Question

In a binary star system, the average separation between the stars is 5.0 AU, and their orbital period is 5.0 years. What is the sum of their two masses?The average distance of Star A from the center of mass is four times that of Star B. What are their individual masses?

Answer 1

Answer:

A. 5.03 solar mass.

B. the masses of Star A and B are 1.01 solar mass and 4.02 solar mass respectively.

Explanation:

A. The sum of their two masses can be found using Kepler's third law:

$\frac{P^{2}}{a^{3}} = \frac{4\pi^{2}}{G(m_{A} + m_{B})}$

Where:

P: is the period = 5 y = 1.58x10⁸ s

a: is the separation between the stars = 5 AU = 7.5x10¹¹ m

G: is the gravitational constant = 6.67x10⁻¹¹ m³kg⁻¹s⁻²

$m_{A}$ and $m_{B}$ are the masses of Star A and Star B respectively.    

$m_{A} + m_{B} = \frac{4\pi^{2}a^{3}}{P^{2}G} = \frac{4\pi^{2}(5 AU*\frac{1.5\cdot 10^{11} m}{1 AU})^{3}}{(1.58 \cdot 10^{8} s)^{2}6.67 \cdot 10^{-11} m^{3}*kg^{-1}*s^{-2}} = 1.00 \cdot 10^{31} kg*\frac{1 M_{\bigodot}}{1.989\cdot 10^{30} kg} = 5.03 M_{\bigodot}$    

Hence, the sum of their two masses is 5.03 solar mass.  

B. Their individual masses can be found using the center of the mass equation:

$a_{B} = (\frac{m_{A}}{m_{A} + m_{B}})a$

Where:

$a_{B}$ is the distance of Star B from the center of the mass

Since, $a_{A}$ is four times $a_{B}$ and a = 5 AU we have:

$a_{A} = 4a_{B}$

$a = a_{A} + a_{B}$

$a_{B} = a - a_{A} = a - 4a_{B} \rightarrow a_{B} = \frac{a}{5}$      

Then, their individual masses are:

$\frac{a}{5} = (\frac{m_{A}}{5.03 M_{\bigodot}})a$  

$m_{A} = \frac{5.03 M_{\bigodot}}{5} = 1.01 M_{\bigodot}$

Now, the mass of Star B is:

$m_{B} = m_{T} - m_{A} = 5.03 M_{\bigodot} - 1.01 M_{\bigodot} = 4.02 M_{\bigodot}$

Therefore, the masses of Star A and B are $1.01 M_{\bigodot}$ and $4.02 M_{\bigodot}$ respectively.

I hope it helps you!