The ratio of escape velocity at earth $(v_e)$ to the escape velocity at a planet $(v_p)$ whose radius and mean density are twice as that of earth is:

The mass of the moon is $1/144$ times the mass of a planet and its diameter is $1/16$ times the diameter of a planet. If the escape velocity on the planet is $v$,the escape velocity on the moon will be:

$A$ space station is at a height equal to the radius of the Earth. If $V_{E}$ is the escape velocity on the surface of the Earth,the escape velocity on the space station is __ times $V_{E}$.

The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius four times that of Earth and the same mass density is:

$A$ body of mass $m$ is situated at a distance equal to $2R$ ($R$ is the radius of the Earth) from the Earth's surface. The minimum energy required to be given to the body so that it may escape out of the Earth's gravitational field is:

Gas escapes from the surface of a planet because it acquires an escape velocity. The escape velocity will depend on which of the following factors :
$I.$ Mass of the planet
$II.$ Mass of the particle escaping
$III.$ Temperature of the planet
$IV.$ Radius of the planet
Select the correct answer from the codes given below :