If the radius of a planet is $R$ and density is $\rho$,then the escape velocity $v_{e}$ of any body from its surface will be proportional to:

$A$ rocket is fired vertically with a speed of $4 \ km/s$ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth (in $km$)? (Take radius of Earth $R = 6.4 \times 10^6 \ m$ and $g = 10 \ m/s^2$)

$A$ particle is kept at rest at a distance $R$ from the surface of the Earth (of radius $R$). The minimum speed with which it should be projected so that it does not return is

$A$ satellite is orbiting close to the Earth and has a kinetic energy $K$. The minimum extra kinetic energy required by it to just overcome the gravitational pull of the Earth is

$A$ body is projected from the earth's surface with thrice the escape velocity. What will be its velocity when it escapes the gravitational pull?

The radius of the planet is double that of the earth,but their average densities are same. $V_{p}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$,the value of '$x$' is