What is the value of acceleration due to gravity at a height equal to half the radius of the Earth,from its surface?

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

If the radius of the earth contracts by $2\%$ and its mass remains the same,then the weight of a body at the earth's surface:

The acceleration due to gravity at a height of $6400 \,km$ from the surface of the earth is $2.5 \,ms^{-2}$. The acceleration due to gravity at a height of $12800 \,km$ from the surface of the earth is (Radius of the earth $= 6400 \,km$) (in $\,ms^{-2}$)

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is [ $R$ = radius of earth,$g$ = acceleration due to gravity,$n=$ integer].

For a body of mass $m$,the acceleration due to gravity at a distance $R$ from the surface of the earth is $\frac{g}{4}$. Its value at a distance $\frac{R}{2}$ from the surface of the earth is ($R = \text{radius of the earth}$,$g = \text{acceleration due to gravity at the surface}$)