Question

If the radius of the earth is $R$ , then the height $h$ at which the value of $g$ becomes one-fourth, will be

• A. $\frac{R}{8}$
• B. $R$
• C. $\frac{3R}{4}$
• D. $\frac{R}{2}$

Answer 1

Answer: (B)

The acceleration due to gravity, $g^{\prime }=\frac{GM}{r^2}$
On rearranging, $g^{\prime }=\frac{GM}{{\left(R+h\right)}^2}$
Taking the $R$ common, $g^{\prime }=\frac{GM}{R^2{\left(1+\frac{h}{R}\right)}^2}$
We know that the value of acceleration due to gravity at height $h$ is given by,
$g^{\prime }=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$
$\Rightarrow \frac{g}{4}=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$
$\Rightarrow 1+\frac{h}{R}=2$
$\Rightarrow h=R$