The Earth is assumed to be a sphere of radius $R$. $A$ platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$,where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $V_{e}$,then the escape velocity from the planet is:

$A$ body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\text{rd}}$ of escape velocity. The maximum height reached by the body is ($R=$ radius of earth).

The escape velocity for a body of mass $1\, kg$ from the earth's surface is $11.2\, km/s$. The escape velocity for a body of mass $100\, kg$ would be:

$A$ rocket is launched straight up from the surface of the earth. When its altitude is $\frac{1}{3}$ of the radius of the earth,its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth,the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is $11.2 \ km/s$.) (in $km/s$)

The escape velocity of a body from the earth's surface is $v_e$. The escape velocity of the same body from a height equal to $R$ from the earth's surface will be