Two particles of equal mass go round a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is
$v = \frac{1}{{2R}}\,\sqrt {\frac{1}{{Gm}}} $
$v = \sqrt {\frac{{Gm}}{{2R}}} $
$v = \frac{1}{2}\,\sqrt {\frac{{Gm}}{R}} $
$v = \sqrt {\frac{{4Gm}}{{R}}} $
Two stars of masses $m_1$ and $m_2$ are parts of a binary star system. The radii of their orbits are $r_1$ and $r_2$ respectively, measured from the centre of mass of the system. The magnitude of gravitational force $m_1$ exerts on $m_2$ is
If potential energy of a body of mass $m$ on the surface of earth is taken as zero then its potential energy at height $h$ above the surface of earth is [ $R$ is radius of earth and $M$ is mass of earth]
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,
A body of mass $m$ falls from a height $R$ above the surface of the earth, where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$ )
At what altitude will the acceleration due to gravity be $25\% $ of that at the earth’s surface (given radius of earth is $R$) ?