Suppose that the matter (stars, gas, dust) of a particular galaxy, of total mass $\mathit{M}$, is distributed uniformly throughout a sphere of radius.A star of mass $\mathit{R}$is revolving about the center of the galaxy in a circular orbit of radius.$\mathit{r}\mathbf{<}\mathit{R}$

(a) Show that the orbital speed$\mathit{v}$ of the star is given by

$\mathit{v}\mathbf{=}\mathit{r}\sqrt{\frac{\mathbf{G}\mathbf{M}}{{\mathbf{r}}^{\mathbf{3}}}}$

And therefore that the star’s period $\mathit{T}$of revolution is

$\mathit{T}\mathbf{=}\mathbf{2}\mathit{\pi }\sqrt{\frac{{\mathbf{R}}^{\mathbf{3}}}{\mathbf{G}\mathbf{M}}}$

Independentof.$\mathit{r}$Ignore any resistive forces.

(b) Next suppose that the galaxy’s mass is concentrated near the galactic center, within a sphere of radius less than . What expression then gives the star’s orbital period?

Radius of the galaxy is.$R$

Orbit radius of the star is.$r$

Total mass of the galaxy is$M$.

Mass of the star is $m$.

The gravitational force between two masses $m_1$ and $m_2$ at a distance $r$ from each other is

${\mathit{F}}_{\mathbf{g}}\mathbf{=}\frac{\mathbf{G}{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$ ..... (I)

Here, $G$is the universal gravitational constant.

The radically outward centripetal force of a mass rotating uniformly with speed $v$ in a circular orbit of radius $r$is

${\mathit{F}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$ ..... (II)

The mass per unit volume of the galaxy is

$\rho =\frac{M}{\frac{4}{3}\pi R^3}$

The mass of the galaxy inside the star orbit is

$\begin{array}{c}{M}^{\text{'}}=\rho \times \frac{4}{3}\pi r^3\\ =\frac{M}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi r^3\\ =\frac{M{r}^3}{R^3}\end{array}$

This center of mass of this portion of the galaxy mass will be at the center of the galaxy and will gravitationally pull the star inside. From equation (I) the strength of this pull is

$\begin{array}{c}{F}_g=\frac{G{M}^{\text{'}}m}{r^2}\\ =\frac{Gm}{r^2}\times \frac{M{r}^3}{R^3}\\ =\frac{GMmr}{R^3}\end{array}$

From equation (II) the centripetal force on the start is

$F_c=\frac{m{v}^2}{r}$

Since the star doesn't have any radial movement, these two forces cancel each other. Thus

$\begin{array}{c}\frac{GMmr}{R^3}=\frac{m{v}^2}{r}\\ v^2=\frac{GM{r}^2}{R^3}\\ v=r\sqrt{\frac{GM}{R^3}}\end{array}$

The period of revolution is

$\begin{array}{c}T=\frac{2\pi r}{v}\\ =\frac{2\pi r}{r\sqrt{\frac{GM}{R^3}}}\\ =2\pi \sqrt{\frac{R^3}{GM}}\end{array}$

Thus, the period is.$2\pi \sqrt{\frac{R^3}{GM}}$

The total mass of the galaxy will pull gravitationally pull the star in this case and the gravitational force from equation (I) will become

$F_g=\frac{GMm}{r^2}$

Equate to the centripetal force to get

$\begin{array}{c}\frac{GMm}{r^2}=\frac{m{v}^2}{r}\\ v^2=\frac{GM}{r}\\ v=\sqrt{\frac{GM}{r}}\end{array}$

The time period is

$\begin{array}{c}T=\frac{2\pi r}{v}\\ =\frac{2\pi r}{\sqrt{\frac{GM}{r}}}\\ =2\pi \sqrt{\frac{r^3}{GM}}\end{array}$

Thus, the period is $2\pi \sqrt{\frac{r^3}{GM}}$.