If the value of $g$ (acceleration due to gravity) at the Earth's surface is $10 \ m/s^2$,what is its value in $m/s^2$ at the center of the Earth,assuming the Earth is a sphere of radius $R$ meters with uniform mass density?

The radius of the earth is $6400 \, km$ and $g = 10 \, m/s^2$. In order that a body of $5 \, kg$ weighs zero at the equator,the angular speed of the earth is

$A$ 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 $g = \pi^2 \ ms^{-2}$]

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{4}$ below the surface of the Earth,what is the new frequency of oscillation? ($R =$ radius of the Earth)

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (both $x$ and $h$ being much smaller than the radius of the earth)

If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)