At what height above the Earth's surface does the acceleration due to gravity become $1\%$ of its value at the Earth's surface (in $R$)? (Radius of Earth $= R$)

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).

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 gravitational pull of the moon is $(1/6)^{\text{th}}$ of the earth and the mass of the moon is $(1/8)^{\text{th}}$ of the earth. This implies that the:

Derive the equation for the variation of acceleration due to gravity $g$ with depth $d$ below the surface of the Earth.

When a body is taken from the equator to the poles,its weight