Question

The height from earth's surface at which acceleration due to gravity becomes $\frac{g}{4}$ is (where $g$ is acceleration due to gravity on the surface of earth and $R$ is radius of earth).

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

Answer 1

The correct option is B $R$

We know that $g=\frac{G{M}_e}{R^2}$

And at height h from the earth surface $g_1=\frac{G{M}_e}{(R+h{)}^2}$

And here $g_1=\frac{g}{4}$

So, $\frac{g}{4}=\frac{G{M}_e}{(R+h{)}^2}$

so, $\frac{G{M}_e}{4R^2}=\frac{G{M}_e}{(R+h{)}^2}$

Hence, $\frac{1}{4R^2}=\frac{1}{(R+h{)}^2}$

$⟹\frac{(R+h{)}^2}{R^2}=4$

$⟹\frac{R+h}{R}=2$

$⟹R+h=2R$

$⟹R=h$

Hence, at height equal to R from earth surface gravity will become $\frac{g}{4}$.

Answer-(B)