Acceleration due to gravity $g$ for a body of mass $m$ on the Earth's surface is proportional to (Radius of Earth $= R$,mass of Earth $= M$)

The radii of two planets are respectively $R_1$ and $R_2$ and their densities are respectively $\rho_1$ and $\rho_2$. The ratio of the accelerations due to gravity at their surfaces is

$A$ spherical planet far out in space has a mass $M_0$ and diameter $D_0$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth (in $N$)?

The height $h$ at which the weight of a body will be the same as that at the same depth $h$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

The acceleration due to gravity on the earth's surface at the poles is $g$ and the angular velocity of the earth about the axis passing through the poles is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be the same,then $h$ is: ($h << R$,where $R$ is the radius of the earth)