If radius of the earth is $6347\, km ,$ then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface ?

The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by ($R =$ Earth's radius)

Write the equation of gravitation acceleration which is used for any height from the surface of earth.

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 (both $x$ and $h$ being much smaller than the radius of the earth)

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

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