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Show that the velocity of a star orbiting its galaxy in a circular orbit is inversely proportional to the square root of its orbital radius, assuming the mass of the stars inside its orbit acts like a single mass at the center of the galaxy. You may use an equation from a previous chapter to support your conclusion, but you must justify its use and define all terms used.

Short Answer

Expert verified

The velocity of a star is inversely proportional to the square root of its orbital radius.

Step by step solution

01

Newton’s law of gravitation and centripetal force

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.

02

Explanation

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.

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