$A$ body weighs $W$ newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

If $R_{E}$ is the radius of the Earth,then the ratio between the acceleration due to gravity at a depth $r$ below and a height $r$ above the Earth's surface is: (Given: $r < R_{E}$)

The weight of a body on the surface of the earth is $100\,N$. The gravitational force on it when taken at a height,from the surface of earth,equal to one-fourth the radius of the earth is $..........\,N$.

Obtain an expression for the acceleration due to gravity of the Earth at a depth $d$ below its surface.

What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$