The ratio of the escape velocity of a planet to the escape velocity of the Earth will be: Given: Mass of the planet is $16$ times the mass of the Earth and the radius of the planet is $4$ times the radius of the Earth.

$A$ 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

Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B,$ then the ratio of the escape velocity $\frac{V_A}{V_B}$ is

$A$ 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).

$A$ body is projected vertically upwards from the Earth's surface with a velocity $2 v_{e}$,where $v_{e}$ is the escape velocity from the Earth's surface. The velocity of the body when it escapes the gravitational pull is

The escape velocity of a body depends upon its mass as: