The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

The mass of a planet is $\frac{1}{9}$ of the mass of the Earth and its radius is half that of the Earth. If a body weighs $9 \ N$ on the Earth,its weight on the planet would be ........ $N$.

The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth,assumed to be a sphere of radius $R$ of uniform density,is as shown in the figure below. The correct figure is:

$A$ person will get more quantity of matter in $kg-wt$ at which of the following locations?

The height at which the weight of a body becomes $\left(\frac{1}{9}\right)^{\text{th}}$ of its weight on the surface of the Earth is $(R = \text{radius of Earth})$: (in $R$)

The ratio of the height $h$ above the surface of the Earth to the depth $d$ below the surface of the Earth,for which the gravitational accelerations are the same (assuming small heights),is: