Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M_{\odot }$ and 0.62 $M_{\odot }$, repectively.

Answer to Problem 42Q

Solution:

$\text{1}{\text{.65×10}}^{\text{6}}\text{km}$

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M_{\odot }$ and 0.62 $M_{\odot }$.

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m_1+m_2)p^2=a^3$

Here, $m_1$ is the mass of the first star, $m_2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $\left(M_{\odot }\right)$ and the period is in years, then the semi-major axis is in astronomical unit $\left(\text{au}\right)$.

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$\begin{array}{c}8\text{hours}=\frac{1}{3}\text{days}\\ =\frac{1}{3\times 365}\text{years}\\ \text{=}\frac{1}{1095}\text{years}\end{array}$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m_1+m_2)p^2=a^3$

Substitute $0.99M_{\odot }$ for $m_1$, $0.62M_{\odot }$ for $m_2$, and $\frac{\text{1}}{1095}\text{years}$ for p and calculate $a$.

$\begin{array}{c}{a}^3=\left(0.99M_{\odot }+0.62M_{\odot }\right){\left(\frac{1}{1095}\right)}^2\\ =1.34\times {10}^{-6}{\left(\text{au}\right)}^3\\ a=1.10\times {10}^{-2}\text{au}\end{array}$

Convert $a$ from au to km using the following conversion,

$1\text{au}=1.496\times {10}^8\text{km}$

So, the distance between the two binary stars is,

$\begin{array}{c}a=\left(1.10\times {10}^{-2}\right)\left(1.496\times {10}^8\right)\\ =1.65\times {10}^6\text{km}\end{array}$

Conclusion:

Hence, the value of average separation between the stars is $1.65\times {10}^6\text{km}$.

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $\text{1}\text{.14}{R}_{\odot }$ and $\text{0}\text{.83}{R}_{\odot }$ and the result obtained in part (a).

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $\text{1}\text{.14}{R}_{\odot }$ and $\text{0}\text{.83}{R}_{\odot }$, respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘$R_t$’.

$\begin{array}{c}{R}_t=1.14R_{\odot }+0.83R_{\odot }\\ =1.97R_{\odot }\end{array}$

Convert unit of $R_t$ to km.

$\begin{array}{c}{R}_t=\left(1.97\right)\left(6.96\times {10}^5\right)\\ =1.37\times {10}^6\text{km}\end{array}$

Now, compare the value of $R_t$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.