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

If earth has a mass nine times and radius twice to the planet P. Then $\dfrac{v_e}{3}\sqrt{x}$ ms$^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of P, where $v_e$ is escape velocity on earth. The value of $x$ is

A body is projected vertically from earth's surface of radius R with velocity equal to half the escape velocity. The maximum height reached by the body is

The density of a planet is three times that of earth and radius $2.5$ times that of earth. If $g_p$ and $g_e$ represents acceleration due to gravity on planet and earth respectively then ratio $g_p$ to $g_e$ is