Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g$,where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \ km/s$,the escape speed on the surface of the planet in $km/s$ will be:

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 value of the escape velocity $v_e$ for a stationary object on the surface of a planet is directly proportional to the mass and radius of the planet." Is this statement correct? If not,correct it.

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:

$A$ body is projected from the earth's surface with a speed $\sqrt{5}$ times the escape speed $(V_{e})$. The speed of the body when it escapes from the gravitational influence of the earth is

Two spherical stars $A$ and $B$ have densities $\rho_A$ and $\rho_B$,respectively. $A$ and $B$ have the same radius,and their masses $M_A$ and $M_B$ are related by $M_B = 2M_A$. Due to an interaction process,star $A$ loses some of its mass,so that its radius is halved,while its spherical shape is retained,and its density remains $\rho_A$. The entire mass lost by $A$ is deposited as a thick spherical shell on $B$ with the density of the shell being $\rho_A$. If $v_A$ and $v_B$ are the escape velocities from $A$ and $B$ after the interaction process,the ratio $\frac{v_B}{v_A} = \sqrt{\frac{10n}{15^{1/3}}}$. The value of $n$ is. . . . .