The ratio of the orbital velocity of a body near the surface of a planet to the escape velocity of a body from the surface of the same planet is:

Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where $R_2 = \frac{R_1}{2}$. The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then $\frac{v_2}{v_1}$ is :

The radius in kilometer to which the present radius of earth $(R = 6400 \ km)$ must be compressed so that the escape velocity is increased $10$ times is ............ $km$.

The radius and mean density of a planet are four times that of the Earth. The ratio of the escape velocity on the Earth to the escape velocity on the planet is:

$A$ spaceship of mass $2 \times 10^4 \, kg$ is launched into a circular orbit close to the Earth's surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be $......$ (given $g = 10 \, m/s^2$ and radius of Earth $R = 6400 \, km$).

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