$A$ 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).

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 . . . . . . $km/s$.

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 $km \ s^{-1}$)

$A$ particle of mass $m$ is kept at rest at a height $3R$ from the surface of the Earth,where $R$ is the radius of the Earth and $M$ is the mass of the Earth. The minimum speed with which it should be projected upward so that it does not return back is ($g$ = acceleration due to gravity on the Earth's surface).

The mass and radius of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a body of mass $m$ should be projected from a distance $\frac{2d}{3}$ from the centre of $M_1$ so as to escape to infinity is:

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: