The value of escape velocity on a certain planet is $2 \, km/s$. Then the value of orbital speed for a satellite orbiting close to its surface is

To project a body of mass $m$ from Earth's surface to infinity,the required kinetic energy is (assume,the radius of Earth is $R_E$,$g =$ acceleration due to gravity on the surface of Earth):

If the Earth has a mass nine times and a radius twice that of a planet $P$. Then $\frac{v_e}{3} \sqrt{x} \; ms^{-1}$ will be the minimum velocity required by a rocket to escape the gravitational force of $P$,where $v_e$ is the escape velocity on Earth. The value of $x$ is

The escape velocity for a body projected vertically upwards from the surface of the Earth is $11 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ........... $km/s$.

The ratio of the kinetic energy required to be given to a satellite to escape from the Earth's surface to the kinetic energy required to be given to the same satellite to revolve around the Earth in an orbit just above the Earth's surface is ...........

$A$ body is projected vertically upwards from the Earth's surface with a velocity $2 v_{e}$,where $v_{e}$ is the escape velocity from the Earth's surface. The velocity of the body when it escapes the gravitational pull is