IPhO 2007 · Theory problem 3

Binary stars

Download PDFIPhO 2007

Two stars rotating around their center of mass form a binary star system. Almost half of the stars in our galaxy are binary star systems. It is not easy to realize the binary nature of most of these star systems from Earth, since the distance between the two stars is much less than their distance from us and thus the stars cannot be resolved with telescopes. Therefore, we have to use either photometry or spectrometry to observe the variations in the intensity or the spectrum of a particular star to find out whether it is a binary system or not.

Photometry of Binary Stars

If we are exactly on the plane of motion of the two stars, then one star will occult (pass in front of) the other star at certain times and the intensity of the whole system will vary with time from our observation point. These binary systems are called ecliptic binaries.

1 Assume that two stars are moving on circular orbits around their common center of mass with a constant angular speed \(\omega\) and we are exactly on the plane of motion of the binary system. Also assume that the surface temperatures of the stars are \(T_1\) and \(T_2\) \((T_1>T_2)\), and the corresponding radii are \(R_1\) and \(R_2\) \((R_1>R_2)\), respectively. The total intensity of light, measured on Earth, is plotted in Figure 1 as a function of time. Careful measurements indicate that the intensities of the incident light from the stars corresponding to the minima are respectively 90 and 63 percent of the maximum intensity, \(I_0\), received from both stars \(\left(I_0=4.8\times10^{-9}\ \mathrm{W/m^2}\right)\). The vertical axis in Figure 1 shows the ratio \(I/I_0\) and the horizontal axis is marked in days.

Figure 1

Figure 1. The relative intensity received from the binary star system as a function of time. The vertical axis has been scaled by \(I_0=4.8\times10^{-9}\ \mathrm{W/m^2}\). Time is given in days.

1.1 — 0.8 points

Find the period of the orbital motion. Give your answer in seconds up to two significant digits.
What is the angular frequency of the system in rad/sec?

To a good approximation, the receiving radiation from a star is a uniform black body radiation from a flat disc with a radius equal to the radius of the star. Therefore, the power received from the star is proportional to \(AT^4\) where \(A\) is area of the disc and \(T\) is the surface temperature of the star.

1.2 — 1.6 points

Use the diagram in Figure 1 to find the ratios \(T_1/T_2\) and \(R_1/R_2\).

Spectrometry of Binary Systems

In this section, we are going to calculate the astronomical properties of a binary star by using experimental spectrometric data of the binary system.

Atoms absorb or emit radiation at their certain characteristic wavelengths. Consequently, the observed spectrum of a star contains absorption lines due to the atoms in the star’s atmosphere. Sodium has a characteristic yellow line spectrum (\(\mathrm{D}_1\) line) with a wavelength \(5895.9\ \text{Å}\) (\(10\ \text{Å}=1\ \mathrm{nm}\)). We examine the absorption spectrum of atomic Sodium at this wavelength for the binary system of the previous section. The spectrum of the light that we receive from the binary star is Doppler-shifted, because the stars are moving with respect to us. Each star has a different speed. Accordingly the absorption wavelength for each star will be shifted by a different amount. Highly accurate wavelength measurements are required to observe the Doppler shift since the speed of the stars is much less than the speed of light. The speed of the center of mass of the binary system we consider in this problem is much smaller than the orbital velocities of the stars. Hence all the Doppler shifts can be attributed to the orbital velocity of the stars. Table 1 shows the measured spectrum of the stars in the binary system we have observed.

Table 1: Absorption spectrum of the binary star system for the Sodium \(\mathrm{D}_1\) line

\(t/\mathrm{days}\) 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4
\(\lambda_1\) (Å) 5897.5 5897.7 5897.2 5896.2 5895.1 5894.3 5894.1 5894.6
\(\lambda_2\) (Å) 5893.1 5892.8 5893.7 5896.2 5897.3 5898.7 5899.0 5898.1
\(t/\mathrm{days}\) 2.7 3.0 3.3 3.6 3.9 4.2 4.5 4.8
\(\lambda_1\) (Å) 5895.6 5896.7 5897.3 5897.7 5897.2 5896.2 5895.0 5894.3
\(\lambda_2\) (Å) 5896.4 5894.5 5893.1 5892.8 5893.7 5896.2 5897.4 5898.7

(Note: There is no need to make a graph of the data in this table)

2 Using Table 1,

2.1 — 1.8 points

Let \(v_1\) and \(v_2\) be the orbital velocity of each star. Find \(v_1\) and \(v_2\).
The speed of light \(c=3.0\times10^8\ \mathrm{m/s}\). Ignore all relativistic effects.

2.2 — 0.7 points

Find the mass ratio of the stars \((m_1/m_2)\).

2.3 — 0.8 points

Let \(r_1\) and \(r_2\) be the distances of each star from their center of mass.
Find \(r_1\) and \(r_2\).

2.4 — 0.2 points

Let \(r\) be the distance between the stars. Find \(r\).

3 The gravitational force is the only force acting between the stars.

3.1 — 1.2 points

Find the mass of each star up to one significant digit.
The universal gravitational constant \(G=6.7\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\).

General Characteristics of Stars

4 Most of the stars generate energy through the same mechanism. Because of this, there is an empirical relation between their mass, \(M\), and their luminosity, \(L\), which is the total radiant power of the star. This relation could be written in the form \(L/L_{\mathrm{Sun}}=(M/M_{\mathrm{Sun}})^\alpha\).

Here, \(M_{\mathrm{Sun}}=2.0\times10^{30}\ \mathrm{kg}\) is the solar mass and, \(L_{\mathrm{Sun}}=3.9\times10^{26}\ \mathrm{W}\) is the solar luminosity. This relation is shown in a log-log diagram in Figure 2.

Figure 2

Figure 2. The luminosity of a star versus its mass varies as a power law. The diagram is log-log. The star-symbol represents Sun with a mass of \(2.0\times10^{30}\ \mathrm{kg}\) and luminosity of \(3.9\times10^{26}\ \mathrm{W}\).

4.1 — 0.6 points

Find \(\alpha\) up to one significant digit.

4.2 — 0.6 points

Let \(L_1\) and \(L_2\) be the luminosity of the stars in the binary system studied in the previous sections. Find \(L_1\) and \(L_2\).

4.3 — 0.9 points

What is the distance, \(d\), of the star system from us in light years?
To find the distance you can use the diagram of Figure 1. One light year is the distance light travels in one year.

4.4 — 0.4 points

What is the maximum angular distance, \(\theta\), between the stars from our observation point?

4.5 — 0.4 points

What is the smallest aperture size for an optical telescope, \(D\), that can resolve these two stars?

Official English text of the problem.

Answers

Later results use unrounded intermediate values.

  • 1.1: \(P=3.0\ \mathrm{days}\simeq2.6\times10^5\ \mathrm{s}\); \(\omega\simeq2.4\times10^{-5}\ \mathrm{rad/s}\).
  • 1.2: \(T_1/T_2\simeq1.39\); \(R_1/R_2\simeq1.56\).
  • 2.1: \(v_1\simeq9.2\times10^4\ \mathrm{m/s}\); \(v_2\simeq1.6\times10^5\ \mathrm{m/s}\).
  • 2.2: \(m_1/m_2\simeq1.7\).
  • 2.3: \(r_1\simeq3.8\times10^9\ \mathrm{m}\); \(r_2\simeq6.5\times10^9\ \mathrm{m}\).
  • 2.4: \(r\simeq1.0\times10^{10}\ \mathrm{m}\).
  • 3.1: To one significant digit, \(m_1\simeq6\times10^{30}\ \mathrm{kg}\), \(m_2\simeq4\times10^{30}\ \mathrm{kg}\).
  • 4.1: \(\alpha\simeq4\).
  • 4.2: \(L_1\simeq3.2\times10^{28}\ \mathrm{W}\); \(L_2\simeq3.7\times10^{27}\ \mathrm{W}\).
  • 4.3: \(d\simeq82\ \text{light years}\simeq7.7\times10^{17}\ \mathrm{m}\).
  • 4.4: \(\theta_{\max}\simeq1.3\times10^{-8}\ \mathrm{rad}\), or \(2.7\times10^{-3}\) arcseconds.
  • 4.5: \(D_{\min}\simeq54\ \mathrm{m}\), using the Rayleigh criterion at the given sodium wavelength, \(589.59\ \mathrm{nm}\).

Solution

1.1

The orbital period is the interval between successive minima of the same depth. Figure 1 gives approximately \(3.0\) days. Adjacent minima of different depths correspond to the two different conjunctions and are separated by half a period.

Thus \[ P=3.0\times86400\ \mathrm{s} =259200\ \mathrm{s} \simeq\boxed{2.6\times10^5\ \mathrm{s}}. \]

The angular frequency is \[ \omega=\frac{2\pi}{P} =2.424\ldots\times10^{-5}\ \mathrm{rad/s} \simeq\boxed{2.4\times10^{-5}\ \mathrm{rad/s}}. \]

1.2

Let the unobscured contributions to the intensity at Earth be \[ I_1=C R_1^2T_1^4,\qquad I_2=C R_2^2T_2^4, \] where \(C\) is common to both stars because they are effectively at the same distance. Outside eclipse, \[ I_0=I_1+I_2. \]

The edge-on geometry produces central eclipses. Since \(R_1>R_2\), the larger star can completely hide the smaller star. During this eclipse the lost intensity is \(I_2\).

At the other eclipse, the smaller star blocks an equal-sized patch of the hotter star. The lost intensity is then \[ B=C R_2^2T_1^4. \] Because \(T_1>T_2\), \(B>I_2\). Therefore, the shallower minimum is the eclipse in which star 2 is hidden.

The shallow minimum gives \[ I_1=0.90I_0,\qquad I_2=0.10I_0. \] The deep minimum gives \[ B=(1-0.63)I_0=0.37I_0. \]

Dividing \(B\) by \(I_2\), \[ \left(\frac{T_1}{T_2}\right)^4 =\frac{C R_2^2T_1^4}{C R_2^2T_2^4} =\frac{0.37}{0.10}=3.7, \] so \[ \boxed{\frac{T_1}{T_2}=3.7^{1/4}\simeq1.39}. \]

Similarly, \[ \left(\frac{R_1}{R_2}\right)^2 =\frac{I_1}{B} =\frac{0.90}{0.37}, \] hence \[ \boxed{\frac{R_1}{R_2} =\sqrt{\frac{0.90}{0.37}}\simeq1.56}. \]

2.1

For nonrelativistic motion, the Doppler shift is \[ \frac{\lambda-\lambda_0}{\lambda_0}=\frac{u}{c}, \] where \(u\) is the line-of-sight velocity, positive for recession, and \[ \lambda_0=5895.9\ \text{Å}. \]

In an edge-on circular orbit, the line-of-sight velocity varies sinusoidally between \(-v_i\) and \(+v_i\). Thus the wavelength amplitude determines the orbital speed: \[ v_i=\frac{c}{\lambda_0} \frac{\lambda_{i,\max}-\lambda_{i,\min}}{2}. \]

Using the tabulated extrema, \[ \Delta\lambda_1 =\frac{5897.7-5894.1}{2}\ \text{Å} =1.8\ \text{Å}, \] \[ \Delta\lambda_2 =\frac{5899.0-5892.8}{2}\ \text{Å} =3.1\ \text{Å}. \]

Therefore, \[ v_1=3.0\times10^8\frac{1.8}{5895.9} =9.159\ldots\times10^4\ \mathrm{m/s}, \] \[ v_2=3.0\times10^8\frac{3.1}{5895.9} =1.577\ldots\times10^5\ \mathrm{m/s}. \]

Thus \[ \boxed{v_1\simeq9.2\times10^4\ \mathrm{m/s},\qquad v_2\simeq1.6\times10^5\ \mathrm{m/s}}. \]

2.2

In the center-of-mass frame, the stars have equal and opposite momenta: \[ m_1v_1=m_2v_2. \] Consequently, \[ \boxed{\frac{m_1}{m_2} =\frac{v_2}{v_1} =\frac{3.1}{1.8} =1.722\ldots\simeq1.7}. \]

2.3

Both stars have the same angular frequency. For uniform circular motion, \[ v_i=\omega r_i, \qquad r_i=\frac{v_iP}{2\pi}. \]

Using the unrounded values from the preceding calculations, \[ r_1=3.778\ldots\times10^9\ \mathrm{m}, \qquad r_2=6.507\ldots\times10^9\ \mathrm{m}. \]

Therefore, \[ \boxed{r_1\simeq3.8\times10^9\ \mathrm{m},\qquad r_2\simeq6.5\times10^9\ \mathrm{m}}. \]

2.4

The center of mass lies between the stars, so their separation is \[ r=r_1+r_2=1.02854\ldots\times10^{10}\ \mathrm{m}. \] Thus, to two significant figures, \[ \boxed{r\simeq1.0\times10^{10}\ \mathrm{m}}. \]

3.1

The gravitational force supplies the centripetal force for each star: \[ \frac{Gm_1m_2}{r^2} =m_1\omega^2r_1 =m_2\omega^2r_2. \]

It follows that \[ m_1=\frac{\omega^2r_2r^2}{G}, \qquad m_2=\frac{\omega^2r_1r^2}{G}. \]

Substitution, retaining intermediate digits, gives \[ m_1=6.037\ldots\times10^{30}\ \mathrm{kg}, \qquad m_2=3.5056\ldots\times10^{30}\ \mathrm{kg}. \]

The requested answers to one significant digit are therefore \[ \boxed{m_1\simeq6\times10^{30}\ \mathrm{kg},\qquad m_2\simeq4\times10^{30}\ \mathrm{kg}}. \]

The unrounded calculated masses will be retained below, since raising prematurely rounded masses to a power would introduce substantial numerical error.

4.1

Taking logarithms of the empirical relation gives \[ \log\!\left(\frac{L}{L_{\mathrm{Sun}}}\right) =\alpha\log\!\left(\frac{M}{M_{\mathrm{Sun}}}\right). \] Thus \(\alpha\) is the slope of the trend in the log-log diagram.

The trend passes approximately through \[ \left(\frac{M}{M_{\mathrm{Sun}}}, \frac{L}{L_{\mathrm{Sun}}}\right) =(1,1) \quad\text{and}\quad (10,10^4). \] Hence \[ \alpha\simeq\frac{\log_{10}(10^4)-\log_{10}(1)} {\log_{10}(10)-\log_{10}(1)} =\boxed{4}. \]

4.2

With \(\alpha=4\), \[ L_i=L_{\mathrm{Sun}} \left(\frac{m_i}{M_{\mathrm{Sun}}}\right)^4. \]

Using the unrounded dynamical masses, \[ L_1 =3.9\times10^{26} \left(\frac{6.037\ldots\times10^{30}} {2.0\times10^{30}}\right)^4 =3.238\ldots\times10^{28}\ \mathrm{W}, \] \[ L_2 =3.9\times10^{26} \left(\frac{3.5056\ldots\times10^{30}} {2.0\times10^{30}}\right)^4 =3.681\ldots\times10^{27}\ \mathrm{W}. \]

Therefore, \[ \boxed{L_1\simeq3.2\times10^{28}\ \mathrm{W},\qquad L_2\simeq3.7\times10^{27}\ \mathrm{W}}. \]

As a consistency check, these give \(L_1/L_2\simeq8.8\), close to the photometric ratio \(I_1/I_2=9\), as expected for the approximate empirical law and measured data.

4.3

For isotropic radiation, luminosity \(L\) produces intensity \(L/(4\pi d^2)\) at distance \(d\). Outside eclipse, both stars contribute their full flux: \[ I_0=\frac{L_1+L_2}{4\pi d^2}. \]

Thus \[ d=\sqrt{\frac{L_1+L_2}{4\pi I_0}} =\sqrt{\frac{3.6065\ldots\times10^{28}} {4\pi(4.8\times10^{-9})}} =7.7325\ldots\times10^{17}\ \mathrm{m}. \]

Taking one year as approximately \(365\) days, \[ 1\ \text{light year} =c(365\times86400\ \mathrm{s}) \simeq9.46\times10^{15}\ \mathrm{m}. \] Therefore, \[ \boxed{d\simeq82\ \text{light years} \simeq7.7\times10^{17}\ \mathrm{m}}. \]

4.4

The projected separation is largest when the line joining the stars is perpendicular to the line of sight. Its value is then \(r\).

Since \(r\ll d\), the small-angle approximation gives \[ \theta_{\max}\simeq\frac{r}{d} =\frac{1.02854\ldots\times10^{10}} {7.7325\ldots\times10^{17}} =1.330\ldots\times10^{-8}\ \mathrm{rad}. \]

Hence \[ \boxed{\theta_{\max}\simeq1.3\times10^{-8}\ \mathrm{rad}} \] or approximately \[ \boxed{\theta_{\max}\simeq2.7\times10^{-3}\ \text{arcseconds}}. \]

4.5

For an ideal diffraction-limited telescope with a circular aperture, the Rayleigh criterion gives the minimum resolvable angular separation: \[ \theta_{\mathrm{R}}=1.22\frac{\lambda}{D}. \]

The smallest aperture capable of resolving the stars at their maximum separation therefore satisfies \[ D_{\min}=\frac{1.22\lambda}{\theta_{\max}}. \]

Using the sodium wavelength supplied in the problem, \[ \lambda=5895.9\ \text{Å} =5.8959\times10^{-7}\ \mathrm{m}, \] we obtain \[ D_{\min} =\frac{1.22(5.8959\times10^{-7})} {1.330\ldots\times10^{-8}}\ \mathrm{m} \simeq\boxed{54\ \mathrm{m}}. \]

More generally, the wavelength dependence is \[ \boxed{D_{\min}(\lambda)\simeq 54\ \mathrm{m}\left(\frac{\lambda}{589.59\ \mathrm{nm}}\right)}. \]

The solution was checked against the official solution.