At what height above the surface of the Earth does the value of $g$ decrease by $2 \%$? [Radius of the Earth is $6400 \, km$]

The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

By approximately how many meters is the radius of the Earth at the equator greater than the radius of the Earth at the poles (in $,000 \ m$)?

The moon's radius is $1/4$ that of the earth and its mass is $1/80$ times that of the earth. If $g$ represents the acceleration due to gravity on the surface of the earth,then the acceleration due to gravity on the surface of the moon is:

The weight of a body at the centre of the earth is

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400 \, km$. At the poles $g = 10 \, m/s^2$)