ above the surface of the earth such that $R + h = r$ where $r > R$. The force acting on the object at a height is given by:
${F_h} = G\dfrac{{Mm}}{{{{\left( {R + h} \right)}^2}}}$
Therefore, the acceleration due to gravity at height$({g_h})$ is given by:
$m{g_h} = G\dfrac{{Mm}}{{{{\left( {R + h} \right)}^2}}}$
$ \Rightarrow {g_h} = G\dfrac{M}{{{{\left( {R + h} \right)}^2}}}$
$ \Rightarrow {g_h} = \dfrac{{GM}}{{{R^2}{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}......(7)$
Equating equations (3) and (7) we get,
$ \Rightarrow {g_h} = \dfrac{g}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}$
$ \Rightarrow {g_h} = \dfrac{{g{R^2}}}{{{{\left( {R + h} \right)}^2}}}$
$ \Rightarrow {g_h} = \dfrac{{g{R^2}}}{{{r^2}}}$
$ \Rightarrow {g_h} \propto \dfrac{1}{{{r^2}}}$
Hence here the value of ${g_h}$ decreases with an increase in $r$.
If the earth stops rotating, then there will be no centripetal force acting on the body, and gravitational force increases. Therefore the value of $g$ will increase in some places and remain the same in other places.
Therefore, options (A) and (D) are correct.

Note:
All the objects experience the same acceleration due to gravity irrespective of their mass as we can infer from the formula. The acceleration due to gravity depends on the radius and mass of the earth which is constant. Also, the value of $g$ varies at poles and the equator of the earth too. The value of $g$ is more at poles than the equator.