If ${v_e}$ and ${v_o}$ represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius $R$,then

The escape velocity for a body projected vertically upwards from the surface of the Earth is $11 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ........... $km/s$.

If an object is projected vertically upwards with a speed equal to half the escape speed of Earth,then the maximum height attained by it is ($R$ is the radius of Earth).

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:

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 condition for a uniform spherical mass $m$ of radius $r$ to be a black hole is [$G=$ gravitational constant and $c=$ speed of light]