Question

At a height equal to earth’s radius, above the earth’s surface, the acceleration due to gravity is:
A) $g$.
B) $\dfrac{g}{2}$.
C) $\dfrac{g}{4}$
D) $\dfrac{g}{5}$.

Answer 1

Due to the large mass of the earth there is a force that pulls the objects towards the centre of the earth and that force is gravitational force. The gravitational force attracts the object towards the centre of the earth with an acceleration known as acceleration due to gravity.

Complete step by step answer:
The formula of the acceleration due to gravity is given by,
$g = G\dfrac{{{M_e}}}{{{d^2}}}$
Where acceleration due to gravity is g the universal gravitational constant is G the mass of earth is ${M_e}$ and the distance of the object from the centre of the earth is d.
Step by step solution:
It is given in the problem that at height R from the surface of the earth an object is placed and we need to find the value of acceleration due to gravity on that object where the radius of the earth is equal to R.
The formula of the acceleration due to gravity is given by,
$g = G\dfrac{{{M_e}}}{{{d^2}}}$
Where acceleration due to gravity is g the universal gravitational constant is G the mass of earth is ${M_e}$ and the distance of the object from the centre of the earth is d.
The acceleration due to gravity at the surface of the earth is equal to,
$\Rightarrow g = G\dfrac{{{M_e}}}{{{d^2}}}$
$\Rightarrow g = G\dfrac{{{M_e}}}{{{R^2}}}$………eq. (1)
As the object is R distance from the surface of the earth where the radius of the earth is R the total distance of the object from the centre of the earth is 2R, then the acceleration due to gravity is given by,
$\Rightarrow {g_1} = G\dfrac{{{M_e}}}{{{d^2}}}$
$\Rightarrow {g_1} = G\dfrac{{{M_e}}}{{{{\left( {2R} \right)}^2}}}$
$\Rightarrow {g_1} = G\dfrac{{{M_e}}}{{4{R^2}}}$………eq. (2)
On comparing the equation (1) and equation (2) we get.
$\Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{{G\left( {\dfrac{{{M_e}}}{{4{R^2}}}} \right)}}{{G\left( {\dfrac{{{M_e}}}{{{R^2}}}} \right)}}$
$\Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{{\left( {\dfrac{1}{4}} \right)}}{1}$
$\Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{1}{4}$
$\Rightarrow {g_1} = \dfrac{g}{4}$.
So the acceleration due to gravity is equal to ${g_1} = \dfrac{g}{4}$.
The correct answer for this problem is option C.

Note: The acceleration due to gravity varies with the distance from the centre of the earth and the acceleration due to gravity is inversely proportional to the distance of the object from the centre of the earth and therefore more far the object is placed from the centre of the earth the less the acceleration due to gravity acts on the object.