Problems for Chapter 8

Builds on: Sections 8.1 to 8.8, 4.6, 4.7, 5.4 and 3.6
Reading time: 45 min
Exercises: 0
Olympiad problems: 2

The past problems in the earlier sections of this chapter each used one or two ideas at a time. The two in this section join them up. The first, from IOAA 2022, finds the temperatures, the separation, the total mass and then the separate masses of a binary, using Wien's law, the resolution of a telescope, Kepler's third law and the light lost in a transit. Each part is short, and together they join several ideas of this chapter to pieces of Chapters 4 and 5. The second, from IOAA 2024, follows a pair of black holes through a field of stars. Each star that passes close by is flung out faster than it came, and the energy it carries away comes from the orbit of the pair, which slowly shrinks.

8.9.1From textbook to olympiad

Part (b) of the first problem below says that a space telescope cannot split a binary for 38 days of its 100-day period, and asks for the distance between the stars. The method of Section S2 takes us from the chapter to the solution in these steps.

We start by drawing it. The orbit is circular and seen edge on, so from Earth the two stars move back and forth along a line, as the stars of Figure 8.3 would look from Earth, far below the page. Their distance apart on the sky is largest when the line joining them is across our view and zero when one is in front of the other.
We name the physics. A telescope splits two points only if they are more than $1.22 λ / D$ apart (5.7). The distance apart on the sky is the true separation $l$ shortened by the cosine of the angle $θ$ that the line joining the stars makes with the plane of the sky, and divided by the distance.
We work backwards from the target. To find $l$ we need the angle at which the stars just merge, since there $l \cos θ / d$ equals the smallest angle the telescope splits.
We use every given. The stars pass in front of each other twice in each period, so the 38 days are two spells of 19 days, each centred on a moment when $θ = 90 °$ . In 19 days the stars turn through $360 ° \times 19 / 100 = 68.4 °$ , so each spell runs from 34.2° before that moment to 34.2° after it, and the stars merge at $θ = 55.8 °$ .
We check the answer. At their greatest separation the stars are $1.45 / 89 = {0.0163}^{″}$ apart, which is 16.3 milliarcseconds and more than the 9.16 milliarcseconds the telescope can split, so they are resolved for most of the period, as the problem says. Part (c) then gives about 40 solar masses, far more than two stars about as hot as the Sun could have, so the numbers of this problem were chosen to exercise the method and do not describe a real binary.

Olympiad problemsPast IOAA problems, restated2 problems

2022Weighing a binary from its photometry20 points

Adapted from IOAA 2022 (Kutaisi), theory problem 6. We have restated it in our own words and kept the original data. The official solution prints the temperature of star A as 5978 K, a slip for 5796 K, which its later steps use. It also rounds the difference of magnitudes to 0.48 and the ratio of fluxes to 0.65, which gives masses of 11.4 and 29.3 solar masses, and names the larger mass M_B where it means M_A. The numbers do not describe a real binary, since stars of these masses and temperatures would be far larger and brighter. The solution was last checked on 23 October 2026.

A binary 89 pc away has a circular orbit seen edge on, with a period of 100 days. A space telescope of diameter $D = 10$ m, working at a wavelength of 364 nm, cannot resolve the two stars for a total of 38 days in each period. The spectrum of star A peaks at 500 nm and that of star B at 600 nm.

(a) Find the temperatures $T_{A}$ and $T_{B}$ of the stars. (2 points)

(b) Find the distance $l$ between the stars. (6 points)

(d) The table gives the combined light of both stars in two configurations, with the stars side by side and with B passing in front of A. The extinction in U towards the system is 1.4 magnitudes per kiloparsec, the bolometric correction is $BC = m_{bol} - V$ , and the ratio of the mean densities of the stars is $ρ_{A} / ρ_{B} = 0.7$ . Find the masses of the two stars. (10 points)

Configuration	U	U − B	B − V	BC
Side by side	6.39	0.2	0.1	0.1
B in front of A	6.86	0.25	0.12	0.175

I solved it without the solution

Answer(a) $T_{A} = 5796$ K and $T_{B} = 4830$ K. (b) The stars are 1.45 au apart. (c) The total mass is 40.7 solar masses. (d) Star A has 29.1 solar masses and star B 11.6 solar masses.

2024The hardening of a binary black hole25 points

Adapted from IOAA 2024 (Rio de Janeiro), theory problem T8. We have restated it in our own words and kept the original data. The solution was last checked on 23 October 2026.

Two black holes, each of mass $M$ , move on a circular orbit of separation $a$ round their centre of mass. The pair moves through a large, uniform field of stars, each of mass $m$ much less than $M$ , with $n$ stars in each unit of volume. In the frame of the centre of mass, a star comes in from far away with speed $v$ and impact parameter $b$ , the distance by which it would miss the centre if it went straight on. It comes closest to the centre at $r_{p} = a / 2$ . In parts (a) and (c), $v^{2}$ is much less than $G M / a$ .

(a) Treating the pair as one mass $2 M$ at the centre, find $b$ in terms of $M$ , $a$ , $v$ and $G$ . (5 points)

(b) Near the pair the star feels mainly one black hole. Suppose it meets that black hole head on, with its own speed small next to the orbital speed of the black hole, and leaves along the line it came in on. Energy and momentum are conserved, and the meeting is much shorter than the period of the orbit. Find the speed $v_{f}$ of the star when it is far away again, in terms of $M$ , $a$ and $G$ . (6 points)

(c) The pair moves through the field at speed $v_{0}$ . Count only the stars that meet it with impact parameters between $b_{0} / 2$ and $3 b_{0} / 2$ , where $b_{0}$ is the impact parameter of (a) for the speed $v_{0}$ , and take each of them to come about $a / 2$ from the centre and leave at the speed $v_{f}$ . The energy they carry away comes from the orbit, which stays circular. Show that

\frac{d}{d t} (\frac{1}{a}) = H \frac{G ρ}{v_{0}},

where $ρ = n m$ is the density of the field of stars, and find the number $H$ . (14 points)

I solved it without the solution

Answer(a) $b \approx \sqrt{2 G M a} / v$ . (b) $v_{f} = \sqrt{2 G M / a}$ . (c) $H = 8 π$ . The pair's separation shrinks, and $1 / a$ grows steadily with time.