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:

Two spherical planets $P$ and $Q$ have the same uniform density $\rho$,masses $M_P$ and $M_Q$,and surface areas $A$ and $4A$,respectively. $A$ spherical planet $R$ also has uniform density $\rho$ and its mass is $(M_P + M_Q)$. The escape velocities from the planets $P, Q,$ and $R$ are $V_P, V_Q,$ and $V_R$ respectively. Then:

The potential energy of a satellite is $-8 \times 10^9 \ J$. What is its binding energy (escape energy)?

The escape velocity of a body from a planet whose mass is $6$ times the mass of Earth and radius is $2$ times the radius of Earth will be (where $V_{e}$ is the escape velocity of a body from the Earth's surface).

Earth has mass $M_1$ and radius $R_1$,and the moon has mass $M_2$ and radius $R_2$. The distance between their centers is $r$. $A$ body of mass $M$ is placed on the line joining them at a distance $r/3$ from the center of the Earth. To project the mass $M$ to escape to infinity,the minimum speed required is:

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 ...... .