Question

The height at which the acceleration due to gravity becomes $g/9$ (where $g$ is acceleration due to gravity on the surface of the earth) in terms of the radius of the earth, $R$ is
(correct answer + 1, wrong answer - 0.25)

• A. $R/\sqrt{2}$
• B. $R/2$
• C. $\sqrt{2}R$
• D. $2R$

Answer 1

The correct option is D $2R$

Acceleration due to gravity $\left(g\right)=\frac{GM}{r^2}$, where $r$ is distance from centre of earth
$g_h=\frac{GM}{(R+h{)}^2}=\frac{GM}{R^2{[1+\frac{h}{R}]}^2}$, where $R=$ radius of earth
Put $\frac{GM}{R^2}=g$
$g_h=\frac{g}{{[1+\frac{h}{R}]}^2}$
Given $g_h=\frac{g}{9}$
$\frac{g}{9}=\frac{g}{{[1+\frac{h}{R}]}^2}$
${[1+\frac{h}{R}]}^2=9$
$1+\frac{h}{R}=3\Rightarrow h=2R$