$A$ simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface,where $R$ is the radius of the earth. The value of $T_2/T_1$ is

The variation of acceleration due to gravity $g$ with distance $x$ from the centre of the Earth is best represented by ($R \rightarrow$ Radius of the Earth):

The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth,is (radius of earth is $6400 \ km$) (in $km$)

Assuming the Earth to be a sphere of uniform mass density,how much would a body weigh (in $N$) halfway down to the centre of the Earth if it weighed $250 \; N$ on the surface?

The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $g/9$, where $g$ is acceleration due to gravity on earth's surface, is . . . . . . .

At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).