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 the earth) in terms of $R$, the radius of the earth is

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

Answer 1

Answer: (A)

$g'=\frac{GM}{\left(R+h\right)^{2}},$ acceleration due to gravity at height $h$
$\Rightarrow \frac{g}{g}=\frac{GM}{R^{2}}. \frac{R^{2}}{\left(R+h\right)^{2}}=g\left(\frac{R}{R+h}\right)^{2}$
$\Rightarrow \frac{1}{g}=\left(\frac{R}{R+h}\right)^{2} \Rightarrow \frac{R}{R+h}=\frac{1}{3}$
$\Rightarrow 3R=R+h \Rightarrow 2R=h$