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

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 $km/s$:

$A$ bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height,its acceleration due to the planet's gravity is $1/4$ of its value at the surface of the planet. If the escape velocity from the planet is $v_{esc} = v \sqrt{N}$,then the value of $N$ is (ignore energy loss due to atmosphere).

$A$ body of mass $m$ is situated at a distance equal to $2R$ ($R$ is the radius of the Earth) from the Earth's surface. The minimum energy required to be given to the body so that it may escape out of the Earth's gravitational field is:

$A$ body is projected vertically upwards from the surface of the earth with a speed of $k{v_e}$,where $k < 1$ and ${v_e}$ is the escape velocity of the earth. What is the maximum height from the center of the earth that the body will reach? (Given: $R$ is the radius of the earth)

$A$ satellite is orbiting close to the Earth and has a kinetic energy $K$. The minimum extra kinetic energy required by it to just overcome the gravitational pull of the Earth is