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 $= 9.8\, m/s^2$)

The linear speed of a particle at the equator of the earth due to its spin motion is $V$. The linear speed of the particle at latitude $30^{\circ}$ is:

The value of $g$ at a particular point is $9.8\,m/s^2$. Suppose the Earth suddenly shrinks uniformly to half its present size without losing any mass. The value of $g$ at the same point (assuming that the distance of the point from the centre of the Earth does not shrink) will now be ......... $m/s^2$.

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

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{1}{6}$ th as much as at present is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is