D10

Problems for Chapter D

Builds on: Sections D1 to D9
Reading time: 1 h
Exercises: 0
Olympiad problems: 3

In the data-analysis round of IOAA 2024 in Rio de Janeiro, every student was handed two tables of faint stars from the same strip of sky along the celestial equator, measured by two different surveys whose labels had been lost. Telling the tables apart took one look at their uncertainties, and part (b) turned those uncertainties into a straight line on semi-log paper. The three problems in this section come from past data-analysis rounds and use the whole of the chapter. As in Section 1.7 and Section 2.8, they are not labelled with the sections they need, because recognising which method a problem calls for is part of the skill. Each one is as long as a problem of the real round, and we suggest doing it with graph paper, a ruler and a calculator in about an hour and a half, as in the exam.

D10.1From textbook to olympiad

The first problem below, from IOAA 2024, gives two tables of positions and magnitudes with their uncertainties and asks which survey each came from, how the uncertainty grows with magnitude, which stars appear in both tables, and which of those could tie the two surveys together. The method of Section S2 gets us from the chapter to the solution in the steps this problem needs, with a check at the end.

We start by drawing it, which in this round means sketching the graph before plotting it. Both tables hold faint stars, between magnitudes 21.1 and 22.0, where the uncertainty of a magnitude grows quickly as the stars get fainter. On a semi-log graph, with the magnitude across and its uncertainty up on a logarithmic scale, each table should give a rising straight line, and the two lines should lie one above the other.
We name the physics. An uncertainty of 0.01 in a magnitude is an uncertainty of about 1% in the flux, so the inverse of the uncertainty is close to the of (5.10). For a star much fainter than the sky in its aperture, the noise comes almost entirely from the sky and is the same for every star, while the signal falls by a factor of $10^{0.4}$ for each magnitude. The uncertainty then grows by the same factor, so its logarithm rises by 0.4 for each magnitude, which is the second form of (D.5) with logarithms to base 10. For a star much brighter than the sky, the noise is the star's own photon noise, and the logarithm rises by 0.2 for each magnitude.
We work backwards from the targets. Part (c) needs only the line of part (b). Part (e) needs the pairs of part (d), and part (d) needs only the positions.
We use every given. The sizes of the two telescopes, 2.5 m and 4 m, answer part (a), because with everything else equal the larger telescope collects $(4 / 2.5)^{2} = 2.56$ times as much light and so measures the same star with a smaller uncertainty. The uncertainties of the tables, used for the fit in part (b), come back as the error bars of part (e).
We check the answer. Both slopes should lie between about 0.2 and 0.4, and the survey with the larger telescope should reach the higher signal-to-noise ratio at any given magnitude.

The second step is the one that unlocks the problem, because the physical picture predicts slopes between about 0.2 and 0.4 and so gives a check on part (b). The third problem also uses the Stefan–Boltzmann law of Section 4.6 and the motion of a star in a binary orbit, which a later chapter covers in full.

Olympiad problemsPast IOAA problems, restated3 problems

2024Two surveys of Stripe 8275 points

Adapted from IOAA 2024 (Rio de Janeiro), data-analysis problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

The Sloan Digital Sky Survey (SDSS) used a 2.5 m telescope at Apache Point in the United States during the 2000s, and the Dark Energy Survey (DES) used a 4 m telescope at Cerro Tololo in Chile from 2013 to 2019. They mostly covered different parts of the sky, but both observed an equatorial strip called Stripe 82, which can be used to compare and calibrate their photometry. The two tables below list stars in Stripe 82, each with an identification number, its right ascension and declination, and its magnitude $m_{g}$ in the g band with its uncertainty $σ_{g}$ . The names of the two files were lost, so we do not know which table came from which survey.

Table 1

ID1	RA (°)	Dec (°)	$m_{g}$	$σ_{g}$
1	0.047255	0.000406	21.7649	0.0120
2	0.064741	0.021568	21.1111	0.0067
3	0.064911	0.026395	21.3931	0.0084
4	0.098343	0.054871	21.3934	0.0088
5	0.022256	0.039129	21.9933	0.0157
6	0.006188	0.066928	21.5490	0.0088
7	0.083945	0.074259	21.9395	0.0126
8	0.076715	0.079496	21.4808	0.0089
9	0.057422	0.076006	21.8897	0.0127
10	0.024412	0.087688	21.8341	0.0126
11	0.044723	0.091782	21.8868	0.0172
12	0.071089	0.091053	21.4390	0.0098

Table 2

ID2	RA (°)	Dec (°)	$m_{g}$	$σ_{g}$
1	0.006167	0.066874	21.9020	0.0576
2	0.018660	0.007450	21.8039	0.0529
3	0.047853	0.061487	21.3007	0.0418
4	0.050870	0.015659	21.1678	0.0388
5	0.051270	0.020812	21.2524	0.0401
6	0.057414	0.075999	21.8884	0.0578
7	0.064745	0.021583	21.3634	0.0422
8	0.064910	0.026419	21.6428	0.0488
9	0.071102	0.091058	21.9259	0.0751
10	0.074946	0.002792	21.3258	0.0410
11	0.076709	0.079474	21.5303	0.0476
12	0.092635	0.077395	21.6995	0.0513
13	0.098343	0.054854	21.6542	0.0499
14	0.099332	0.093711	21.8802	0.0577

(a) Which table comes from SDSS and which from DES? Take the two surveys to be alike in their detectors, exposure times and sites. (5 points)

(b) For each table, plot $σ_{g}$ on a logarithmic scale against $m_{g}$ on an ordinary scale, and find the slope $A$ and intercept $B$ of the straight line $\log σ_{g} = A m_{g} + B$ . The uncertainties of $A$ and $B$ are not needed. (35 points)

(c) The signal-to-noise ratio is about $1 / σ_{g}$ . From the lines of part (b), what signal-to-noise ratio does each survey reach at a magnitude of 21.5? (5 points)

(d) A star of Table 1 that lies within 1″ of a star of Table 2 can be taken to be the same star. Find the stars that appear in both tables and list their pairs of identification numbers. (15 points)

(e) For the stars in both tables, plot the magnitude from Table 2 against the magnitude from Table 1 on ordinary graph paper, with error bars of twice $σ_{g}$ in both directions. Which stars would be suitable for calibrating the photometry of one survey against the other? Give their numbers in Table 1. (15 points)

I solved it without the solution

At the same magnitudes, the uncertainties of Table 1 run from 0.0067 to 0.0172, about 0.011 on average, and those of Table 2 from 0.0388 to 0.0751, about 0.050 on average. The 4 m telescope of DES collects 2.56 times as much light as the 2.5 m telescope of SDSS, so it measures the same star with the smaller uncertainty. Table 1 comes from DES, and Table 2 from SDSS. With everything else equal, the signal-to-noise ratio of a faint star grows in proportion to the diameter of the telescope, so the uncertainties should differ by a factor of about 1.6, while the tables differ by about 4.5. The sign of the difference answers part (a), and its size shows that the two surveys differ in more than their mirrors.
The plotted points of each table lie close to a straight line. The least-squares line of (D.6) through the points $(m_{g}, \log σ_{g})$ has a slope of $A = 0.278$ for SDSS and $A = 0.406$ for DES. The intercept $B$ is the value of the line at $m_{g} = 0$ , far off the graph, so it is found from the slope and the centre of the points. For SDSS the centre is at $m_{g} = 21.596$ and $\log σ_{g} = - 1.307$ , so $B = - 1.307 - 0.278 \times 21.596 = - 7.31$ . For DES the centre is at $m_{g} = 21.640$ and $\log σ_{g} = - 1.967$ , and with the slope to one more figure, $B = - 1.967 - 0.4063 \times 21.640 = - 10.76$ . A best line drawn by hand gives values close to these, and the official solution accepted any within 10% of them.
At $m_{g} = 21.5$ the SDSS line gives $\log σ_{g} = - 1.333$ , so $σ_{g} = 0.046$ and the signal-to-noise ratio is about 22. The DES line gives $σ_{g} = 0.0095$ and a signal-to-noise ratio of about 106. The SDSS slope lies between 0.2 and 0.4, and the DES slope of 0.406 is as close to the 0.4 of a sky-limited star as the scatter of its points allows, as the physical picture expects. The larger telescope reaches the higher signal-to-noise ratio.
All the stars lie within 0.1° of the celestial equator, where $\cos δ$ differs from 1 by less than two parts in a million, so the separation of two stars on the sky is $\sqrt{Δ α^{2} + Δ δ^{2}}$ , and 1″ is 0.000278°. Comparing each star of Table 1 with the stars of Table 2 near its right ascension gives seven pairs closer than 1″. As (ID1, ID2) they are (2, 7), (3, 8), (4, 13), (6, 1), (8, 11), (9, 6) and (12, 9), all within 0.21″ of each other. The next closest pair, (3, 7), is 17″ apart.
The differences between the magnitudes of Table 2 and Table 1 for the stars 2, 3, 4, 6, 8, 9 and 12 of Table 1 are 0.25, 0.25, 0.26, 0.35, 0.05, 0.00 and 0.49. The error bars of each difference, twice the two uncertainties combined in quadrature by (S.16), are 0.09 to 0.15 long either side.
Stars 8 and 9 lie on the line where the two magnitudes are equal, within their error bars. Stars 2, 3 and 4 lie together about 0.25 above that line, and star 6, at 0.35, agrees with them within its error bar of 0.12. Star 12, at 0.49, lies away from both groups. A calibration measures the offset between the magnitudes of two surveys, so the useful stars are those that agree with one another up to one offset.

AnswerTable 1 comes from DES and Table 2 from SDSS. The lines are $A = 0.278$ , $B = - 7.31$ for SDSS and $A = 0.406$ , $B = - 10.76$ for DES, which reach signal-to-noise ratios of about 22 and 106 at magnitude 21.5. Seven stars appear in both tables, and stars 8 and 9 of Table 1 agree between the surveys with no offset, while stars 2, 3, 4 and 6 agree with an offset of about 0.28 magnitudes, their mean difference. The official solution accepted either group, 8 and 9 for a calibration with no offset, or 2, 3, 4 and 6 for one with an offset.

2016Standardising type Ia supernovae50 points

Adapted from IOAA 2016 (Bhubaneswar), data-analysis problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

A type Ia supernova brightens and then fades along a light curve whose shape is nearly the same for all of them, which makes them useful for measuring the distances of faraway galaxies. To compare their light curves, we first turn every observed interval of time $Δ t_{obs}$ into the interval $Δ t_{gal}$ that passed in the host galaxy, by dividing it by $1 + z$ , where $z$ is the , because the expansion of the Universe stretches the observed intervals by that factor. The light curve fades by 2 magnitudes from its peak in a time $Δ t_{0}$ after the peak. Stretching the time of each supernova by a factor $s$ , so that $Δ t_{s} = s Δ t_{gal}$ and the stretched $Δ t_{0}$ is the same for all of them, makes all the light curves the same shape. The factor $s$ turns out to depend linearly on the absolute magnitude $M_{peak}$ of the supernova at its peak, as $s = a + b M_{peak}$ , where $a$ and $b$ are constants.

The table gives three supernovae, with the $μ$ of the first two, their speeds of recession $c z$ , and their apparent magnitudes $m$ at times $Δ t_{obs}$ from their peaks. The magnitudes have already been corrected for extinction in our Galaxy and in the air.

	SN 2006TD	SN 2006IS	SN 2005LZ
$μ$ (mag)	34.27	35.64
$c z$ (km/s)	4515	9426	12 060
$Δ t_{obs}$ (days)	$m$ (mag)	$m$ (mag)	$m$ (mag)
−15	19.41	18.35	20.18
−10	17.48	17.26	18.79
−5	16.12	16.42	17.85
0	15.74	16.17	17.58
5	16.06	16.41	17.72
10	16.72	16.82	18.24
15	17.53	17.37	18.98
20	18.08	17.91	19.62
25	18.43	18.39	20.16
30	18.64	18.73	20.48

(a) Find $Δ t_{gal}$ for every observation and plot the three light curves in the frames of their galaxies. (15 points)

(b) Taking the factor $s_{2}$ of SN 2006IS to be 1.00, find the factors $s_{1}$ of SN 2006TD and $s_{3}$ of SN 2005LZ from their values of $Δ t_{0}$ . (5 points)

(c) Find $Δ t_{s}$ for every observation and plot the three stretched light curves, to show that they now have the same shape. (14 points)

(d) Find the absolute magnitudes at peak of SN 2006TD and SN 2006IS, and from them the constants $a$ and $b$ . (6 points)

(e) Find the absolute magnitude at peak and the distance modulus of SN 2005LZ. (4 points)

(f) From the distance modulus of SN 2005LZ, estimate the Hubble constant $H_{0}$ and the characteristic age of the Universe $1 / H_{0}$ . (6 points)

I solved it without the solution

AnswerThe stretch factors are $s_{1} = 1.322$ and $s_{3} = 1.163$ , the constants are $a = 7.678$ and $b = 0.343$ per magnitude, and SN 2005LZ has $M_{peak, 3} = - 18.99$ and $μ_{3} = 36.57$ , which gives $H_{0} = 58.5$ km/s per Mpc and $1 / H_{0} = 16.7$ billion years. The official solution reads $Δ t_{0}$ as 16.4, 22.0 and 18.8 days from smooth curves drawn by hand, within the day that it accepted either way. It then divides 22.2 days instead of 22.0 by the others, a slip that changes $s_{1}$ , $s_{3}$ , $a$ and $b$ but hardly changes $M_{peak, 3}$ , and it finds the same $H_{0}$ of 58.5 km/s per Mpc. It also accepted an age of two thirds of $1 / H_{0}$ , the age of a flat universe that holds only matter.

2018An unseen companion in Auriga66 points

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.

The figure below holds the radial velocities of the third problem. It is meant for checking a period found by hand, after drawing the sine on graph paper as the exam asks.

Figure D.10Fitting a sine to radial velocitiesinteractive

Plot

Trial period

Semi-amplitude of the best sine
Typical distance of the points from it

The figure shows the eleven radial velocities of the red giant 2M05215658+4359220 measured with the TRES spectrograph between September 2017 and January 2018, which are the data of the third problem, with time counted from MJD 58000. Their error bars are smaller than the dots. For each trial period set by the slider, the thin line is the sine of that period that fits the points best by weighted least squares, and the table gives its semi-amplitude and the typical distance of the points from it. At the best sine misses the points by about 15 km/s, and at by about 4 km/s. At it passes through every point, with a semi-amplitude of 44.6 km/s, and the points trace one cycle of the sine.