Question

At a place, value of acceleration due to gravity $g$ is reduced by $2$% of its value on the surface of the earth (Radius of earth = $6400km$). The place is:-

• A. $64km$ below the surface of the earth
• B. $64km$ above the surface of the earth
• C. $32km$ above the surface of the earth
• D. $32km$ below the surface of the earth

Answer 1

The correct option is B $64km$ above the surface of the earth

At the surface of earth acceleration due to gravity is given by relation $g=\frac{GM}{R^2}$

Here $G=$ Gravitational constant

$M=$ mass of earth

$R=$ Radius of earth

At height $h$ from earth's surface acceleration due to gravity $g^1$ is

$g^1=\frac{GM}{{(R+h)}^2}$

from above two relations

$\frac{g^1}{g}=\frac{R}{{(R+h)}^2}$

$\frac{g^1}{g}=\frac{2}{100}=\frac{R}{{(R+h)}^2}$

$\frac{R}{R+h}=\frac{\sqrt{2}}{10}$

$\frac{1+h}{R}=\frac{\sqrt{2}}{10}$

$\frac{h}{R}=\frac{\sqrt{2}}{10}-1$

$h=(\frac{\sqrt{2}}{10}-1)R\frac{\sqrt{2}-10}{10}\times 6400$

$=640(\sqrt{2}-10)$

$=640\times 1.732-6400$

$=1108.48-6400$

$=64km$ from surface of earth.