Chapter 9, Problem 18P

To determine

The circumferences of the two orbits, the separation between the stars, the total mass of the system and the individual mass of each star.

Answer to Problem 18P

The circumferences of the two orbits are $2.84\text{AU}$ and $\text{1}\text{.42AU}$. The separation between the stars is $\text{0}\text{.678 AU}$. The total mass of the system is $\text{40}\text{.6 solar mass}$. The individual masses of each star are  $13.5\text{ solar mass}$ and $\text{27}\text{.1solar mass}$.

Explanation of Solution

Write the equation for the circumference of first orbit.

$C_1=v_{1}T$        (I)

Here, $C_1$ is the circumference of first star. $v_1$ is the speed of first star and $T$ is the orbital period.

Write the equation for the circumference of second orbit.

$C_2=v_{2}T$        (II)

Here, $C_2$ is the circumference of second star, $v_2$ is the speed of second star and $T$ is the orbital period.

Write the expression for the separation between two stars.

$r=\frac{v_1+v_2}{\omega }$        (III)

Here, $r$ is the separation between two stars and $\omega$ is the angular speed.

Write the expression for the angular speed.

$\omega =\frac{2\pi }{T}$        (IV)

Rewrite the expression for the separation between the stars.

$r=\frac{v_1+v_2}{\left(2\pi /T\right)}$        (V)

Write the expression for the distance of first star by using equation (IV).

$\begin{array}{c}{r}_1=\frac{v_1}{\omega }\\ =\frac{v_1}{\left(2\pi /T\right)}\end{array}$        (VI)

Here, $r_1$ is the distance of first star.

Write the expression for the distance of second star by using equation (IV).

$\begin{array}{c}{r}_2=\frac{v_2}{\omega }\\ =\frac{v_2}{\left(2\pi /T\right)}\end{array}$        (VII)

Here, $r_2$ is the distance of second star.

Write the equation for the total mass of a visual binary system by using (V).

$M=\frac{r^3}{T^2}$        (VIII)

Here, $M$ is the total mass.

Write the expression for the total mass of the system.

$M=M_A+M_B$        (IX)

Here, $M_A$ is the mass of first star and $M_B$ is the second star.

Write the expression for the center of mass by using equation (VI) and (VII).

$\begin{array}{c}{x}_{cm}=\frac{M_A{r}_1-M_B{r}_2}{M_A+M_B}\\ =\frac{M_A\left(\frac{v_1}{\left(2\pi /T\right)}\right)-M_B\left(\frac{v_2}{\left(2\pi /T\right)}\right)}{M_A+M_B}\end{array}$        (X)

Here, $x_{cm}$ is the center of mass.

Conclusion:

Substitute, $154\text{km/s}$ for $v_1$, $32\text{days}$ for $T$ in equation (I) to find $C_1$.

$\begin{array}{c}{C}_1=\left[\left(154\text{km/s}\right)\left(\frac{1\times {10}^3\text{m}}{1\text{km}}\right)\right]\left[\left(32\text{days}\right)\left(\frac{24\text{hrs}}{1\text{day}}\right)\left(\frac{3600\text{s}}{1\text{hr}}\right)\right]\\ =\left(4.26\times {10}^{11}\text{m}\right)\left(\frac{6.68459\times {10}^{-12}\text{AU}}{1\text{m}}\right)\\ \approx 2.84\text{AU}\end{array}$

Substitute, $77\text{km/s}$ for $v_1$, $32\text{days}$ for $T$ in equation (II) to find $C_2$.

$\begin{array}{c}{C}_2=\left[\left(77\text{km/s}\right)\left(\frac{1\times {10}^3\text{m}}{1\text{km}}\right)\right]\left[\left(32\text{days}\right)\left(\frac{24\text{hrs}}{1\text{day}}\right)\left(\frac{3600\text{s}}{1\text{hr}}\right)\right]\\ =\left(2.13\times {10}^{11}\text{m}\right)\left(\frac{6.68459\times {10}^{-12}\text{AU}}{1\text{m}}\right)\\ =1.42\text{AU}\end{array}$

Substitute, $154\text{km/s}$ for $v_1$, $77\text{km/s}$ for $v_2$, $32\text{days}$ for $T$ in equation (V) to find $r$.

$\begin{array}{c}r=\frac{\left[\left(154\text{km/s}\right)\left(\frac{1\times {10}^3\text{m}}{1\text{km}}\right)\right]+\left[\left(77\text{km/s}\right)\left(\frac{1\times {10}^3\text{m}}{1\text{km}}\right)\right]}{\left(2\pi /\left[\left(32\text{days}\right)\left(\frac{24\text{hrs}}{1\text{day}}\right)\left(\frac{3600\text{s}}{1\text{hr}}\right)\right]\right)}\\ =\left(1.01\times {10}^{11}\text{m}\right)\left(\frac{6.68459\times {10}^{-12}\text{AU}}{1\text{m}}\right)\\ =0.678\text{AU}\end{array}$

Substitute, $0.678\text{AU}$ for $r$, $32\text{days}$ for $T$ in equation (VIII) to find $M$.

$\begin{array}{c}M=\frac{{\left(0.678\text{AU}\right)}^3}{{\left[\left(32\text{days}\right)\left(\frac{0.00273973\text{year}}{1\text{days}}\right)\right]}^2}\\ =40.6\text{ solar mass}\end{array}$

Substitute, $154\text{km/s}$ for $v_1$, $32\text{days}$ for $T$ in equation (IX) to find $M_A$.

$\begin{array}{c}{M}_A=\frac{{\left[\left(154\text{km/s}\right)\left(\frac{0.210805\text{AU/yr}}{1\text{km/s}}\right)\right]}^3\left[\left(32\text{days}\right)\left(\frac{0.00273973\text{year}}{1\text{day}}\right)\right]}{{\left(2\pi \right)}^3}\\ =12.1\end{array}$

Substitute, $0$ for $x_{cm}$, $154\text{km/s}$ for $v_1$, $77\text{km/s}$ for $v_2$ in equation (X) to find relation between $M_A,M_B$.

$\begin{array}{l}0=\frac{M_A\left(\frac{v_1}{\left(2\pi /T\right)}\right)-M_B\left(\frac{v_2}{\left(2\pi /T\right)}\right)}{M_A+M_B}\\ M_A{v}_1-M_B{v}_2=0\\ M_A\left[\left(154\text{km/s}\right)\left(\frac{0.210805\text{AU/yr}}{1\text{km/s}}\right)\right]-M_B\left[\left(77\text{km/s}\right)\left(\frac{0.210805\text{AU/yr}}{1\text{km/s}}\right)\right]=0\\ 32.46M_A-16.23M_B=0\end{array}$        (XI)

Substitute, $40.6\text{ solar mass}$ for $M$ in equation (IX) to find relation between $M_A,M_B$.

  $M_A+M_B=40.6\text{ solar mass}$        (XII)

Solving equation (XI) and (XII), value of the individual masses of star.

$M_A=13.5\text{ solar mass , }{M}_B=27.1\text{ solar mass}$

Thus, the circumferences of the two orbits are $2.84\text{AU}$ and $\text{1}\text{.42AU}$. The separation between the stars is $\text{0}\text{.678 AU}$. The total mass of the system is $\text{40}\text{.6 solar mass}$. The individual masses of each star are  $13.5\text{ solar mass}$ and $\text{27}\text{.1solar mass}$.