What should be the angular speed of the Earth so that a body lying on the equator may appear weightless? $(g = 10\,m/s^2, R = 6400\,km)$

If the angular speed of the earth is doubled,the value of acceleration due to gravity $(g)$ at the north pole:

Acceleration due to gravity on the moon is $\frac{1}{6}$ of the acceleration due to gravity on the earth. If the ratio of densities of earth $(\rho_e)$ and moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,then the radius of the moon $R_m$ in terms of $R_e$ will be:

If the acceleration due to gravity experienced by a point mass at a height $h$ above the surface of the Earth is the same as the acceleration due to gravity at a depth $d = \alpha h$ $(h \ll R_{e})$ from the Earth's surface,then the value of $\alpha$ will be: (use $R_{e} = 6400 \ km$)

Two planets $A$ and $B$ have densities $\varrho_1$ and $\varrho_2$ and have radii $r_1$ and $r_2$ respectively. The ratio of acceleration due to gravity on $A$ to that of $B$ is:

The mass of a planet and its diameter are three times those of the Earth. Then the acceleration due to gravity on the surface of the planet is ....... $m/s^2$.