If $g$ is the acceleration due to gravity at the earth's surface and $r$ is the radius of the earth,the escape velocity for a body to escape out of the earth's gravitational field is

$Assertion$: The escape speed does not depend on the direction in which the projectile is fired.
$Reason$: Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the Earth rotates about its axis.

$A$ body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body from the surface of the earth is

The escape velocity from the earth is about $11 \, km/s$. The escape velocity from a planet having twice the radius and the same mean density as the earth is ......... $km/s$.

$A$ mass of $6 \times 10^{24} \ kg$ is to be compressed into a sphere in such a way that the escape velocity from the sphere is $3 \times 10^8 \ m/s$. What should be the radius of the sphere? (Given: $G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

$A$ particle released at a large distance from a planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre. If $v_e$ is the escape velocity of a body at the planet,then the particle's speed at the centre of the planet is