91Ó°ÊÓ

The force acting between two particles is given by Newton’s law of gravitation,

\(F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\)

Here\(F\)is the force acting between two particles,\({m_1}\)and\({m_2}\)are the masses of the particle,\(G\)is the universal gravitational constant and\(r\)is the distance between the masses.

The centripetal force is given by,

\({F_c} = m\frac{{{v^2}}}{r}\)

Here\(m\)is the mass,\(v\)is the velocity and\(r\)is the radius of the circular path.

The gravitational pull on a star with mass \({\rm{m}}\) circling a galaxy with mass \({\rm{M}}\) in a circular orbit of radius \({\rm{r}}\) with velocity \({\rm{v}}\) must be equal to the centripetal force necessary to maintain it in motion.

Then we can write

\(\begin{array}{c}\overbrace {m\frac{{{v^2}}}{r}}^{{F_{c\,p}}} = \overbrace {G\frac{{mM}}{{{r^2}}}}^{{F_g}}\frac{m}{r}\\{v^2} = \frac{{GM}}{r}\\v = \sqrt {\frac{{GM}}{r}} \end{array}\)

From the above expression it is clear that the velocity of a star is inversely proportional to the square root of its orbital radius.