The ratio of the height $h$ above the surface of the Earth to the depth $d$ below the surface of the Earth,for which the gravitational accelerations are the same (assuming small heights),is:

The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $g/9$, where $g$ is acceleration due to gravity on earth's surface, is . . . . . . .

How does the acceleration due to gravity $(g)$ at a location on Earth change with latitude?

Explain the variations of acceleration due to gravity inside and outside the Earth and draw the graph.

$T$ is the time period of a simple pendulum on the Earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the Earth's radius) above the Earth's surface. Then,the value of $x$ will be:

When one moves from a point $16 \text{ km}$ below the earth's surface to a point $16 \text{ km}$ above the earth's surface, the change in $g$ is approximately $\alpha \%$. The value of $\alpha$ is . . . . . . . (Take radius of the earth $R = 6400 \text{ km}$.)