Radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth than change in potential energy will be
$3mgR$
$\frac {3}{2}\,mgR$
$mgR$
$\frac {3}{4}\,mgR$
In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g = 10\,m{s^{ - 2}}$ and radius of earth is $6400 \,kms)$
In a satellite if the time of revolution is $T$, then $K.E.$ is proportional to
The potential energy of a satellite of mass $m$ and revolving at a height $R_e$ above the surface of earth where $R_e =$ radius of earth, is
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
A particle of mass $m$ is placed at the centre of a uniform spherical shell of mass $3\,m$ and radius $R$. The gravitational potential on the surface of the shell is