$A$ satellite revolving around a planet has an orbital velocity of $10 \ km/s$. The additional velocity required for the satellite to escape from the gravitational field of the planet is: (in $km/s$)

$Assertion$: 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.

$A$ 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 $g = 10 \ m/s^2$)

For a planet having mass equal to the mass of the Earth but a radius that is one-fourth of the radius of the Earth,the escape velocity for this planet will be (in $km/s$): (in $km/s$)

The ratio of accelerations due to gravity $g_{1}:g_{2}$ on the surfaces of two planets is $5:2$ and the ratio of their respective average densities $\rho_{1}:\rho_{2}$ is $2:1$. What is the ratio of respective escape velocities $v_{1}:v_{2}$ from the surface of the planets?

$A$ body of mass $m$ is thrown vertically upward with speed $\sqrt{3} v_e$, where $v_e$ is the escape velocity of a body from the Earth's surface. The final velocity of the body is