The escape velocity of a sphere of mass $m$ is given by ($G =$ Universal gravitational constant; $M_e =$ Mass of the earth and $R_e =$ Radius of the earth).

$A$ 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?

$A$ 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

If the escape velocity on Earth is $11.2 \text{ km/s}$, what is its value for a planet having double the radius and $8$ times the mass of Earth (in $\text{ km/s}$)?

$A$ 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

$A$ satellite is moving with a constant speed '$V$' in a circular orbit about the Earth. An object of mass '$m$' is ejected from the satellite such that it just escapes from the gravitational pull of the Earth. At the time of its ejection,the kinetic energy of the object is