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BAAO 2023, Теория — Задача 2. Wolf-Rayet Dust Shells

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British Astronomy and Astrophysics Olympiad 2022-2023 — Astronomy & Astrophysics Competition Paper — Monday 6th February 2023 · 6 февруари 2023 г. · 40 т.

Указания

This question paper must not be photographed or taken out of the exam room

Time: 3 hours ( $\sim 65$ minutes for Q1, $\sim 70$ minutes for Q2 and $\sim 45$ minutes for Q3).

Questions: All three questions should be attempted. Each question contains independent parts so that later parts can be attempted even if earlier parts are incomplete.

Solutions: Answers and calculations are to be written on loose paper. YOU WILL NEED GRAPH PAPER. Students should ensure their name and school is clearly written on the first answer sheet and that all pages are numbered. EACH QUESTION ANSWERED must be started on a new page. A standard formula booklet with standard physical constants may be used if desired.

Instructions: To accommodate students sitting the paper at different times, please do not discuss any aspect of the paper on the internet until 8 am Saturday 11th February.

Clarity: Solutions must be written legibly, in black pen, and working down the page. Scribble will not be marked and overall clarity is an important aspect of this exam.

Eligibility: The International Olympiad will be held in August 2023; all sixth form students are eligible to participate.

Calculators: Any standard calculator may be used, but calculators cannot be programmable and must not have symbolic algebra capability.

Training Dates and the IOAA (Chorzów, Poland, 10th - 20th August 2023)

The team will be selected from students taking this paper. The best students that are eligible to represent the UK at the IOAA will be invited to attend the Training Camp to be held in Oxford from Saturday 1st to Wednesday 5th April 2023. Astronomy material will be covered; problem solving skills and observational skills (telescope and naked eye observations) will be developed. At the Training Camp a Data Analysis exam along with a Round 3 theory paper will be sat. A team of five students (plus one reserve) will be selected for further training, including additional training camps in the summer.

Important Constants

Constant	Symbol	Value
Speed of light	$c$	$3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$
Earth’s rotation period	1 day	24 hours
Earth’s orbital period	1 year	365.25 days
parsec	pc	$3.09 \times 10^{16}\ \mathrm{m}$
Astronomical Unit	au	$1.50 \times 10^{11}\ \mathrm{m}$
Radius of the Sun	$R_{\odot}$	$6.96 \times 10^{8}\ \mathrm{m}$
Radius of the Earth	$R_{\oplus}$	$6.37 \times 10^{6}\ \mathrm{m}$
Mass of the Sun	$M_{\odot}$	$1.99 \times 10^{30}\ \mathrm{kg}$
Mass of the Earth	$M_{\oplus}$	$5.97 \times 10^{24}\ \mathrm{kg}$
Luminosity of the Sun	$L_{\odot}$	$3.85 \times 10^{26}\ \mathrm{W}$
Stephan-Boltzmann constant	$\sigma$	$5.67 \times 10^{-8}\ \mathrm{J\,m^{-2}\,K^{-4}}$
Gravitational constant	$G$	$6.67 \times 10^{-11}\ \mathrm{m^{3}\,kg^{-1}\,s^{-2}}$
Boltzmann constant	$k_{\mathrm{B}}$	$1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}$
Permittivity of free space	$\varepsilon_0$	$8.85 \times 10^{-12}\ \mathrm{F\,m^{-1}}$
Permeability of free space	$\mu_0$	$4\pi \times 10^{-7}\ \mathrm{H\,m^{-1}}$
Planck’s constant	$h$	$6.63 \times 10^{-34}\ \mathrm{J\,s}$
Elementary charge	$e$	$1.60 \times 10^{-19}\ \mathrm{C}$
Proton rest mass	$m_{\mathrm{p}}$	$1.67 \times 10^{-27}\ \mathrm{kg}$
Electron rest mass	$m_{\mathrm{e}}$	$9.11 \times 10^{-31}\ \mathrm{kg}$
Wien’s displacement law	$\lambda_{\max}T$	$2.90 \times 10^{-3}\ \mathrm{m\,K}$
Avagadro’s constant	$N_A$	$6.02 \times 10^{23}\ \mathrm{mol^{-1}}$

Basic calculus formulae:

Chain rule	$\frac{\mathrm{d}}{\mathrm{d}x}f(g(x)) = f'(g(x))g'(x)$
Product rule	$\frac{\mathrm{d}}{\mathrm{d}x}(uv) = \frac{\mathrm{d}u}{\mathrm{d}x}v + u\frac{\mathrm{d}v}{\mathrm{d}x}$
Quotient rule	$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u}{v}\right) = \frac{\frac{\mathrm{d}u}{\mathrm{d}x}v - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$
Integration by parts	$\int u\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x$
Standard integral	$\int \frac{1}{x}\,\mathrm{d}x = \ln\vert x\vert + C$

Important Formulae

You might find the diagram of an elliptical orbit below useful in solving some of the questions:

Elements of an elliptic orbit:

$a = \mathbf{OA}\ (= \mathbf{OP})$	semi-major axis
$b = \mathbf{OB}\ (= \mathbf{OC})$	semi-minor axis
$e = \sqrt{1 - \frac{b^2}{a^2}}$	eccentricity
$\mathbf{F}$	focus
$\mathbf{PF} = a(1-e)$	periapsis distance (shortest distance from $\mathbf{F}$ )
$\mathbf{AF} = a(1+e)$	apoapsis distance (longest distance from $\mathbf{F}$ )

Kepler’s Third Law: For an elliptical orbit, the square of the period, $T$ , of an object about the focus is proportional to the cube of the semi-major axis, $a$ (as defined above), such that

T^2 = \frac{4\pi^2}{GM}a^3\,,

where $M$ is the total mass of the system (typically dominated by the central object) and $G$ is the universal gravitational constant.

Vis-Viva Equation: For an elliptical orbit, the speed $v$ of an object at a distance $r$ from the focus is related to the semi-major axis, $a$ , total mass of the system, $M$ , and universal gravitational constant, $G$ , (as defined above), such that

v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)\,.

Magnitudes: The apparent magnitudes of two objects, $m_1$ and $m_0$ , are related to their apparent brightnesses, $b_1$ and $b_0$ , via the formula

\frac{b_1}{b_0} = 10^{-0.4(m_1 - m_0)}\,.

The absolute magnitude of an object, $\mathcal{M}$ , is the same as its apparent magnitude when viewed from 10 pc, hence the relationship between apparent and absolute magnitude and distance is

m - \mathcal{M} = 5\log\left(\frac{d}{10}\right),

where $d$ is measured in parsecs.

Note for markers

BAAO 2022/23 Solutions and Marking Guidelines

Note for markers:

Answers to two or three significant figures are generally acceptable. The solution may give more in order to make the calculation clear. Units should be present on final answers when appropriate.
There are multiple ways to solve some of the questions; please accept all good solutions that arrive at the correct answer. Students getting the answer in a box will get all the marks available for that calculation / part of the question (as indicated in red), so long as there are no unphysical / nonsensical steps or assumptions made (students may not explicitly calculate the intermediate stages and should not be penalised for this so long as their argument is clear).

Условие

Wolf-Rayet (WR) stars are some of the hottest stars known, with very strong stellar winds causing considerable mass to the be lost to the interstellar medium (ISM). In binary systems between a WR star and a very large O or B spectral class star, where their strong stellar winds collide can create the conditions for the formation of dust which goes on to enrich the ISM.

Left: JWST infrared image of WR140 — a bright central star with many nested, roughly circular pinkish dust shells on a dark star field. Right: log plot of Normalized flux (0.0001 to 1,000) against Projected distance from WR 140 (″) from 0 to 50. Three peaked curves: F770W (×10) blue, F1500W (×100) green, F2100W (×1,000) red-brown, plus a dashed black Model curve; grey vertical lines mark the median shell separation. On the model curve the peak near 4″ is labelled 2 and the peak near 44″ is labelled 17. — Figure 3: *Left:* A view of the WR140 binary system taken with the James Webb Space Telescope (JWST) in July 2022, showing clearly at least 17 nested dust shells. Credit: NASA/ESA/CSA/STScI/JPL-Caltech. *Right:* A radial plot along the image in the three mid infrared JWST filters used corresponding to 7.7, 15 and 21 µm, as well as the model of the dust production. The peaks correspond to each shell. Shells 2 and 17 are indicated on the model and the median shell separation is shown by the grey vertical lines. The projected distance is given in arcseconds. Credit: Lau et al. (2022).

The WR140 system consists of a WR and an O star which produce dust very regularly when the two stars are close together, around periastron. They are in a highly elliptical orbit ( $e = 0.8993$ ) with a period of $2895$ days. Once far from the stars, these dust shells move through space at a remarkably constant speed as indicated by the regularity of the shells in the recent image taken with the James Webb Space Telescope (JWST), shown above in Figure 3. An artist’s impression of the two stars in the system and the orbit (in the reference frame of the WR star) is shown in Figure 4 below.

Left: artist's impression on a black background of a tiny Sun (upper left, labelled Sun), a bright white Wolf-Rayet Star surrounded by blue turbulent wind, and a larger pale blue O Star. Right: dashed elliptical orbit tilted on the page; the WR star (small light-blue dot labelled WR) sits near the upper-left end, where a red solid arc marks part of the orbit; the O star (blue dot labelled O) lies at the lower right with a dark arrow showing the direction of motion. — Figure 4: *Left:* The relative size of the Sun, upper left, compared to the two stars in the system WR140. The O-type star is $\sim 30\,M_{\odot}$ , while its companion is $\sim 10\,M_{\odot}$ . Credit: NASA/JPL-Caltech. *Right:* The projected orbital configuration of WR 140 in the reference frame of the WR star. The red solid region around the periastron passage is where the O star is when dust is being formed. Credit: Lau et al. (2022).

Diagram of an elliptical orbit from the Important Formulae page: ellipse with centre O, focus F (blue dot), periapsis P and apoapsis A at the ends of the major axis, B and C at the ends of the minor axis; dashed semi-axes labelled a and b.

a. (i) Take the distance to the system to be $1.64$ kpc.

By taking measurements from Figure 3, show that the average radial expansion velocity between shells 2 to 17 is $\sim 2600$ km s $^{-1}$ . [1 arcsecond $= 1''$ is a measurement of angle, where $3600'' = 1^{\circ}$ .] [3 т.]

a. (ii) Shell 1 was observed by JWST on 27th July 2022 to be $1.63''$ away from the central stars. It was formed during the last periastron passage of the O star, which (as viewed from Earth) took place in December 2016. Taking light travel time into account, in what year was the periastron passage responsible for shell 17? [3 т.]

The average radial expansion velocity of shell 1 is less than that seen for shells 2 to 17, suggesting a mechanism accelerates the dust up to that speed from some lower speed resultant from the bow shock between the two colliding stellar winds. A prime candidate is radiation pressure, where photons deposit all of their momentum onto an optically thick dust cloud, before it moves too far from the original stars and becomes optically thin (so the transfer of momentum becomes less and less efficient) as shown in the right panel of Figure 5.

Panel a: points WR and OB on a dashed horizontal line; a blue shaded hyperbolic cone opens to the right around OB, labelled Conical shock front, with an ellipse around it marked Azimuthal direction; a dashed curved path with an arrow is labelled Orbital direction. Panel b: WR and OB close together on a dashed line, with a blue curved band to the right; regions labelled No dust (inside), Optically thick dust (the band) and Optically thin dust (beyond a dotted outer arc). — Figure 5: *Left:* The collision of stellar winds between the WR and O star creates a conical shock front in which the conditions are right for dust production. This cone is shown when at periastron in the reference frame of the WR star, although dust production is not limited to periastron so the cone can have a wide range of orientations. The place where the shock front is closest to the WR star is known as the stagnation point. *Right:* Close to the stars no dust is made in the conical shock fronts, however once far enough away the dust forms and is optically thick so experiences strong radiation pressure accelerating it. Eventually a distance is reached where the dust becomes optically thin and the acceleration decreases. Credit: Han et al. (2022).

The radiation force, $F_{\mathrm{rad}}$ , experienced by a single dust grain at a distance $r$ is given as

F_{\mathrm{rad}} = \sigma_g\frac{L_{\mathrm{bin}}}{4\pi r^2 c}\,,

where $\sigma_g$ is the cross-sectional area of the dust grain, $c$ is the speed of light and $L_{\mathrm{bin}}$ is the combined luminosity of the binary system. Han et al. (2022) suggest the following acceleration model to match the observations:

a(t) = \begin{cases} 0 & 0 \le r < r_{inner} \\ a_{\max} & r_{inner} \le r < r_{outer} \\ \frac{a_{\max}r_{outer}^2}{r^2} & r \ge r_{outer} \end{cases}

Here, $r_{inner}$ is the inner radius of the optically thick dust shell and $r_{outer}$ is the distance at which it becomes optically thin. $a_{\max}$ is the constant acceleration experienced by the dust whilst optically thick, and is equal to the acceleration of a dust grain by radiation pressure at $r = r_{outer}$ .

b. (i) In the Han et al. (2022) model they use $r_{inner} = 50$ au, $r_{outer} = 220$ au, and assume spherical dust grains with a radius of $40$ nm and density $1.6$ g cm $^{-3}$ , as well as a dust-to-gas ratio of $0.019$ . Take the apparent magnitude of the system (after correcting for substantial extinction by the dust shells and the general ISM) as $m = 0.426$ , and you are given the absolute magnitude of the Sun is $\mathcal{M}_{\odot} = 4.74$ .

Show that $a_{\max}$ is $\sim 940$ km s $^{-1}$ yr $^{-1}$ . [5 т.]

b. (ii) Hence, sketch $a(r)$ from $0 \le r \le 2500$ au. [2 т.]

b. (iii) Given shell 1 is measured to be moving at $2540$ km s $^{-1}$ , calculate the initial speed of the gas streaming away from the stars in the conical shock front, $v_0$ , before it experiences the acceleration due to radiation pressure. [5 т.]

The stellar winds from each star are initially thermally accelerated up to a terminal velocity, $v_{\infty}$ , following the relationship $v(r) = v_{\infty}(1 - R_{\star}/r)$ , where $R_{\star}$ is the radius of the star. The shape of the conical shock front is well approximated as a hyperbolic cross-section so long as the speed of each star’s wind at the shock front is $\approx v_{\infty}$ where they meet. This is true for much of the orbit and the cone, however due to the extreme eccentricity of the WR140 system at periapsis it may not hold close to the stagnation point (the closest the cone gets to the WR star, as seen in Figure 5).

c. (i) The measured semi-major axis of the WR140 system from radio interferometry is $8.82$ milliarcseconds. Investigation of spectral lines shows that for the O star $v_{\infty,O} = 3100$ km s $^{-1}$ and for the WR star $v_{\infty,WR} = 2800$ km s $^{-1}$ , and measurements of X-ray absorption lead to suggested mass loss rates of $\dot{M}_O = 3.7\times10^{-7}\,M_{\odot}\ \mathrm{yr}^{-1}$ and $\dot{M}_{WR} = 1.7\times10^{-5}\,M_{\odot}\ \mathrm{yr}^{-1}$ respectively for each star. In the shock front, the pressure from each stellar wind is balanced. Considering the system at periapsis:

Show that the distance from the O star to the stagnation point is

r_O = \frac{\sqrt{\eta}}{1+\sqrt{\eta}}r_{peri}\,,

where $\eta$ is the ratio of the momenta of the two stellar winds, $\eta \equiv \frac{p_O}{p_{WR}}$ and $r_{peri}$ is the separation between the stars at periapsis. [2 т.]

c. (ii) Hence, calculate $r_O$ in au. [4 т.]

c. (iii) Taking the WR star to be approximately 2 times more luminous than the O star, and the effective surface temperatures of the stars to be $T_{\mathrm{eff,O}} = 35$ kK and $T_{\mathrm{eff,WR}} = 60$ kK, calculate the radius of each star in solar units. [3 т.]

c. (iv) Hence, determine for which star(s) the assumption of $v \approx v_{\infty}$ at the stagnation point is reasonable, and suggest (if any) the change in the position of the stagnation point taking in the real values of $v$ at $r_O$ . [3 т.]

d. The conditions for dust production turn on when the stars get within $\approx 7.9$ au of each other, corresponding to the solid red region in the right panel of Figure 4. A line from the WR star to the start of dust production region makes an angle of $45^{\circ}$ with the semi-major axis. By using Kepler’s second law, or otherwise, estimate the rough duration of dust production during each orbit. Give your answer in years to 2 decimal places, making any simplifying assumptions clear. [Hint: Kepler’s second law states that the area swept out by one object orbiting another is equal in the same amount of time, such that $\mathrm{d}A/\mathrm{d}t = \mathrm{constant} = \pi ab/T$ where $T$ is the period] [10 т.]

Отговори

Решение

Q2 – Wolf-Rayet Dust Shells [40 marks]

a. (i)

Measuring from the graph, the average spacing between shells is 2.67” [1]

[Expect to see candidates realise that 40” covers the 15 intervals from shell 2 to 17, and hence the average separation is 40”/15 = 2.67”. Allow ±0.03” and ecf on all subsequent parts of the question accordingly]

Long thin right-angled triangle: the horizontal base is labelled d_system, the short vertical side on the right is labelled d_shell, and the small angle θ is at the left vertex.

Using the small angle approximation (with the angle converted into radians), the distance between shells implied by their average angular spacing is

d_{shell} = d_{system}\tan\theta \approx d_{system}\theta = 1.64\times\left(\frac{2.67}{3600}\times\frac{2\pi}{360}\right)

= 2.12\times10^{-5}\ \mathrm{kpc} = 6.56\times10^{11}\ \mathrm{km}

[1]

Given the period is 2895 days and converting it to seconds,

v_{shell} = \frac{d_{shell}}{T} = \frac{6.56\times10^{11}}{2895\times24\times60\times60} = 2622.6\ldots = \boxed{2620\ \mathrm{km\,s^{-1}}}

[1] [3]

[Given it is a ‘show that’, it must have $\geq$ 3 s.f. in final answer]

a. (ii)

From shell 1 to shell 17 is 16 intervals, so in years

\Delta t = 16\times\left(\frac{2895}{365}\right) = 126.9\ \mathrm{yrs}

[1]

[Accept use of 365.25 days in a year]

The light travel time from the system is

t_{system} = \frac{d_{system}}{c} = \frac{1.24\times1000\times3.09\times10^{16}}{3.00\times10^{8}} = 1.69\times10^{11}\ \mathrm{s} = 5356.4\ \mathrm{yrs}

[1]

Given that December 2016 is approximately decimalised as 2016.9, and that 1 BC goes to 1 AD with no year zero, then the year it was created is

2016.9 - (126.9 + 5356.4) - 1 = -3467.4 = (\text{July})\ \boxed{3467\ \text{BC}}

[1] [3]

[Accept use of 2016 instead of 2016.9, and not skipping year 0, giving an allowance of ±2 years on the final answer]

b. (i)

First we can find the absolute magnitude of the system

\mathcal{M} = m - 5\log\left(\frac{d_{system}}{10}\right) = 0.426 - 5\log\left(\frac{1.64\times10^{3}}{10}\right) = -10.65

[1]

By comparison with the Sun we can turn this into a luminosity

L = 10^{\frac{\mathcal{M}-\mathcal{M}_{\odot}}{-2.5}}L_{\odot} = 10^{\frac{-10.65-4.74}{-2.5}}L_{\odot} = 1.43\times10^{6}\,L_{\odot} = 5.50\times10^{32}\ \mathrm{W}

[1]

From Newton’s 2nd Law

a_{max} = \frac{F_{rad}}{m_{tot}} = \frac{F_{rad}}{m_{dust}}\times\frac{m_{dust}}{m_{tot}} \approx \frac{F_{rad}}{m_{dust}}\times\frac{m_{dust}}{m_{gas}} = \frac{\sigma_g}{m_{dust}}\frac{L_{bin}}{4\pi r_{outer}^2c}\times\frac{m_{dust}}{m_{gas}}

= \frac{\pi r_{dust}^2}{\rho_{dust}\times\frac{4}{3}\pi r_{dust}^3}\frac{L_{bin}}{4\pi r_{outer}^2c}\times\frac{m_{dust}}{m_{gas}} = \frac{3L_{bin}}{16\rho_{dust}r_{dust}\pi r_{outer}^2c}\times\frac{m_{dust}}{m_{gas}}

[1]

\therefore a_{max} = \frac{3\times5.50\times10^{32}}{16\times1600\times40\times10^{-9}\times\pi\times(220\times1.50\times10^{11})^2\times3.00\times10^{8}}\times0.019

= 2.99\times10^{-2}\ \mathrm{m\,s^{-2}}

[1]

= \boxed{941\ \mathrm{km\,s^{-1}\,yr^{-1}}}

[1] [5]

[Given it is a ‘show that’, it must have $\geq$ 3 s.f. in final answer. Final mark is for correct unit conversion. Give full credit for the use of the exact corrective factor $\frac{m_{dust}}{m_{tot}} = \frac{m_{dust}}{m_{dust}+m_{gas}} = \frac{1}{1+m_{gas}/m_{dust}} = 0.01865$ , leading to $a_{\max} = 2.93\times10^{-2}\ \mathrm{m\,s^{-2}} = \boxed{924\ \mathrm{km\,s^{-1}\,yr^{-1}}}$ . Forgetting about the use of the dust-to-gas ratio gives $a_{\max} = 1.57\ \mathrm{m\,s^{-2}} = 49554\ \mathrm{km\,s^{-1}\,yr^{-1}}$ and scores 4 marks max. If no attempt at the algebraic simplification, the third mark can be gained for either a correct $m_{\mathrm{dust}}$ (= $4.29\times10^{-19}$ kg) or $\sigma_{\mathrm{g}}$ (= $5.03\times10^{-15}\ \mathrm{m^2}$ )]

b. (ii)

Zero from $0 \le r < 50$ au

and flat section at ~940 km s $^{-1}$ yr $^{-1}$ from $50 \le r < 220$ au [1]

Following 1/r $^2$ shape from $220 \le r \le 2500$ au [1] [2]

[Lose 1 mark if the axes are not labelled, or a scale on either axis is missing. The horizontal axis does not need to be to scale, so long as key values indicated. Expect to see a (not to scale) sketch like the one below]

Sketch of a (km s^-1 yr^-1) against r (au): zero from 0 to 50, a vertical jump at 50 to a flat plateau at 940 until 220 (dashed vertical line at 220), then a curve decaying as 1/r^2 towards the axis by 2500.

b. (iii)

The work done by the radiative force on the dust is equal to its change in KE

\Delta KE = \frac{1}{2}m_{dust}v_{shell1}^2 - \frac{1}{2}m_{dust}v_0^2

= \int_{r_{inner}}^{r_{shell1}}F(r)\,\mathrm{d}r = m_{dust}\int_{r_{inner}}^{r_{shell1}}a(r)\,\mathrm{d}r

\therefore \frac{1}{2}v_0^2 = \frac{1}{2}v_{shell1}^2 - \int_{r_{inner}}^{r_{shell1}}a(r)\,\mathrm{d}r

[1]

[This formula could also be arrived at by generalising the SUVAT equation $v^2 = u^2 + 2as$ into $v^2 = u^2 + 2\int a(s)\,\mathrm{d}s$ ]

In a. ii we were told the angular separation of shell 1 from the star, so

r_{shell1} = 1.64\times\left(\frac{1.63}{3600}\times\frac{2\pi}{360}\right) = 1.30\times10^{-5}\ \mathrm{kpc} = 4.00\times10^{14}\ \mathrm{m}

[1]

[This corresponds to 2670 au. If they use $r_{\mathrm{shell1}}$ = 2500 au in the subsequent integration lose only this mark (so get max 4 marks)]

Evaluating the integral with everything in SI units

\int_{r_{inner}}^{r_{shell1}}a(r)\,\mathrm{d}r = \int_{r_{inner}}^{r_{outer}}a_{max}\,\mathrm{d}r + \int_{r_{outer}}^{r_{shell1}}\frac{a_{max}r_{outer}^2}{r^2}\,\mathrm{d}r

= [a_{max}\times(r_{outer}-r_{inner})] + \left[-a_{max}r_{outer}^2\left(\frac{1}{r_{shell1}^2}-\frac{1}{r_{outer}^2}\right)\right]

[1]

= [2.99\times10^{-2}\times(220-50)\times1.50\times10^{11}]

+[-2.99\times10^{-2}\times(220\times1.50\times10^{11})^2\times\left(\frac{1}{(4.00\times10^{14})^2}-\frac{1}{(220\times1.50\times10^{11})^2}\right)

= 7.61\times10^{11} + 9.04\times10^{11}

= 1.67\times10^{12}\ \mathrm{m^2\,s^{-2}}

[1]

[0.5 marks for either correctly evaluated integral (flat section or curved section)]

We can now evaluate $v_0$

\therefore v_0 = \sqrt{v_{shell1}^2 - 2\int_{r_{inner}}^{r_{shell1}}a(r)\,\mathrm{d}r} = \sqrt{(2540\times10^3)^2 - 2\times1.67\times10^{12}}

= 1.77\times10^{6}\ \mathrm{m\,s^{-1}} = \boxed{1770\ \mathrm{km\,s^{-1}}}

[1] [5]

[Accept answer in m s $^{-1}$ or km s $^{-1}$ for final mark. Allow ecf from their $a_{\max}$ ]

c. (i)

Line diagram: WR star (small dark-blue circle) on the left and O star (larger pale circle) on the right, with the stagnation point marked by a dot between them, closer to the O star. Arrows above mark r_peri − r_O (WR star to stagnation point) and r_O (stagnation point to O star); an arrow below spans the whole separation r_peri.

The ram pressure is $p_{ram} = \frac{F}{A} = \frac{\dot{p}}{A}$ where $\dot{p}$ is the rate of change of momentum, so at the stagnation point (assuming spherically symmetric stellar winds)

\frac{\dot{p}_O}{4\pi r_O^2} = \frac{\dot{p}_{WR}}{4\pi(r_{peri}-r_O)^2}

[1]

\therefore \frac{\dot{p}_O}{\dot{p}_{WR}} = \frac{r_O^2}{(r_{peri}-r_O)^2}

But since $\eta = \frac{p_O}{p_{WR}} = \frac{\dot{p}_O}{\dot{p}_{WR}}$ then $\eta = \frac{r_O^2}{(r_{peri}-r_O)^2} \therefore \sqrt{\eta} = \frac{r_O}{r_{peri}-r_O}$ [1] [2]

This can then be rearranged to give the required formula

\sqrt{\eta}(r_{peri}-r_O) = r_O \therefore \sqrt{\eta}r_{peri} = (1+\sqrt{\eta})r_O \therefore r_O = \frac{\sqrt{\eta}}{1+\sqrt{\eta}}r_{peri}

[The first mark is for setting up the problem algebraically or for a helpful diagram. The second mark is for correctly introducing $\eta$ and having convincing algebra to the final answer (since it was given, no standalone mark for the final expression)]

c. (ii)

First, we can calculate $\eta$

\eta = \frac{p_O}{p_{WR}} = \frac{\dot{p}_O}{\dot{p}_{WR}} = \frac{\dot{M}_Ov_{\infty,O}}{\dot{M}_{WR}v_{\infty,WR}} = \frac{3.7\times10^{-7}\times3100}{1.7\times10^{-5}\times2800} = 0.024

[1]

We can also calculate the semi-major axis, and then use the given eccentricity to get the periapsis distance

a = 1.64\times\left(\frac{8.82\times10^{-3}}{3600}\times\frac{2\pi}{360}\right) = 7.01\times10^{-8}\ \mathrm{kpc} = 14.44\ \mathrm{au}

[1]

r_{peri} = a(1-e) = 14.44\times(1-0.8993) = 1.45\ \mathrm{au}

[1]

Putting it all into the formula of the previous part

r_O = \frac{\sqrt{\eta}}{1+\sqrt{\eta}}r_{peri} = \frac{\sqrt{0.024}}{1+\sqrt{0.024}}\times1.45 = \boxed{0.195\ \mathrm{au}}

[1] [4]

[Must be in au for the final mark, although a and $r_{\mathrm{peri}}$ might be given in metres ( $2.17\times10^{12}$ m and $2.18\times10^{11}$ m respectively)]

c. (iii)

We can split up the luminosity calculated earlier in b. i.

L_{WR} = \frac{2}{3}L_{bin} = \frac{2}{3}\times1.43\times10^{6} = 9.53\times10^{5}\,L_{\odot} = 3.67\times10^{32}\ \mathrm{W}

[0.5]

L_O = \frac{1}{3}L_{bin} = \frac{1}{3}\times1.43\times10^{6} = 4.77\times10^{5}\,L_{\odot} = 1.83\times10^{32}\ \mathrm{W}

[0.5]

Using the Stephan-Boltzmann law

R_{WR} = \sqrt{\frac{L_{WR}}{4\pi\sigma T_{WR}^4}} = \sqrt{\frac{3.67\times10^{32}}{4\pi\times5.67\times10^{8}\times(60\times10^3)^4}} = 6.30\times10^{9}\ \mathrm{m} = \boxed{9.06\,R_{\odot}}

[1] [1.5]

R_O = \sqrt{\frac{L_O}{4\pi\sigma T_O^4}} = \sqrt{\frac{1.83\times10^{32}}{4\pi\times5.67\times10^{8}\times(35\times10^3)^4}} = 1.31\times10^{10}\ \mathrm{m} = \boxed{18.82\,R_{\odot}}

[1] [1.5]

[Score max 2 out of 3 if their final answer is not in $R_{\odot}$ .

Give full credit if they assume $T_{\odot} \approx 5780\ \mathrm{K}$ and so use $R = \sqrt{\frac{L/L_{\odot}}{(T/T_{\odot})^4}}R_{\odot}$ for calculations]

[This matches up well with the scale of the stars in Figure 4 in the paper, showing these are both large, very hot stars.]

c. (iv)

Using the given formula:

v_{r_O} = v_{\infty,O}\left(1-\frac{R_O}{r_O}\right) = \left(1-\frac{1.31\times10^{10}}{0.195\times1.50\times10^{11}}\right)v_{\infty,O} = 0.55v_{\infty,O}

[1]

v_{r_{WR}} = v_{\infty,WR}\left(1-\frac{R_{WR}}{r_{peri}-r_O}\right) = \left(1-\frac{6.30\times10^{9}}{(1.45-0.195)\times1.50\times10^{11}}\right)v_{\infty,WR} = 0.97v_{\infty,WR}

[1]

So $v \approx v_{\infty}$ is only a reasonable assumption for the WR star (at periapsis) [0.5]

This means the stagnation point will be even closer to the O star [0.5] [3]

[Do not allow bald statement that it is a reasonable assumption for the WR star without evidence of calculation]

Ellipse with a dashed major axis; the WR star (blue dot) at the left focus. Two lines from the focus at 45° above and below the axis, each 7.9 au long, meet the ellipse; the region of the ellipse to the left of these lines, around periastron, is shaded pink.

From Kepler’s 2nd law, $\frac{\text{area of shaded section}}{\text{total area of the ellipse}} = \frac{\text{time spent in shaded section}}{\text{period of the orbit}}$ [1]

We can use approximate geometry to calculate this area e.g. breaking the non-shaded section up into a triangle (ABS) and trapezium (ABDC) sat within a half-ellipse: [1]

The same ellipse with the focus S on the dashed major axis. Lines of length 7.9 au from S at 45° meet the ellipse at A (above) and B (below). A red trapezium ABDC has vertical side AB and vertical side CD at the centre of the ellipse (C on top, D at the bottom); its width is marked h with a red double-headed arrow.

Given the symmetry, angle $\angle ASB = 2\times45^{\circ} = 90^{\circ}$

\therefore \text{area of triangle } ABS = \frac{1}{2}\times7.9^2 = 31.2\ \mathrm{au}^2

[1]

We can also use Pythagoras’ theorem to get the length AB

AB = \sqrt{7.9^2+7.9^2} = 11.2\ \mathrm{au}

[1]

The height of the trapezium

h = a - r_{peri} - 7.9\cos45^{\circ} = 14.44 - 1.45 - 5.59 = 7.41\ \mathrm{au}

[1]

The length of CD is the minor axis (= 2b), which can be calculated from rearranging the formula given at the beginning of the paper

e = \sqrt{1-\frac{b^2}{a^2}} \therefore b = a\sqrt{1-e^2} = 14.44\times\sqrt{1-0.8993^2} = 6.3\ \mathrm{au}

[1]

Hence, the area of the trapezium

\text{area } ABDC = \frac{1}{2}(AB+CD)h = \frac{1}{2}(11.2+(2\times6.3))\times7.41 = 88.2\ \mathrm{au}^2

[1]

Finally, the area of the half-ellipse (to the left of CD)

\text{area half-ellipse} = \frac{1}{2}\pi ab = \frac{1}{2}\pi\times14.44\times6.3 = 143.4\ \mathrm{au}^2

[1]

The area of the shaded section is therefore

\text{shaded area} = \text{area of half-ellipse} - (\text{area ABS} + \text{area ABDC})

= 143.4 - (31.2+88.2) = 24.0\ \mathrm{au}^2

[1]

Finally, applying Kepler’s 2nd law

t = \frac{\text{shaded area}}{\text{total area}}\times T = \frac{24.0}{2\times143.4}\times2895 = 0.084\times2895 = 242\ \text{days} = \boxed{0.66\ \text{yrs}}

[1] [10]

[Must be in years and to 2 d.p. for the final mark. A wide variety of approaches are possible, so award marks based upon sensible choices of shapes to evaluate the area of. The first mark is for realising how to apply Kepler’s 2nd law in this context, the second mark is for a clear approach to break the ellipse up into shapes (ideally with a diagram – and not necessarily the triangle and trapezium used here), and the final mark is for the final answer – all other marks are available for sensible areas and lengths relevant to their method. Accept any method that gives a value of 0.60-0.70 years so long as the justification is clear]

[The more exact way to work this out is using our knowledge of anomalies, although some approximations were made to simplify the geometric case in terms of the given length and angle. Assuming the angle to be correct, leads to a true anomaly $\nu = 180^{\circ} - 45^{\circ} = 135^{\circ}$ . Given this and the quoted eccentricity and semi-major axis, the real length AS is 7.6 au, and utilising Kepler’s equation to get the time for $-135^{\circ} \le \nu \le 135^{\circ}$ gives 0.633 years. Alternatively, assuming the length AS = 7.9 au to be correct, leads to a real true anomaly of 136.3°, and when looking at the time to travel from $-136.3^{\circ} \le \nu \le 136.3^{\circ}$ gives 0.671 years. Students using this approach (rather than areas of shapes) get full credit. However the problem is approached, it is clear that dust is only produced on a relatively short time frame, which is why each shell has such contrast.]

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Surbiton High School)

Josh Brown (University of Cambridge)

Оригинал в Архива: BAAO_2023.pdf · официални решения: BAAO_2023_solutions.pdf