As we move from the poles of the Earth towards the equator,what happens to the value of the acceleration due to gravity $(g)$?

Derive the equation for the variation of acceleration due to gravity $g$ with depth $d$ below the surface of the Earth.

If all objects on the equator of the earth feel weightless,then the duration of the day will nearly become ....... $hr$.

$A$ simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4 \ m$,then the time period of small oscillations will be . . . . . . $s$. [Take $g = \pi^2 \ ms^{-2}$]

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 $R = 6400 \text{ km}$.)

Two spherical planets $A$ and $B$ have the same mass,but their densities are in a ratio $8:1$. For these planets,the ratio of acceleration due to gravity at the surface of $A$ to its value at the surface of $B$ is: