Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g$,where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \ km/s$,the escape speed on the surface of the planet in $km/s$ will be:

The mass of a planet is half that of the Earth and the radius of the planet is one-fourth that of the Earth. If we plan to send an artificial satellite from the planet,the escape velocity will be (escape velocity on Earth $v_e = 11 \ km \ s^{-1}$): (in $km \ s^{-1}$)

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ($R$ - Radius of the earth).

An object $A$ of mass '$m$' is located at a point '$P$' at distances '$r$' and '$2r$' from two planets $B$ and $C$ of masses '$M$' and '$6M$' respectively,as shown in the figure. If the escape speed of the object $A$ from point '$P$' due to the gravitational influence of only planet $B$ is $5 \ km/s$,then the escape speed of the object $A$ from point '$P$' due to the gravitational influence of both the planets is . . . . . . $km/s$.

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):

$A$ particle of mass $m_0$ is projected from the midpoint of the line joining two fixed particles,each of mass $m$. If the separation between the fixed particles is $l$,the minimum velocity of projection of the particle so as to escape is equal to: