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 escape velocity from the moon's surface is less than that on the earth's surface,because:

$A$ spaceship of mass $2 \times 10^4 \, kg$ is launched into a circular orbit close to the Earth's surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be $......$ (given $g = 10 \, m/s^2$ and radius of Earth $R = 6400 \, km$).

The escape velocity from the surface of the Earth is $V_e$. The escape velocity from the surface of a planet whose mass and radius are $3$ times those of the Earth will be:

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g$,where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \ km/s$,the escape speed on the surface of the planet in $km/s$ will be:

The acceleration due to gravity on a planet is the same as that on Earth,and its radius is four times that of Earth. What will be the value of the escape velocity on that planet if it is $v_e$ on Earth?