Earth binds the atmosphere because of

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?

$A$ bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height,its acceleration due to the planet's gravity is $1/4$ of its value at the surface of the planet. If the escape velocity from the planet is $v_{esc} = v \sqrt{N}$,then the value of $N$ is (ignore energy loss due to atmosphere).

The radius of the planet is double that of the earth,but their average densities are same. $V_{p}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$,the value of '$x$' is

The escape velocity for a planet whose radius is $1.7 \times 10^6 \ m$ and acceleration due to gravity is $1.7 \ m s^{-2}$ is

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):