Two particles of equal mass $m$ each move in a circle of radius $r$ under the action of their mutual gravitational attraction. The time period of each particle is

$A$ satellite is revolving in a circular orbit at a height $h$ from the Earth's surface,such that $h \ll R$,where $R$ is the radius of the Earth. Assuming that the effect of the Earth's atmosphere can be neglected,the minimum increase in the speed required so that the satellite could escape from the gravitational field of the Earth is:

Four identical particles of mass $M$ are located at the corners of a square of side $a$. What should be their speed if each of them revolves under the influence of the others' gravitational field in a circular orbit circumscribing the square?

$A$ large spherical mass $M$ is fixed at one position and two identical point masses $m$ are kept on a line passing through the centre of $M$ (see figure). The point masses are connected by a rigid massless rod of length $\ell$ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to $M$ is at a distance $r = 3\ell$ from $M$,the tension in the rod is zero for $m = k\left(\frac{M}{288}\right)$. The value of $k$ is

The relative uncertainty in the period of a satellite orbiting around the Earth is $10^{-2}$. If the relative uncertainty in the radius of the orbit is negligible,the relative uncertainty in the mass of the Earth is:

$A$ satellite is orbiting the Earth at a height $h$ above the surface (where $R$ is the radius of the Earth and $h \ll R$). What is the minimum increase in the orbital velocity required for the satellite to escape the Earth's gravitational field? (Ignore the effect of the atmosphere.)