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

An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches up to a height of $\frac{4}{5} R_E$ from the earth's surface,where $R_E$ is the radius of the earth. If the escape velocity of the object is $v_E$,then the value of $\frac{v}{v_E}$ is:

The radius of a planet is twice the radius of the Earth. Both have almost equal average mass densities. If $V_P$ and $V_E$ are the escape velocities of the planet and the Earth respectively,then:

$A$ body of mass $m$ is situated at a distance $4R_e$ above the Earth's surface,where $R_e$ is the radius of Earth. How much minimum energy must be given to the body so that it may escape?

$A$ rocket is fired vertically with a speed of $4 \ km/s$ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth (in $km$)? (Take radius of Earth $R = 6.4 \times 10^6 \ m$ and $g = 10 \ m/s^2$)

To project a body of mass $m$ from Earth's surface to infinity,the required kinetic energy is (assume,the radius of Earth is $R_E$,$g =$ acceleration due to gravity on the surface of Earth):