2018

An unseen companion in Auriga

Round: Data analysis
Points: 66
Level: Silver
Taught in: Section D10

Adapted from IOAA 2018 (Beijing), data-analysis problem 2. We have restated it in our own words and kept the original data. The original was worth 75 points. We leave out one part, worth 9 points, which asked for a rough first estimate of the companion's mass from the three APOGEE measurements, and our part (g) asks directly for the range of the companion's mass, which the original read from a graph of the companion's mass against the star's. The solution was last checked on 20 October 2026.

The red giant 2M05215658+4359220 in Auriga was found by the APOGEE survey to change its radial velocity, which suggests an unseen companion. Its proper motion can be ignored. Three APOGEE spectra gave the following radial velocities.

MJD	Radial velocity (km/s)	Uncertainty (km/s)
56 204.9537	−37.417	0.011
56 229.9213	34.846	0.010
56 233.8732	42.567	0.010

(a) Using a straight line between measurements, estimate the largest acceleration of the star along the line of sight, in km/s per day. (6 points)

Follow-up spectra with the TRES spectrograph on the 1.5 m telescope of the Fred Lawrence Whipple Observatory gave these radial velocities.

MJD	Radial velocity (km/s)	Uncertainty (km/s)
58 006.9760	0	0.075
58 023.9823	−43.313	0.075
58 039.9004	−27.963	0.045
58 051.9851	10.928	0.118
58 070.9964	43.782	0.075
58 099.8073	−30.033	0.054
58 106.9178	−42.872	0.135
58 112.8188	−44.863	0.088
58 123.7971	−25.810	0.115
58 136.6004	15.691	0.146
58 143.7844	34.281	0.087

(b) Plot the radial velocity against time, draw a sine curve that fits the points, and estimate from it the orbital period $P$ and the $K$ . (14 points)

(c) For a circular orbit, find the smallest possible radius of the star's orbit, in solar radii and in astronomical units. (4 points)

(d) The mass function of a binary is $f = (M_{2} \sin i)^{3} / (M_{1} + M_{2})^{2}$ , where $M_{1}$ is the mass of the visible star, $M_{2}$ that of its companion and $i$ the inclination of the orbit. Express it in terms of quantities that can be observed, and find its value in solar masses. (7 points)

Studies of the spectra and a parallax from Gaia gave the visible star an effective temperature of $4890 \pm 130$ K, a surface gravity of $\log g = 2.2 \pm 0.1$ with $g$ in cm/s², a parallax of $0.272 \pm 0.049$ mas, a rotation speed of $v \sin i = 14.1 \pm 0.6$ km/s, and a bolometric flux of $(1.1 \pm 0.1) \times 10^{- 12}$ W/m². Its brightness varies with the orbital period, so we may take its rotation period to be equal to $P$ and its equator to lie in the plane of the orbit.

(e) Find the luminosity $L_{1}$ , radius $R_{1}$ and mass $M_{1}$ of the visible star, and $\sin i$ , each with its uncertainty. (16 points)

(f) Is the visible star a blue giant, a yellow main-sequence star, a red giant, a red main-sequence star or a white dwarf? (4 points)

(g) Using the mass function, find the range of $M_{2}$ allowed by the values of $M_{1}$ and $\sin i$ within their uncertainties. A white dwarf can have at most 1.4 solar masses and a neutron star about 2.5. What kind of object could the companion be? (15 points)

I solved it without the solution

Between the first two APOGEE measurements the radial velocity rises by $34.846 + 37.417 = 72.263$ km/s in 24.968 days, which is 2.89 km/s per day. Between the second and third it rises by 7.721 km/s in 3.952 days, only 1.95 km/s per day. The largest acceleration is therefore about 2.89 km/s per day. The curve can be steeper between the measurements, so the true largest acceleration may be larger.
Plotted against time, the TRES velocities trace a sine through a little more than one and a half cycles. The period is best found from two times at which the curve rises through zero, because the velocity changes fastest there. Joining neighbouring points by straight lines, it rises through zero at MJD 58 048.6 and MJD 58 131.8, which gives $P = 83.2$ days. The extreme measured velocities are 43.8 and −44.9 km/s, so $K$ is about 44.3 km/s, and the sine drawn through the points peaks a little beyond the measured extremes. A weighted least-squares fit of a sine, as in Figure D.10, gives $P = 83.14$ days and $K = 44.6$ km/s. The official solution accepted any period from 81 to 85 days and any semi-amplitude from 44 to 46 km/s. As a check, the largest acceleration of the sine is $2 π K / P = 3.37$ km/s per day, more than the 2.89 of part (a), as it should be.
The star's orbital speed is $K / \sin i$ , so its orbit has the radius $r = P K / (2 π \sin i)$ , which is smallest for $\sin i = 1$ . Then $r = 83.14 \times 86 400 \times 44.6 / (2 π) = 5.10 \times 10^{7}$ km, which is 73 solar radii or 0.34 au.
The star goes round the centre of mass at the distance $r_{1} = D M_{2} / (M_{1} + M_{2})$ , where $D$ is the distance between the two bodies, with the speed $v_{1} = 2 π r_{1} / P$ . Kepler's third law, which (S.19) gives for a circular orbit, gives $G (M_{1} + M_{2}) = 4 π^{2} D^{3} / P^{2}$ , so
$v_{1}^{3} = \frac{8 π^{3} r_{1}^{3}}{P^{3}} = \frac{2 π G}{P} \frac{M_{2}^{3}}{(M_{1} + M_{2})^{2}} .$
With $K = v_{1} \sin i$ , the mass function is $f = P K^{3} / (2 π G)$ , which is $1.52 \times 10^{30}$ kg, or 0.77 solar masses. Since $f$ is never more than $M_{2}$ , the companion has at least 0.77 solar masses.
The distance is $1 / 0.272$ mas, which is 3.68 kpc, uncertain by $0.049 / 0.272 = 0.18$ of itself, or 0.66 kpc. The luminosity is $L_{1} = 4 π d^{2} F = 465 L_{⊙}$ , taking $L_{⊙} = 3.828 \times 10^{26}$ W, with a fractional uncertainty of $\sqrt{(0.1 / 1.1)^{2} + (2 \times 0.18)^{2}} = 0.37$ , so $L_{1} = 465 \pm 173 L_{⊙}$ .
From the Stefan–Boltzmann law of (4.12), $L_{1} = 4 π R_{1}^{2} σ T^{4}$ , the radius is $R_{1} = 30.0 R_{⊙}$ , with a fractional uncertainty of $\sqrt{(2 \times 130 / 4890)^{2} + (0.37 / 2)^{2}} = 0.19$ , so $R_{1} = 30 \pm 6 R_{⊙}$ .
The equator turns at $2 π R_{1} / P$ , so $\sin i = P (v \sin i) / (2 π R_{1}) = 0.77$ , with a fractional uncertainty of $\sqrt{{0.19}^{2} + (0.6 / 14.1)^{2}} = 0.20$ , so $\sin i = 0.77 \pm 0.15$ .
The surface gravity is $g = 10^{2.2}$ cm/s², which is $1.58$ m/s², uncertain by $g \ln 10 \times 0.1 = 0.36$ m/s². The mass is $M_{1} = g R_{1}^{2} / G = 5.2 M_{⊙}$ , with a fractional uncertainty of $\sqrt{(0.36 / 1.58)^{2} + (2 \times 0.19)^{2}} = 0.45$ , so $M_{1} = 5.2 \pm 2.3 M_{⊙}$ .
A star at 4890 K with 465 times the Sun's luminosity, 30 times its radius and a surface gravity 170 times weaker than the Sun's is a red giant.
Solving $M_{2}^{3} \sin^{3} i = f (M_{1} + M_{2})^{2}$ for $M_{2}$ with $M_{1} = 5.2$ and $\sin i = 0.77$ gives $M_{2} = 5.9 M_{⊙}$ . The lightest companion comes with the lightest star and the largest $\sin i$ , $M_{1} = 2.87$ and $\sin i = 0.924$ , which give 3.3 solar masses, and the heaviest with $M_{1} = 7.56$ and $\sin i = 0.619$ , which give 10.0 solar masses.
Even the lightest of these is more than the 2.5 solar masses a neutron star can have. An ordinary star of three or more solar masses would be a hot main-sequence star, bright enough for its own lines to show in the spectrum, so the companion is most likely a black hole.

AnswerThe largest acceleration is about 2.89 km/s per day, the orbit has $P = 83.1$ days and $K = 44.6$ km/s, the star's orbit is at least 73 solar radii, or 0.34 au, in radius, and the mass function is 0.77 solar masses. The visible star is a red giant with $L_{1} = 465 \pm 173 L_{⊙}$ , $R_{1} = 30 \pm 6 R_{⊙}$ , $\sin i = 0.77 \pm 0.15$ and $M_{1} = 5.2 \pm 2.3 M_{⊙}$ . The companion has between about 3.3 and 10 solar masses, most likely 5.9, and is probably a black hole. The official solution gives $P = 83.26$ days, $K = 45.02$ km/s and a mass function of 0.79 solar masses from a fit that it does not show, and the same results for the star and a companion of about 6 solar masses.

This problem also appears among the olympiad problems of Section D10, Problems for Chapter D, which teaches the ideas it uses. Equation and figure numbers refer to the course.