The value of acceleration due to gravity at a depth of $ 1600 \,km $ is equal to: (Radius of Earth $ = 6400 \,km $) (in $\,ms^{-2}$)

The rotation of the Earth (of radius $R$) about its axis speeds up to a value such that a man at latitude angle $45^{\circ}$ feels weightlessness. The duration of a day in such a case is

What should be the angular speed of the Earth about its axis so that the acceleration due to gravity at the equator becomes zero?

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

The value of acceleration due to gravity $g$ is maximum at

$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth (in $N$)?