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)

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