$A$ satellite revolving around a planet has an orbital velocity of $10 \ km/s$. The additional velocity required for the satellite to escape from the gravitational field of the planet is: (in $km/s$)

$A$ 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:

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):

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 $km/s$:

$A$ body of mass $m$ is dropped from a height $h = \frac{R}{2}$ above the surface of the Earth,where $R$ is the radius of the Earth. Find its speed when it hits the Earth's surface. (Given: $v_e$ is the escape velocity from the Earth's surface).

The ratio of accelerations due to gravity $g_{1}:g_{2}$ on the surfaces of two planets is $5:2$ and the ratio of their respective average densities $\rho_{1}:\rho_{2}$ is $2:1$. What is the ratio of respective escape velocities $v_{1}:v_{2}$ from the surface of the planets?