The escape velocity for a planet whose mass is six times the mass of Earth and whose radius is twice the radius of Earth will be (where $V_{e}$ is the escape velocity from the Earth).

$A$ particle of mass $m$ is projected with a velocity $v = k V_{e}$ $(k < 1)$ from the surface of the earth. $(V_{e} = \text{escape velocity})$. The maximum height above the surface reached by the particle is:

If ${v_e}$ and ${v_o}$ represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius $R$,then

Escape velocity on Earth is $11.2 \, km/s$. What would be the escape velocity on a planet whose mass is $1000$ times and radius is $10$ times that of Earth?

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches up to a height of $\frac{4}{5} R_E$ from the earth's surface,where $R_E$ is the radius of the earth. If the escape velocity of the object is $v_E$,then the value of $\frac{v}{v_E}$ is: