The acceleration due to gravity is $g$ at a point distant $r$ from the centre of the Earth of radius $R$. If $r < R$,then:

When one moves from a point $16 \text{ km}$ below the earth's surface to a point $16 \text{ km}$ above the earth's surface,the change in $g$ is approximately $\alpha \%$. The value of $\alpha$ is . . . . . . . (Take radius of the earth $R = 6400 \text{ km}$.)

If the radius of earth shrinks by $2 \%$ while its mass remains the same,the acceleration due to gravity on the earth's surface will approximately:

The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth).

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $0.5\, m$. If the mean density of the moon is two-thirds that of the earth and the radius is one-quarter that of the earth,what is the maximum vertical distance through which he can jump on the moon and the ratio of the duration of the jump on the moon to that on the earth?

$A$ body weighs $48 \ N$ on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is (in $N$)