Q. The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g = \ $the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is
A) $2R$
B) $\frac{R}{\sqrt{3}}$
C) $\frac{R}{2}$
D) $\sqrt{2}R$
Explanation
(a) $g^{'} = \frac{GM}{(R + h)^{2}},$ acceleration due to gravity at height$\ h$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \frac{g}{9} = \frac{GM}{R^{2}} \bullet \frac{R^{2}}{(R + h)^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = g\left( \frac{R}{R + h} \right)^{2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \frac{1}{9} = \left( \frac{R}{R + h} \right)^{2}$
$\Longrightarrow \ \ \frac{R}{R + h} = \frac{1}{3}$
$\Longrightarrow \ \ \ \ \ \ \ 3R = R + h$
$\Longrightarrow \ \ \ \ \ \ \ 2R = h$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.