$A$ body is projected vertically upwards from the surface of the earth with a speed of $k{v_e}$,where $k < 1$ and ${v_e}$ is the escape velocity of the earth. What is the maximum height from the center of the earth that the body will reach? (Given: $R$ is the radius of the earth)

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A :$ The escape velocities of planet $A$ and $B$ are same. But $A$ and $B$ are of unequal mass.
Reason $R :$ The product of their mass and radius must be same,$M_{1}R_{1} = M_{2}R_{2}$.
In the light of the above statements,choose the most appropriate answer from the options given below.

The radius of a planet is $\frac{1}{4}$ of earth's radius and its acceleration due to gravity is double that of earth's acceleration due to gravity. How many times will the escape velocity at the planet's surface be as compared to its value on earth's surface?

The mass and radius of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a body of mass $m$ should be projected from a distance $\frac{2d}{3}$ from the centre of $M_1$ so as to escape to infinity is:

$A$ 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 ratio of the escape velocity of a planet to the escape velocity of the Earth will be: Given: Mass of the planet is $16$ times the mass of the Earth and the radius of the planet is $4$ times the radius of the Earth.