$A$ body weighs $W$ newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?

If the earth were to cease rotating about its own axis,the increase in the value of $g$ in the $C.G.S.$ system at a place of latitude of $45^{\circ}$ will be ........ $cm/sec^{2}$.

Consider two spherical planets of same average density. The second planet is $8$ times as massive as the first planet. The ratio of the acceleration due to gravity of the second planet to that of the first planet is

At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).

What is the effect on the value of $g$ at the equator if the Earth stops its rotation?