The condition for a uniform spherical mass $m$ of radius $r$ to be a black hole is [$G=$ gravitational constant and $c=$ speed of light]

The escape velocity of a planet having mass $6$ times and radius $2$ times as that of Earth is

$A$ rocket is launched straight up from the surface of the earth. When its altitude is $\frac{1}{3}$ of the radius of the earth,its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth,the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is $11.2 \ km/s$.) (in $km/s$)

The escape velocity from the surface of Earth of mass $M$ and radius $R$ is $V_{e}$. The escape velocity from the surface of a planet whose mass and radius are $3$ times that of the Earth will be:

$A$ projectile is projected with velocity $k{v_e}$ in the vertically upward direction from the ground into space. (${v_e}$ is the escape velocity and $k < 1$). If air resistance is considered to be negligible,then the maximum height from the center of the Earth to which it can go will be: ($R$ = radius of Earth)

The least velocity required to throw a body away from the surface of a planet so that it may not return is (radius of the planet is $6.4 \times 10^6 \ m$,$g = 9.8 \ m/s^2$).