The diameters of two planets are in the ratio $4 : 1$ and their mean densities in the ratio $1 : 2$. The acceleration due to gravity on the planets will be in the ratio:

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 $\omega$)

If the Earth is assumed to be a sphere of radius $R$,and $g_{30}$ is the value of acceleration due to gravity at a latitude of $30^\circ$ and $g$ is the value at the equator,the value of $g - g_{30}$ is:

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:

The value of gravitational acceleration $g'$ at a height $h$ above the earth's surface is $\frac{g}{4}$. Then,what is the value of $h$ in terms of the earth's radius $R$?

At which location on Earth is the effective acceleration due to gravity maximum?