Starting from the centre of the earth having radius $R$,the variation of $g$ (acceleration due to gravity) is shown by:

$A$ $90 \,kg$ body placed at $2R$ distance from the surface of the Earth experiences a gravitational pull of: ($R=$ Radius of Earth,$g=10 \,ms^{-2}$) (in $\,N$)

The gravitational pull of the moon is $(1/6)^{\text{th}}$ of the earth and the mass of the moon is $(1/8)^{\text{th}}$ of the earth. This implies that the:

In the provided figure of the Earth,the value of acceleration due to gravity is the same at points $A$ and $C$,but it is smaller than its value at point $B$ (surface of the Earth). The value of $OA : AB$ is $x : y$. The value of $x$ is $\ldots \ldots \ldots$

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?

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 $R$?