The escape velocity of a body on an imaginary planet,which has thrice the radius of the Earth and double the mass of the Earth,is (where $v_e$ is the escape velocity of Earth):

The masses and radii of the Earth and the 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 particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is

The escape velocity for a body projected vertically upwards from the surface of the Earth is $11 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ........... $km/s$.

$A$ 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 escape velocity for a rocket from Earth is $11.2 \ km/s$. Its value on a planet where the acceleration due to gravity is double that on the Earth and the diameter of the planet is twice that of Earth will be in $km/s$:

The escape velocity from Earth is $v_{es}$. $A$ body is projected with a velocity of $2v_{es}$. With what constant velocity will it move in the interplanetary space?