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 $\text{ km}$)

Calculate the difference in the value of $g$ at the equator and at the poles due to the rotation of the Earth.

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

Imagine Earth to be a solid sphere of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth $d$ below Earth's surface is the same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of Earth),the ratio of $\frac{h}{d}$ will be

At what altitude will the acceleration due to gravity be $25\%$ of that at the earth's surface (given radius of earth is $R$)?

Assuming the earth to be a sphere of uniform density,the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to