The ratio of the radius of the earth to that of the moon is $10$. The ratio of acceleration due to gravity on the earth and on the moon is $6$. The ratio of the escape velocity from the earth's surface to that from the moon is

The value of the acceleration due to gravity is $g _{1}$ at a height $h =\frac{ R }{2}( R =$ radius of the earth) from the surface of the earth. It is again equal to $g _{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{ d }{ R }\right)$ equals

A rocket is launched with velocity $10\, km/s$. If radius of earth is $R$, then maximum height attained by it will be

Obtain general equation of gravitation force at distance $r$ from the centre of earth and derive the equation of acceleration due to gravity on the surface of earth.

Imagine earth to be a solid sphece of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth d below earth's surface is same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of earth), the ratio of $\frac{h}{d}$ will be

The value of acceleration due to gravity at Earth’s surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$, is close to: (Radius of earth $= 6.4\times10^6\, m$)