Starting from the centre of the earth having radius $R$,the variation of $g$ (acceleration due to gravity) is shown by:

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

The percentage decrease in the weight of a body when taken to a height of $48 \text{ km}$ above the surface of the earth is (Radius of the earth is $6400 \text{ km}$) (in $\%$)

Two planets of radii in the ratio $2 : 3$ are made from the material of density in the ratio $3 : 2$. Then the ratio of acceleration due to gravity $g_1/g_2$ at the surface of the two planets will be

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 gravitational force of the Earth on a rocket at a height $h$ from the surface of the Earth is $\frac{1}{3}$ of the gravitational force on the surface of the Earth. Find the relationship between $h$ and $R_e$ (the radius of the Earth).