$A$ body of weight $W$ is projected vertically upwards from the Earth's surface to reach a height above the Earth which is equal to nine times the radius of the Earth. The weight of the body at that height will be

The acceleration due to gravity at a height $1\, km$ above the earth is the same as at a depth $d$ below the surface of earth. Then $d = $ ......... $km$

Imagine a new planet having the same density as that of Earth but it is $3$ times bigger than the Earth in size. If the acceleration due to gravity on the surface of Earth is $g$ and that on the surface of the new planet is $g^{\prime}$,then

$A$ mine is located at depth $\frac{R}{3}$ below the earth's surface. The acceleration due to gravity at that depth in the mine is ($R = \text{radius of earth}$,$g = \text{acceleration due to gravity at surface}$).

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

The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is $9 : 4$. The mass of the planet is $\frac{1}{9}^{th}$ of that of the Earth. If $R$ is the radius of the Earth,what is the radius of the planet? (Assume the planets have the same mass density)