What is the gravitational force on a particle at the centre of the Earth?

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is [$R =$ radius of the earth].

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is

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 value of acceleration due to gravity at a height of $10 \,km$ from the surface of the Earth is $x$. At what depth inside the Earth is the value of the acceleration due to gravity the same value $x$ (in $\,km$)?

What is the effect on the value of $g$ at the equator if the Earth stops its rotation?