Question

The height above the surface of the earth at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ is 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. $\sqrt{2}R$

Answer 1

Answer: (A)

$g^{′}=\frac{GM}{\left(R + h\right)^{2}}$ (acceleration due to gravity at height h)
$\Longrightarrow \, \frac{g}{9}=\frac{GM}{R^{2}}\cdot \frac{R^{2}}{\left(R + h\right)^{2}}$
$ \, \, =g\left(\frac{R}{R + h}\right)^{2}$
$\Longrightarrow \, \frac{1}{9}=\left(\frac{R}{R + h}\right)^{2}$
$\Longrightarrow \, \, \frac{R}{R + h}=\frac{1}{3}$
$\Longrightarrow \, 3R=R+h$
$\Longrightarrow \, 2R=h$