Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where $R_2 = \frac{R_1}{2}$. The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then $\frac{v_2}{v_1}$ is :

$A$ satellite revolving around a planet has an orbital velocity of $10 \ km/s$. The additional velocity required for the satellite to escape from the gravitational field of the planet is: (in $km/s$)

The radius of a planet is twice the radius of the Earth. Both have almost equal average mass densities. If $V_P$ and $V_E$ are the escape velocities of the planet and the Earth respectively,then:

$A$ body of mass $m$ is situated at a distance $4R_e$ above the Earth's surface,where $R_e$ is the radius of Earth. How much minimum energy must be given to the body so that it may escape?

If the escape velocity of a body of mass $1\,kg$ from the surface of the Earth is $11.2\,km/s$,then what is the escape velocity for a body of mass $10\,kg$?

$A$ rocket is launched straight up from the surface of the earth. When its altitude is one-fourth of the radius of the earth,its fuel runs out and it coasts. What is the minimum velocity the rocket must have when it starts to coast if it is to escape from the gravitational pull of the earth? (Escape velocity on the surface of the earth is $11.2 \ km/s$)