Question

A body weighs \[63N\] on the surface of the earth. What is the gravitational force (in \[N\] ) on it due to the earth at a height equal to half the radius of the earth?

Answer 1

Hint: The weight of a body is the product of the mass of the object and the acceleration due to gravity. The acceleration due to gravity has to be calculated. The formula of acceleration due to gravity at the given height from the earth's surface is to be used here. The gravitational force is also the product of mass and gravitational acceleration. Using these relations the gravitational force can be found at the given height.

Formula used:
The weight of a body of mass $m$, $W = mg$
$g$ is the acceleration due to the gravity of the earth.
The acceleration due to gravity at the height $h$ from the earth is, $g' = \dfrac{g}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}$
$R$ is the radius of the earth.
The gravitational force (in \[N\]) due to the earth, $W' = mg'$

Complete step by step solution:
Let, A body of mass $m$weights on the surface $W$ ,
The weight of a body of mass $m$, $W = mg$
Given, $W = 63N$
$g$ is the acceleration due to the gravity of the earth. $g = 10m/{s^2}$
$\therefore m = \dfrac{W}{g}$
$ \Rightarrow m = \dfrac{{63}}{{10}} = 6.3kg$
The value of the gravitational acceleration changes with the change of the height from the earth's surface.
The acceleration due to gravity at the height $h$ from the earth is, $g' = \dfrac{g}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}$
Given, $h = \dfrac{R}{2}$ , $R$ is the radius of the earth.
\[ \Rightarrow g' = \dfrac{g}{{{{\left( {1 + \dfrac{R}{{2R}}} \right)}^2}}}\]
\[ \Rightarrow g' = \dfrac{g}{{{{\left( {\dfrac{3}{2}} \right)}^2}}}\]
\[ \Rightarrow g' = \dfrac{{4g}}{9}\]
The gravitational force, $W' = mg'$
$ \Rightarrow W' = 6.3 \times \dfrac{{4g}}{9}$
$ \Rightarrow W' = \dfrac{{63}}{{10}} \times \dfrac{{4 \times 10}}{9}$
$ \Rightarrow W' = 28N$
The gravitational force due to the earth is $28N$

Note:
The gravitation acceleration i.e the acceleration due to gravity also changes with the depth of the earth. The gravitational acceleration decreases with the increase of the depth. That means as we move from the center of the earth to the surface of the earth, the gravity increases. But, in moving from the surface to some height the gravity decreases. The gravity is stronger i.e. maximum at the earth’s surface.