The masses and radii of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centers are at a distance $d$ apart. The minimum velocity with which a particle of mass $m$ should be projected from a point midway between their centers so that it escapes to infinity is:

$A$ 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)

Explain escape energy and provide its definition. Also, explain escape speed.

The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is four times and density is nine times that of the earth,is (in $,v_e$)

$A$ small asteroid is orbiting around the sun in a circular orbit of radius $r_0$ with speed $v_0$. $A$ rocket is launched from the asteroid with speed $v = \alpha v_0$,where $v$ is the speed relative to the sun. The highest value of $\alpha$ for which the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effects of other planets).

If the radius of a planet is four times that of Earth and the value of $g$ is the same for both,the escape velocity on the planet will be ......... $km/s$.