Question

The height at which the acceleration due to gravity decreases by $36\%$ of its value on the surface of the earth. (The radius of the earth is $R$ ).
(A) $\dfrac{R}{6}$
(B) $\dfrac{R}{4}$
(C) $\dfrac{R}{2}$
(D) $\dfrac{{2R}}{3}$

Answer 1

First of all, we will find an expression which gives the apparent acceleration due to gravity. After that we will relate the actual and the apparent acceleration as we are told that it reduces by $36\%$ of the actual value. We will substitute the required values in the main equation and followed by manipulation to obtain the result.

Complete step by step answer:
In the given question, we are supplied the following data:
We are given the surface of the earth with the acceleration due to gravity into account.
The radius of the earth is given as $R$.
We are asked to find out the height above the surface of the earth where the acceleration due to gravity is $36\%$ of its value on the surface of the earth. To begin with, we know that the acceleration due to gravity decreases as we move up the surface of the earth. Higher we move up the surface of the earth, the smaller the effect of gravity becomes at that height. To solve this problem, we will have two acceleration due to gravity, one will be actual and the other will be apparent.
Let us proceed to solve the problem.
We have a formula which gives the acceleration due to gravity at a certain height above the surface of the earth, which is given by:
$g' = \dfrac{{g{R^2}}}{{{{\left( {R + h} \right)}^2}}}$ …… (1)
Where,
$g'$ indicates the acceleration due to gravity at a certain height.
$g$ indicates the acceleration due to gravity on the surface of the earth.
$R$ indicates the radius of the earth.
$h$ indicates height.
Since, the acceleration due to gravity decreases by $36\%$ of its value on the surface of the earth.
We can write the relation between actual acceleration due to gravity and the apparent acceleration due to gravity.
g' = g - \dfrac{{36g}}{{100}} \\\ \Rightarrow g' = \dfrac{{100g - 36g}}{{100}} \\\ \Rightarrow g' = \dfrac{{64g}}{{100}} \\\
So, we substitute the values of apparent acceleration due to gravity in the equation (1) and we get:
\dfrac{{64g}}{{100}} = \dfrac{{g{R^2}}}{{{{\left( {R + h} \right)}^2}}} \\\ \Rightarrow \dfrac{{64}}{{100}} = \dfrac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}} \\\ \Rightarrow {\left( {\dfrac{8}{{10}}} \right)^2} = \dfrac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}} \\\ \Rightarrow \dfrac{8}{{10}} = \dfrac{R}{{R + h}} \\\
Again, we manipulate further and we get:
\Rightarrow 10R = 8R + 8h \\\ \Rightarrow h = \dfrac{{2R}}{8} \\\ \therefore h = \dfrac{R}{4} \\\
Hence, the height above the surface of the earth where the acceleration due to gravity is reduced by $36\%$ of its value on the surface of the earth is $\dfrac{R}{4}$ .

The correct option is (B).

Note: While solving this problem, most of the students tend to make mistakes by assuming the second acceleration to be $36\%$ of the acceleration due to gravity on the surface of the earth, which is wrong. We are rather told that the acceleration decreases by $36\%$ . This will cause a great variation in the result if not solved properly.