The escape speed of an object on the surface of the earth is $V$. If the object is thrown out with speed $4V$ from the surface of the earth,what will be the speed of the object far away from the earth?

There is no atmosphere on the moon because

$A$ rocket is launched normal to the surface of the Earth,away from the Sun,along the line joining the Sun and the Earth. The Sun is $3 \times 10^5$ times heavier than the Earth and is at a distance $2.5 \times 10^4$ times larger than the radius of the Earth. The escape velocity from the Earth's gravitational field is $v_e = 11.2 \text{ km s}^{-1}$. The minimum initial velocity $(v_s)$ required for the rocket to be able to leave the Sun-Earth system is closest to:
(Ignore the rotation and revolution of the Earth and the presence of any other planet)

If the radius of a planet is $R$ and its density is $\rho$,the escape velocity from its surface will be

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

$A$ 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: