A satellite is orbiting just above the surface of the planet of density ' $\rho$ ' with periodic time ' $T$ '. The quantity $\mathrm{T}^2 \rho$ is equal to ( $\mathrm{G}=$ universal gravitational constant)

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ( $\mathrm{g}=$ gravitational acceleration, $\mathrm{R}=$ equatorial radius of the earth)

A simple pendulum has a periodic time ' $\mathrm{T}_1$ ' when it is on the surface of earth of radius ' $R$ '. Its periodic time is ' $\mathrm{T}_2$ ' when it is taken to a height ' $R$ ' above the earth's surface. The value of $\frac{T_2}{T_1}$ is

The minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ is