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)

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