At what height from the surface of the earth will the value of acceleration due to gravity fall to half of its value on the surface of the earth?

The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth,is (radius of earth is $6400 \ km$) (in $km$)

How much is the radius of the Earth at the equator greater than the radius at the poles?

Consider a particle of mass $m$ suspended by a string at the equator. Let $R$ and $M$ denote the radius and mass of the Earth. If $\omega$ is the angular velocity of rotation of the Earth about its own axis,then the tension on the string will be $(\cos 0^{\circ}=1)$

The acceleration due to gravity at a height $h$ above the surface of the $Earth$ is $g_h$. At a depth $d = 90 \text{ km}$ below the $Earth$'s surface, the acceleration due to gravity is also $g_h$. The value of $h$ is: (in $\text{ km}$)

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)