Question

Find out the height from earth's surface at which acceleration due to gravity becomes $\dfrac{g}{4}$(where g is acceleration due to gravity on the surface of earth and R is radius of earth)-
(A) $\sqrt{2}{R}$
(B) ${R}$
(C) $\dfrac{R}{\sqrt{2}}$
(D) ${2}{R}$

Answer 1

Hint
The gravitational acceleration is a falling acceleration of an object without any drag,the change in gravitational acceleration with distance from the centre of Earth follows an inverse-square law. This means that gravitational acceleration is inversely proportional to the square of the distance from the centre of Earth.

Complete step by step answer
Given that, acceleration due to gravity is g, and R is the radius of the earth.
We know that, $g = \dfrac{{G{M_e}}}{{{R^2}}}$
Now let us assume that h is the height from the earth surface
So,${g_1} = \dfrac{{G{M_e}}}{{{{(R + h)}^2}}}$ … (1)
In equation 1 this is the gravity in h height,
So, $\dfrac{g}{4} = \dfrac{{G{M_e}}}{{{{(R + h)}^2}}}$
Now putting the value of $\dfrac{g}{4}$ we get,
$\Rightarrow \dfrac{{G{M_e}}}{{4{R^2}}} = \dfrac{{G{M_e}}}{{{{(R + h)}^2}}}$
Hence,
$\Rightarrow \dfrac{1}{{4{R^2}}} = \dfrac{1}{{{{(R + h)}^2}}}$
$\begin{gathered} \Rightarrow \dfrac{{{{(R + h)}^2}}}{{{R^2}}} = 4 \\\ \Rightarrow \dfrac{{R + h}}{R} = 2 \\\ \end{gathered}$
$\begin{gathered} \Rightarrow R + h = 2R \\\ \Rightarrow R = h \\\ \end{gathered}$
$\therefore R = h$
Hence, at height equal to R from earth surface gravity will become $\dfrac{g}{4}$.
Option (B) is correct.

Note
Newton's three laws of motion may be stated as follows:
A. Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.
B. Force equals mass times acceleration.
C. For every action there is an equal and opposite reaction.