The escape velocity of a planet having mass $6$ times and radius $2$ times as that of Earth is

What is the escape velocity for the surface of a black hole?

Two stars of masses $3\times10^{31} \ kg$ each,and at distance $2\times10^{11} \ m$ rotate in a plane about their common centre of mass $O$. $A$ meteorite passes through $O$ moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star,the minimum speed that the meteorite should have at $O$ is: (Take Gravitational constant $G = 6.67\times10^{-11} \ Nm^2 \ kg^{-2}$)

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 :

Does the escape speed of a body from the Earth depend on:
$(a)$ the mass of the body,
$(b)$ the location from where it is projected,
$(c)$ the direction of projection,
$(d)$ the height of the location from where the body is launched?

If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for an orbit close to the earth's surface,then these are related by: