Four particles,each of mass $M$ and equidistant from each other,move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

$A$ satellite of mass $m$ is at a distance of $a$ from a star of mass $M$. The speed of the satellite is $u$. Suppose the law of universal gravity is $F = -G \frac{Mm}{r^{2.1}}$ instead of $F = -G \frac{Mm}{r^2}$. Find the speed of the satellite when it is at a distance $b$ from the star.

Mean solar day is the time interval between two successive noons when the sun passes through the zenith point (meridian). $A$ sidereal day is the time interval between two successive transits of a distant star through the zenith point (meridian). By drawing an appropriate diagram showing the earth's spin and orbital motion,show that the mean solar day is $4\,\text{min}$ longer than the sidereal day. In other words,distant stars would rise $4\,\text{min}$ early every successive day.

Consider a spherical gaseous cloud of mass density $\rho(r)$ in free space,where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common center with the same kinetic energy $K$. The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant in time,the particle number density $n(r) = \rho(r) / m$ is:
[$G$ is the universal gravitational constant]

$A$ satellite of mass $5M$ orbits the Earth in a circular orbit. At one point in its orbit,the satellite explodes into two pieces,one of mass $M$ and the other of mass $4M$. After the explosion,the mass $M$ ends up travelling in the same circular orbit,but in the opposite direction. After the explosion,the mass $4M$ is in:

How many times is the escape velocity $(V_e)$ of the orbital velocity $(V_0)$ for a satellite revolving near the Earth?