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

The escape velocity of a body on a planet $A$ is $12 \, km/s$. The escape velocity of the body on another planet $B$,whose density is four times and radius is half of the planet $A$,is ................... $km/s$.

An object is kept at rest at a distance of $3R$ above the earth's surface,where $R$ is the earth's radius. The minimum speed with which it must be projected so that it does not return to earth is (Assume $M =$ mass of earth,$G =$ Universal gravitational constant).

$A$ particle is projected from the surface of the earth with a velocity equal to twice the escape velocity. When the particle is very far from the earth,its speed would be

$A$ rocket is launched normal to the surface of the Earth,away from the Sun,along the line joining the Sun and the Earth. The Sun is $3 \times 10^5$ times heavier than the Earth and is at a distance $2.5 \times 10^4$ times larger than the radius of the Earth. The escape velocity from the Earth's gravitational field is $v_e = 11.2 \text{ km s}^{-1}$. The minimum initial velocity $(v_s)$ required for the rocket to be able to leave the Sun-Earth system is closest to:
(Ignore the rotation and revolution of the Earth and the presence of any other planet)

$A$ planet has twice the radius but the mean density is $\frac{1}{4}^{th}$ as compared to Earth. What is the ratio of the escape velocity from Earth to that from the planet?