The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$. $A$ man jumps to a height of $2 \ m$ on the surface of $A$. What is the height of jump by the same person on the planet $B$?

The diameters of two planets are in the ratio $4:1$. Their mean densities have the ratio $1:2$. The ratio of gravitational acceleration on the surface of the planets will be:

The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

Assume there are two identical simple pendulum clocks. Clock-$1$ is placed on the Earth and Clock-$2$ is placed on a space station located at a height $h$ above the Earth's surface. Clock-$1$ and Clock-$2$ operate with time periods of $4\,s$ and $6\,s$ respectively. Then the value of $h$ is $....\,km$ (consider the radius of the Earth $R_E = 6400\,km$ and $g$ on Earth $10\,m/s^2$).

The acceleration due to gravity at the pole $(g_p)$ and at the equator $(g_e)$ are related as:

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (where $x$ and $h$ are much smaller than the radius of the earth $R_e$):