Acceleration due to gravity is $g$ on the surface of the earth. The value of acceleration due to gravity at a height of $32 \, km$ above the earth's surface is ........ $g$. (Radius of the earth $= 6400 \, km$)

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

$Assertion$ : $A$ tennis ball bounces higher on hills than in plains.
$Reason$ : Acceleration due to gravity on the hill is greater than that on the surface of earth.

Two spherical planets $A$ and $B$ have the same mass,but their densities are in a ratio $8:1$. For these planets,the ratio of acceleration due to gravity at the surface of $A$ to its value at the surface of $B$ is:

The height above the surface of the earth where acceleration due to gravity becomes $\frac{g}{9}$ is ( $R$ is the radius of the earth,$g$ is the acceleration due to gravity at the surface).

If the earth stops rotating,the value of $g$ at the equator will