Hunting a black hole.Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed$\mathbf{v}\mathbf{=}\mathbf{270}\mathbf{\text{km/s}}$, orbital period$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{.70}\mathbf{\text{days}}$, and approximate mass${\mathbf{\text{m}}}_{\mathbf{1}}\mathbf{=}\mathbf{6}{\mathbf{\text{M}}}_{\mathbf{\text{s}}}$, where${\mathbf{\text{M}}}_{\mathbf{\text{s}}}$ is the Sun’s mass,$\mathbf{1.99}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathbf{\text{kg}}$. Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. 13-47). What integer multiple of${\mathbf{\text{M}}}_{\mathbf{s}}$ gives the approximate mass${\mathbf{\text{m}}}_{\mathbf{2}}$of the dark star?

The velocity of the visible star is $\text{v}=270\text{\hspace{0.17em}km}/\text{s}=2.7\times {10}^5\text{\hspace{0.17em}m}/\text{s}$

The orbital period of the visible star is,

$\mathrm{T}=1.70\text{\hspace{0.17em}days}=\left(1.70\mathrm{days}\right)\left(\frac{86400\mathrm{s}}{1\mathrm{day}}\right)=146880\text{s}$

The approximate mass of the visible star is, ${\text{m}}_{\text{1}}={\text{6M}}_{\text{s.}}$

The mass of the sun is, ${\text{M}}_{\text{s}}=1.99\times {10}^{30}\text{\hspace{0.17em}s}$

Equate the gravitational force between the planet and the star to the centripetal force. Put the distance of the star (r₁) from C.O.M in this equation. Using the period of the star, find the massm₂of the dark star. Comparing it with the mass of the Sun, write it in terms of M_s.

Formulae are as follows:

The Centripetal force is$\mathrm{F}=\frac{{\mathrm{Mv}}^2}{\mathrm{r}}$

The Gravitational force of attraction between two bodies of masses M and m separated by distance d is, $\mathrm{F}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$

where F is force, G is gravitational constant, M, and m are masses, v is velocity and r is the radius.

The binary star system is orbiting about its center of mass. The gravitational force of attraction between them provides the centripetal force to keep their motion in the circular orbit. So,

$\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}=\frac{{\mathrm{m}}_1{\mathrm{v}}^2}{{\mathrm{r}}_1}$

As both stars revolve about the center of mass,

$\begin{array}{c}{\mathrm{r}}_1=\frac{({\mathrm{m}}_1\left(0\right)+{\mathrm{m}}_2\mathrm{r})}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ =\frac{{\mathrm{m}}_2\mathrm{r}}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ =\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}{\mathrm{r}}_1\end{array}$

The orbital speed of m₁ in terms of T is written as,

$\mathrm{v}=\frac{2{\mathrm{\pi r}}_1}{\mathrm{T}}$

So,

${\mathrm{r}}_1=\frac{\mathrm{vT}}{2\mathrm{\pi }}$

Substituting r₁in r,

$\mathrm{r}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}\left(\frac{\mathrm{vT}}{2\mathrm{\pi }}\right)$

Putting r and r₁ in F₁,

$\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\left[\frac{({\mathrm{m}}_1+{\mathrm{m}}_2)}{{\mathrm{m}}_2}\left(\frac{\mathrm{vT}}{2\mathrm{\pi }}\right)\right]}^2}=\frac{{\mathrm{m}}_1{\mathrm{v}}^2}{\frac{\mathrm{vT}}{2\mathrm{\pi }}}$

$\mathrm{F}=\frac{4{\mathrm{\pi }}^2{\mathrm{Gm}}_1{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2{\mathrm{v}}^2{\mathrm{T}}^2}=\frac{2{\mathrm{\pi m}}_1\mathrm{v}}{\mathrm{T}}$

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{{\mathrm{v}}^3\mathrm{T}}{2\mathrm{\pi G}}$

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{{(2.7\times {10}^5\mathrm{m})}^3\left(146880\mathrm{s}\right)}{2\left(3.142\right)(6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2)}$

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=6.9\times {10}^{30}\text{\hspace{0.17em}kg}$

But, the mass of the Sun is${\mathrm{M}}_{\mathrm{s}}=1.99\times {10}^{30}\text{\hspace{0.17em}kg}$

So,

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{6.9\times {10}^{30}}{1.99\times {10}^{30}}\text{kg}$

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=3.467{\mathrm{M}}_{\mathrm{s}}$

But,

${\mathrm{M}}_1=6{\mathrm{M}}_{\mathrm{s}}=6(1.99\times {10}^{30}\mathrm{kg})$

${\mathrm{M}}_1=11.94\times {10}^{30}\text{kg}$

So,

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{{\mathrm{m}}_2^3}{{(6{\mathrm{M}}_{\mathrm{s}}+{\mathrm{m}}_2)}^2}=3.467{\mathrm{M}}_{\mathrm{s}}$

${\mathrm{m}}_2^3-3.467{\mathrm{M}}_{\mathrm{s}}{(6{\mathrm{M}}_{\mathrm{s}}+{\mathrm{m}}_2)}^2=0$

After solving this equation,

${\mathrm{m}}_2=9.3{\mathrm{M}}_{\mathrm{s}}~9{\mathrm{M}}_{\mathrm{s}}$

Hence,the approximate mass of m₂ of the dark star is 9M_s.

Therefore, using the gravitational force of attraction between two objects and the centripetal force, the mass of one of the objects can be found.