If the radius of the earth is $6347 \, km$,then what will be the difference between the acceleration of free fall and the acceleration due to gravity near the earth's surface?

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 height at which the weight of a body becomes $\frac{1}{16}^{th}$ of its weight on the surface of the Earth (radius $R$) is: (in $R$)

The acceleration due to gravity on the moon is $\frac{1}{6}$ times the acceleration due to gravity on the earth. If the ratio of the density of the earth $\rho_e$ to the density of the moon $\rho_m$ is $\frac{5}{3}$,then the radius of the moon $R_m$ in terms of the radius of the earth $R_e$ is:

The percentage decrease in the weight of a body when taken to a height of $48 \text{ km}$ above the surface of the earth is (Radius of the earth is $6400 \text{ km}$) (in $\%$)

If acceleration due to gravity at distance $d < R$ from the centre of the Earth is $\beta$,then its value at distance $d$ above the surface of the Earth will be [where $R$ is the radius of the Earth].