$A$ satellite with kinetic energy $E_k$ is revolving around the Earth in a circular orbit. How much more kinetic energy should be given to it so that it may just escape into outer space?

$A$ 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)

$A$ mass of $6 \times 10^{24} \,kg$ is to be compressed in the form of a solid sphere such that the escape velocity from its surface is $3 \times 10^4 \,ms^{-1}$. The radius of the sphere is (Universal gravitational constant $G = 6.66 \times 10^{-11} \,N \,m^2 \,kg^{-2}$) (in $\,km$)

$A$ body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\text{rd}}$ of escape velocity. The maximum height reached by the body is ($R=$ radius of earth).

$A$ 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).

The mass of the Earth is $81$ times that of the Moon and the radius of the Earth is $3.5$ times that of the Moon. The ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon will be: