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. $R/2$
• D. $\sqrt{2}R$

Answer 1

The correct option is A $2R$

$\frac{GM}{9R^2}=\frac{GM}{(R+h{)}^2}$
$\Rightarrow 3R=R+h$
$\Rightarrow h=2R$