The depth from the surface of the earth of radius $R$,at which the acceleration due to gravity will be $60 \%$ of its value on the earth's surface,is:

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (where $x$ and $h$ are much smaller than the radius of the earth $R_e$):

The radius of the earth is $6400 \, km$ and $g = 10 \, m/s^2$. In order that a body of $5 \, kg$ weighs zero at the equator,the angular speed of the earth is

The acceleration due to gravity at the surface of the planet is the same as that at the surface of the earth,but the density of the planet is thrice that of the earth. If $R$ is the radius of the earth,the radius of the planet will be:

If the angular speed of the earth is doubled,the value of acceleration due to gravity $(g)$ at the north pole: