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:

The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius four times that of Earth and the same mass density is:

The masses and radii of the earth and moon are $(M_1, R_1)$ and $(M_2, R_2)$ respectively. Their centers are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the midpoint between these two masses.

The escape velocity for the Earth is $11 \ km/s$. The escape velocity for a planet having twice the radius and the same density as the Earth is .......... $km/s$.

$A$ rocket is fired vertically with a speed of $4 \ km/s$ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth (in $km$)? (Take radius of Earth $R = 6.4 \times 10^6 \ m$ and $g = 10 \ m/s^2$)

The initial velocity $v_{i}$ required to project a body vertically upward from the surface of the earth to reach a height of $10 R$,where $R$ is the radius of the earth,may be described in terms of escape velocity $v_{e}$ such that $v_{i} = \sqrt{\frac{x}{y}} \times v_{e}$. The value of $x$ will be ...... .