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 $= 6400 \, km$)

$g_e$ and $g_p$ denote the acceleration due to gravity on the surface of the Earth and another planet whose mass and radius are twice that of the Earth. Then:

Imagine a new planet having the same density as that of Earth but it is $3$ times bigger than the Earth in size. If the acceleration due to gravity on the surface of Earth is $g$ and that on the surface of the new planet is $g^{\prime}$,then

The time period of a simple pendulum increases by a factor of $\sqrt{2}$ at a height $h$ from the surface of the Earth. Then the value of $h$ is:

Where will it be profitable to purchase $1$ kilogram of sugar?

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 . . . . . . $g$.