What should be the angular speed with which the Earth must rotate on its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as their present weight?

Acceleration due to gravity $g$ for a body of mass $m$ on the Earth's surface is proportional to (Radius of Earth $= R$,mass of Earth $= M$)

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

The angular speed of the Earth,such that an object on the equator may appear weightless,is ($g = 10\,m/s^2$,radius of the Earth $R = 6400\,km$).

The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

The ratio of the radii of a planet and the earth is $1: 2$, and the ratio of their mean densities is $4: 1$. If the acceleration due to gravity on the surface of the earth is $9.8 \,ms^{-2}$, then the acceleration due to gravity on the surface of the planet is: (in $\,ms^{-2}$)