Question

The height at which the acceleration due to gravity becomes $\dfrac{g}{9}$ (where $g =$ the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is:
(A) $2R$
(B) $\dfrac{R}{{\sqrt 2 }}$
(C) $\dfrac{R}{2}$
(D) $\sqrt 2 R$

Answer 1

Hint
We can use the formula for the variation of the acceleration due to gravity with the height above the surface of the earth. Then we can substitute the given value of the acceleration due to gravity in the formula and we will be able to calculate the height above the surface of earth in the terms of the radius of the earth.
Formula Used: In this solution we will be using the following formula,
$\Rightarrow g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$
where $g'$ is the acceleration due to gravity at the height $h$ above the surface.
$R$ is the radius of the earth and $g$ is the acceleration due to gravity on the surface of the earth.

Complete step by step answer
The value of the acceleration due to gravity varies with the change in the height above the surface of the earth. This variation in the value of the acceleration due to gravity is given by the formula,
$\Rightarrow g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$
Now in the question we are given that the value of the acceleration due to gravity at the height $h$ above the surface of the earth is given as, $\dfrac{g}{9}$. So on substituting this value in the equation we get,
$\Rightarrow \dfrac{g}{9} = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$
Now we can cancel the $g$ from both the sides of the equation. Therefore, we get
$\Rightarrow \dfrac{1}{9} = {\left( {\dfrac{R}{{R + h}}} \right)^2}$
Now we can take the square root on both sides of the equation. Hence we get,
$\Rightarrow \dfrac{R}{{R + h}} = \dfrac{1}{3}$
On doing cross multiplication we get,
$\Rightarrow 3R = R + h$
Now we can bring the $h$ to the LHS of the equation and take the terms containing $R$ to the RHS. So we get,
$\Rightarrow h = 3R - R$
So the height is, $h = 2R$
This is the height above the surface of the earth at which the value of the acceleration due to gravity becomes $\dfrac{g}{9}$.
So the correct option is A.

Note
The value of the acceleration due to gravity decreases with the increase of the altitude from the surface of the earth because, as the height increases, the distance from the center of the earth also increases. In the formula we used, we have taken earth to be a perfect sphere with symmetrically distributed mass about the center.