The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $V_{e}$,then the escape velocity from the planet is:

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

Planet $A$ has mass $M$ and radius $R$. Planet $B$ has half the mass and half the radius of Planet $A$. If the escape velocities from the Planets $A$ and $B$ are $v_{A}$ and $v_{B}$ respectively,then $\frac{v_{A}}{v_{B}}=\frac{n}{4}$. The value of $n$ 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:

Does the escape speed of a body from the Earth depend on:
$(a)$ the mass of the body,
$(b)$ the location from where it is projected,
$(c)$ the direction of projection,
$(d)$ the height of the location from where the body is launched?

$A$ 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)