Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ($R$ - Radius of the earth).

$A$ particle of mass $m$ is projected with a velocity $v = k V_{e}$ $(k < 1)$ from the surface of the earth. $(V_{e} = \text{escape velocity})$. The maximum height above the surface reached by the particle is:

The escape velocity of a body from a planet of mass $M$ and radius $R$ is $14 \,km \,s^{-1}$. The escape velocity of the body from another planet having same mass and diameter $8R$ (in $km \,s^{-1}$) is

The escape velocity of a sphere of mass $m$ is given by ($G =$ Universal gravitational constant; $M_e =$ Mass of the earth and $R_e =$ Radius of the earth).

The angular velocity of rotation of a star (of mass $M$ and radius $R$) at which the matter starts to escape from its equator will be:

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 :