The ratio of escape velocity at earth $(v_e)$ to the escape velocity at a planet $(v_p)$ whose radius and mean density are twice as that of earth is:

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).

$A$ body is projected vertically upwards from the surface of the Earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is $....\,S.$

The radius of the planet is double that of the earth,but their average densities are same. $V_{p}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$,the value of '$x$' is

The escape velocity of an object from the Earth depends upon the mass of the Earth $(M)$,its mean density $(\rho)$,its radius $(R)$,and the gravitational constant $(G)$. Thus,the formula for escape velocity is:

$A$ body is projected vertically from the Earth's surface of radius $R$ with a velocity equal to half the escape velocity. The maximum height reached by the body is