What should be the angular velocity of earth due to rotation about its own axis so that the weight at the equator becomes $(3/5)^{th}$ of its initial value? ($g$ = acceleration due to gravity, $R$ = radius of earth)

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

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

An astronaut on a strange planet finds that the acceleration due to gravity is twice that on the surface of the Earth. Which of the following could explain this?

The average density of the Earth is [ $g$ is acceleration due to gravity]:

The acceleration due to gravity on the moon is $\frac{1}{6}$ times the acceleration due to gravity on the earth. If the ratio of the density of the earth $\rho_e$ to the density of the moon $\rho_m$ is $\frac{5}{3}$,then the radius of the moon $R_m$ in terms of the radius of the earth $R_e$ is: