91Ó°ÊÓ

The velocity of the visible star is $\text{v}=270\text{ k³¾}/\text{s}=2.7×{10}^5\text{″¾}/\text{s}$

The orbital period of the visible star is,

$\mathrm{T}=1.70\text{ d²¹²â²õ}=\left(1.70\text{ }\mathrm{days}\right)\left(\frac{86400\text{ }\mathrm{s}}{1\text{ }\mathrm{day}}\right)=146880\text{s}$

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

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

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,

$\mathrm{v}=\frac{2{\mathrm{Ï€°ù}}_1}{\mathrm{T}}$

So,

${\mathrm{r}}_1=\frac{\mathrm{vT}}{2\mathrm{Ï€}}$

Substituting r₁in r,

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

Putting r and r₁ in F₁,

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

$\mathrm{F}=\frac{4{\mathrm{Ï€}}^2{\mathrm{Gm}}_1{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2{\mathrm{v}}^2{\mathrm{T}}^2}=\frac{2{\mathrm{Ï€³¾}}_1\mathrm{v}}{\mathrm{T}}$

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{{\mathrm{v}}^3\mathrm{T}}{2\mathrm{Ï€³Ò}}$

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{{(2.7×{10}^5\text{ }\mathrm{m})}^3\left(146880\text{ }\mathrm{s}\right)}{2\left(3.142\right)(6.67×{10}^{−11}\text{ }\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2)}$

$\frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=6.9×{10}^{30}\text{ k²µ}$

But, the mass of the Sun is${\mathrm{M}}_{\mathrm{s}}=1.99×{10}^{30}\text{ k²µ}$

So,

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

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

But,

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

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

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.