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

$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

$A$ particle of mass $m$ is thrown upwards from the surface of the earth with a velocity $u$. The mass and the radius of the earth are $M$ and $R$, respectively. $G$ is the gravitational constant and $g$ is the acceleration due to gravity on the surface of the earth. The minimum value of $u$ so that the particle does not return back to earth 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)

$A$ body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body from the surface of the earth is

The escape velocity from the surface of the Earth is $V_e$. The escape velocity from the surface of a planet whose mass and radius are $3$ times those of the Earth will be: