Question

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where g = the acceleration due to gravity on the surface of th earth) in terms of $R$, radius of the earth is

• A. $2R$
• B. $\frac{R}{\sqrt{3}}$
• C. $\frac{R}{2}$
• D. $\sqrt{2}R$

Answer 1

The correct option is A $2R$

Acceleration due to gravity at height $h$ $g^{\prime }=\frac{GM}{(R+h{)}^2}$
Acceleration due to gravity on surface of earth $g=\frac{GM}{R^2}$
Where $R=$ radius of earth.
$\therefore \frac{g^{\prime }}{g}={\left(\frac{R}{R+h}\right)}^2$
As $g^{\prime }=\frac{g}{9}$ we get
$\Rightarrow \frac{\frac{g}{9}}{g}={\left(\frac{R}{R+h}\right)}^2$
$\Rightarrow \frac{1}{9}={\left(\frac{R}{R+h}\right)}^2\Rightarrow \frac{R}{R+h}=\frac{1}{3}$
$\Rightarrow 3R=R+h\Rightarrow 2R=h$