The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

Consider a particle of mass $m$ suspended by a string at the equator. Let $R$ and $M$ denote the radius and mass of the Earth. If $\omega$ is the angular velocity of rotation of the Earth about its own axis,then the tension on the string will be $(\cos 0^{\circ}=1)$

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

If the radius of earth shrinks by $2 \%$ while its mass remains the same,the acceleration due to gravity on the earth's surface will approximately:

The angular velocity of the Earth at present is $\omega$. With what angular velocity should it rotate so that the weight of a body at the equator appears to be zero (in $\omega$)? (Express the answer as a multiple of $\omega$)

Two planets $A$ and $B$ have the same mass $M$ and radius $R$. The variation of density $\rho$ of the planets with distance $r$ from the center is shown in the following diagrams. The ratio of acceleration due to gravity at the surface of the planets $A$ and $B$ will be: