The escape velocity for a body projected vertically upwards from the surface of earth is $11\, km/s$. If the body is projected at an angle of $45^o$ with the vertical, the escape velocity will be ........... $km/s$
$22$
$11$
$\frac{{11}}{{\sqrt 2 }}$
$11\sqrt 2 $
The radius of a planet is $R$. A satellite revolves around it in a circle of radius $r$ with angular velocity $\omega _0.$ The acceleration due to the gravity on planet’s surface is
A geostationary satellite is orbiting the earth at a height of $6\, R$ from the earth’s surface ($R$ is the earth’s radius ). What is the period of rotation of another satellite at a height of $2.5\, R$ from the earth’s surface
The two planets have radii $r_1$ and $r_2$ and their densities $p_1$ and $p_2$ respectively. The ratio of acceleration due to gravity on them will be
Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$, then,
If $M$ is mass of a planet and $R$ is its radius then in order to become black hole [ $c$ is speed of light]