Chapter 7, Problem 7.6P

(a)

To determine

The ratio of stellar masses.

Answer to Problem 7.6P

The ratio of stellar masses is $4.15$.

Explanation of Solution

Write the expression for ratio of masses of two stars in term of radial velocity.

    $\frac{m_A}{m_B}=\frac{v_B}{v_A}$        (1)

Here, $v_A$ and $v_B$ are the radial velocity of star A and star B respectively.

Conclusion:

Substitute $5.4\text{km}/\text{s}$ for $v_A$ and $22.4\text{km}/\text{s}$ for $v_B$ in the Equation (1).

    $\begin{array}{c}\frac{m_A}{m_B}=\frac{22.4\text{km}/\text{s}}{5.4\text{km}/\text{s}}\\ =4.15\end{array}$

Thus, the ratio of stellar masses is $4.15$.

(b)

To determine

The sum of the masses.

Answer to Problem 7.6P

The sum of two masses is $101.29\times {10}^{22}\text{kg}$.

Explanation of Solution

Write the expression for sum of two masses.

    $m_A+m_B=\frac{P}{2\pi G}\frac{{\left(v_A+v_B\right)}^3}{{\sin }^{3}i}$        (2)

Here, $P$ is the period, $v_A$ and $v_B$ is the radial velocity, $m_A$ and $m_B$ are the masses of two stars, $i$ is the angle of inclination and $G$ is the universal gravitational constant.

Conclusion:

Substitute $6.31\text{yr}$ for $P$, $\text{6}\text{.67}\times {\text{10}}^{-11}{\text{m}}^{\text{3}}{\text{s}}^{-2}{\text{kg}}^{-1}$ for $G$, $5.4\text{km}/\text{s}$ for $v_A$, $22.4\text{km}/\text{s}$ for $v_B$ and $90°$ for $i$ in equation (2).

    $\begin{array}{c}{m}_A+m_B=\frac{6.31\text{yr}}{2\pi \left(\text{6}\text{.67}\times {\text{10}}^{-11}{\text{m}}^{\text{3}}{\text{s}}^{-2}{\text{kg}}^{-1}\right)}\frac{{\left(5.4\text{km}/\text{s}+22.4\text{km}/\text{s}\right)}^3}{{\sin }^{3}90°}\\ =\frac{\left(6.31\text{yr}\times \frac{3.14\times {10}^{17}\text{s}}{1\text{yr}}\right)}{2\pi \left(\text{6}\text{.67}\times {\text{10}}^{-11}{\text{m}}^{\text{3}}{\text{s}}^{-2}{\text{kg}}^{-1}\right)}\frac{{\left(27.8\text{km}/\text{s}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^3}{{\sin }^{3}90°}\\ =101.29\times {10}^{22}\text{kg}\end{array}$

Thus, the sum of two masses is $101.29\times {10}^{22}\text{kg}$.

(c)

To determine

The individual masses of two stars.

Answer to Problem 7.6P

The mass of star A is $81.62\times {10}^{22}\text{kg}$ and the mass of star B is $19.66\times {10}^{22}\text{kg}$.

Explanation of Solution

Write the expression for sum of two masses.

    $m_A+m_B=101.29\times {10}^{22}\text{kg}$        (3)

Here, $m_A$ and $m_B$ are the masses of two stars.

Write the expression of mass of star A.

    $m_A=4.15m_B$        (4)

Conclusion:

Substitute $4.15m_B$ for $m_A$, in equation (3).

    $\begin{array}{c}\left(4.15m_B\right)+m_B=101.29\times {10}^{22}\text{kg}\\ {\text{m}}_B=\frac{101.29\times {10}^{22}\text{kg}}{5.15}\\ =19.66\times {10}^{22}\text{kg}\end{array}$

Substitute $19.66\times {10}^{22}\text{kg}$ for $m_B$ in Equation (4).

    $\begin{array}{c}{m}_A=4.15\left(19.66\times {10}^{22}\text{kg}\right)\\ =81.62\times {10}^{22}\text{kg}\end{array}$

Thus, the mass of star A is $81.62\times {10}^{22}\text{kg}$ and the mass of star B is $19.66\times {10}^{22}\text{kg}$.

(d)

To determine

The individual radii.

Answer to Problem 7.6P

The radius of star A is $1.5\times {10}^9\text{m}/\text{s}$ and the radius of star B is $0.7\times {10}^9\text{m}/\text{s}$.

Explanation of Solution

Write the expression for smallest star

    $r_B=\frac{\left(v_A+v_B\right)}{2}\left(t_B-t_A\right)$        (5)

Here, $r_B$ is the radius of star B, $v_A$ and $v_B$ is the radial velocity and $t_B-t_A$ is the time period between first contact and minimum light.

Write the expression for radius of largest star

    $r_A=r_B+\frac{\left(v_A+v_B\right)}{2}\left(t_C-t_B\right)$        (6)

Here, $r_A$ is the radius of star A, $v_A$ and $v_B$ is the radial velocity and $t_C-t_B$ is the time period between $t_C$ and $t_B$.

Conclusion:

Substitute $0.58\text{d}$ for $t_B-t_A$, $5.4\text{km}/\text{s}$ for $v_A$, $22.4\text{km}/\text{s}$ for $v_B$ in equation (5).

    $\begin{array}{c}{r}_B=\frac{\left(5.4\text{km}/\text{s}+22.4\text{km}/\text{s}\right)}{2}\left(0.58\text{d}\right)\\ =\frac{\left(27.8\text{km}/\text{s}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}{2}\left(0.58\text{d}\times \frac{86400\text{s}}{1\text{d}}\right)\\ =0.7\times {10}^9\text{m}/\text{s}\end{array}$

Substitute $0.64\text{d}$ for $t_C-t_B$, $5.4\text{km}/\text{s}$ for $v_A$, $22.4\text{km}/\text{s}$ for $v_B$ and $0.7\times {10}^9\text{m}/\text{s}$ for $r_B$ in equation (6).

    $\begin{array}{c}{r}_A=0.7\times {10}^9\text{m}/\text{s}+\frac{\left(5.4\text{km}/\text{s}+22.4\text{km}/\text{s}\right)}{2}\left(0.64\text{d}\right)\\ =0.7\times {10}^9\text{m}/\text{s}+\frac{\left(27.8\text{km}/\text{s}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}{2}\left(0.64\text{d}\times \frac{86400\text{s}}{1\text{d}}\right)\\ =1.5\times {10}^9\text{m}/\text{s}\end{array}$

Thus, the radius of star A is $1.5\times {10}^9\text{m}/\text{s}$ and the radius of star B is $0.7\times {10}^9\text{m}/\text{s}$.

(e)

To determine

The ratio of the effective temperatures of the two stars.

Answer to Problem 7.6P

The ratio of the effective temperatures of the two stars is $2.2$.

Explanation of Solution

Write the expression for ratio of brightness of $B_P$ and $B_{\text{O}}$.

    $\frac{B_P}{B_{\text{O}}}={100}^{\left(m_{bol,0}-m_{biol,\text{P}}/5\right)}$        (7)

Here, $B_P$ and $B_{\text{O}}$ is the brightness of primary maximum and minimum and $m_{bol,O}$ and $m_{bol,P}$ is the apparent bolometric magnitude of primary maximum and minimum.

Write the expression for ratio of brightness of $B_S$ and $B_{\text{O}}$.

    $\frac{B_S}{B_{\text{O}}}={100}^{\left(m_{bol,0}-m_{biol,S}/5\right)}$        (8)

Here, $B_S$ and $B_{\text{O}}$ is the brightness of secondary maximum and minimum and $m_{bol,O}$ and $m_{bol,S}$ is the apparent bolometric magnitude of secondary maximum and minimum.

Write the expression of effective temperature of two stars.

    $\frac{T_B}{T_A}={\left[\frac{1-\frac{B_P}{B_{\text{O}}}}{1-\frac{B_S}{B_{\text{O}}}}\right]}^{1/4}$        (9)

Conclusion:

Substitute $5.40$ for $m_{bol,O}$ and $9.20$ for $m_{bol,P}$ in equation (7).

    $\begin{array}{c}\frac{B_P}{B_{\text{O}}}={100}^{\left(5.40-9.20/5\right)}\\ =0.03\end{array}$

Substitute $5.40$ for $m_{bol,O}$ and $5.44$ for $m_{bol,S}$ in equation (8).

    $\begin{array}{c}\frac{B_P}{B_{\text{O}}}={100}^{\left(5.40-5.44/5\right)}\\ =0.96\end{array}$

Substitute $0.03$ for $\frac{B_P}{B_{\text{O}}}$ and $0.96$ for $\frac{B_S}{B_{\text{O}}}$ in Equation (9).

    $\begin{array}{c}\frac{T_B}{T_A}={\left[\frac{1-0.03}{1-0.96}\right]}^{1/4}\\ ={\left[\frac{0.97}{0.04}\right]}^{1/4}\\ =2.2\end{array}$

Thus, the ratio of the effective temperatures of the two stars is $2.2$.