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

An object is kept at rest at a distance of $3R$ above the earth's surface,where $R$ is the earth's radius. The minimum speed with which it must be projected so that it does not return to earth is (Assume $M =$ mass of earth,$G =$ Universal gravitational constant).

The mass and radius of the earth and moon are $M, R$ and $m, r$ respectively. The distance between their centers is $d$. The minimum velocity with which a particle of mass $m_0$ should be projected from the midpoint between them so that it could reach infinity is

Planet $A$ has mass $M$ and radius $R$. Planet $B$ has half the mass and half the radius of Planet $A$. If the escape velocities from the Planets $A$ and $B$ are $v_{A}$ and $v_{B}$ respectively,then $\frac{v_{A}}{v_{B}}=\frac{n}{4}$. The value of $n$ is

If a body of mass $1\text{ kg}$ falls on the earth from infinity, it attains velocity $(v)$ and kinetic energy $(k)$ on reaching the surface of earth. The values of $v$ and $k$ respectively are . . . . . . . (Take radius of earth to be $6400\text{ km}$ and $g = 9.8\text{ m/s}^2$)

$A$ spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $V$. Due to the rotation of the planet about its axis,the acceleration due to gravity $g$ at the equator is $1/2$ of $g$ at the poles. Find the escape velocity of a particle on the planet in terms of $V$.