Question

At what height above the earth surface the value of $g$ is $\dfrac{1}{2}$ of its value on the earth surface. Given its radius $6400km$?

Answer 1

Hint: In order to solve this question, first we will find the expression for the acceleration due to gravity and then find the expression which gives the variation of acceleration due to gravity with the height. Then we need to put the required values and arrive at the solution.

Complete step by step answer:
To solve this question, we need to find the relation for the variation of acceleration due to gravity with height.According to Newton’s second law of motion,
\[F = ma\]
In this case of gravity, $a = g$
So, $F = mg........(1)$
According to Newton’s law of gravitation, the force acting between the two objects $M$ and $m$, separated by a distance $R$ is,
$F = \dfrac{{GMm}}{{{R^2}}}........(2)$
On comparing equation (1) and (2), we get,
$mg = \dfrac{{GMm}}{{{R^2}}}$
$\Rightarrow g = \dfrac{{GM}}{{{R^2}}}.......(3)$

Now, to find the variation in the value of acceleration due to gravity with height, we will assume the shape of earth is a complete sphere instead of being elliptical.Let us suppose that the object is at a height $h$ from the surface of the earth. In this case, the distance between the centre of the earth and the position of the body is $R + h$. So, the acceleration due to gravity acting on the body changes to $g'$ which is given by,
$g' = \dfrac{{GM}}{{{{(R + h)}^2}}}........(4)$

Now, we will find the ratio of $g'$ to $g$,
$g' = g\dfrac{{{R^2}}}{{{{(R + h)}^2}}}$
$\Rightarrow g' = \dfrac{g}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}$
$\Rightarrow g' = g{\left( {1 + \dfrac{h}{R}} \right)^{ - 2}}$
On assuming $h < < R$ and then applying Bernoulli’s theorem,
$g' = g\left( {1 - \dfrac{{2h}}{R}} \right)........(5)$
In this question, we are given that $g' = \dfrac{g}{2}$.

On putting $g' = \dfrac{g}{2}$ in equation (5), we get,
$\dfrac{g}{2} = g\left( {1 - \dfrac{{2h}}{R}} \right)$
$\Rightarrow \dfrac{1}{2} = 1 - \dfrac{{2h}}{R}$
On further solving,
$\dfrac{1}{2} = \dfrac{{2h}}{R}$
$h = \dfrac{R}{4}$
In this question, we are given that $R = 6400km$,
$h = \dfrac{{6400}}{4}$
$\therefore h = 1600\,km$

So, the height above the earth's surface where the value of $g$ is $\dfrac{1}{2}$ of its value on earth's surface is $1600\,km$.

Note:From the expression given in equation (5), we can come to the conclusion that as we move above the surface of the earth, the value of acceleration due to gravity decreases. Similarly, as we go deeper into the earth the acceleration due to gravity gradually decreases.