$A$ body is projected vertically from the Earth's surface with a velocity equal to half the escape velocity. The maximum height reached by the body is ($R =$ radius of the Earth).

The masses of two fixed spheres are $M$ and $2M$ and the radius of each sphere is $R$. Their centres are $10R$ apart. The minimum speed with which a particle of mass $\frac{M}{10}$ be projected from the mid-point of the line joining the centres of the two spheres so that it escapes to infinity is . . . . . . .

An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches up to a height of $\frac{4}{5} R_E$ from the earth's surface,where $R_E$ is the radius of the earth. If the escape velocity of the object is $v_E$,then the value of $\frac{v}{v_E}$ is:

The velocity with which a projectile must be fired so that it escapes Earth's gravitation does not depend on

$A$ 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)

The kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$) to infinity is: