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BAAO 2018, Теория — Задача 1. Stellar Mass Loss

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British Astronomy and Astrophysics Olympiad 2017-2018 — Astronomy & Astrophysics Competition Paper — Monday 22nd January 2018 · 22 януари 2018 г. · 25 т.

Указания

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

Time: 3 hours plus 15 minutes reading time (no writing permitted). Approx 45 minutes per question.

Questions: All four questions should be attempted.

Marks: The questions carry similar marks.

Solutions: Answers and calculations are to be written on loose paper or in examination booklets. Students should ensure their name and school is clearly written on all answer sheets and pages are numbered. A standard formula booklet with standard physical constants should be supplied.

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 27th January.

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

Eligibility: The International Olympiad will be held during September 2018; all sixth form students are eligible to participate, even if they will be attending university from October.

Training Dates and the International Astronomy and Astrophysics Olympiad (IOAA)

The team will be selected from sixth form students taking this paper and Y12 students taking the AS Challenge in March. The best students eligible to represent the UK at the IOAA will be invited to attend the Training Camp to be held in the Physics Department at the University of Oxford, (Monday 9th April to Thursday 12th April 2018). 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 and a short theory paper will be sat. Five students (plus one reserve) will be selected for further training. From May there will be mentoring by email to cover some topics and problems, followed by 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 Earth	$R_{\oplus}$	$6.37 \times 10^{6}\ \mathrm{m}$
Semi-major axis of the Earth’s orbit		1 au
Radius of the Sun	$R_{\odot}$	$6.96 \times 10^{8}\ \mathrm{m}$
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	$G$	$6.67 \times 10^{-11}\ \mathrm{m^{3}\,kg^{-1}\,s^{-2}}$

You might find the diagram of an elliptical orbit below useful in solving some of the questions (the diagram is shown with Question 3).

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
F	focus
P	periapsis (point nearest to F)
A	apoapsis (point furthest from F)

Keplers Third Law: For an elliptical orbit, the square of the period, $T$ , of orbit of an object about the focus is proportional to the cube of the semi-major axis, $a$ (the average of the minimum and maximum distances from the Sun). The constant of proportionality is $4\pi^2/GM$ , where $M$ is the mass of the central object and $G$ is the universal gravitational constant.

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)}

Note for markers

BAAO 2017/18 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.
There are multiple ways to solve some of the questions; please accept all good solutions that arrive at the correct answer. If a candidate gets the final (numerical) answer then allow them all the marks for that part of the question (as indicated in red), so long as there are no unphysical / nonsensical steps or assumptions made.

Условие

All stars lose mass during their lifetimes due to two main routes: particles escaping their surface (referred to as the stellar wind), and the mass defect of the nuclear reactions occurring in their cores.

a. (i) a. Given that, at the distance of the Earth, the proton flux from the Sun’s stellar wind (which is assumed to be radiated equally in all directions) is $3.0 \times 10^{12}\ \mathrm{m^{-2}\,s^{-1}}$ , and that the luminosity of the Sun is solely due to the fusion of hydrogen to helium:

(i) Show that the rate at which the Sun is losing mass, $\dot{M} \equiv \frac{\mathrm{d}M}{\mathrm{d}t}$ , due to its stellar wind is $\sim 10^{-14}\,M_{\odot}\ \mathrm{yr^{-1}}$ . [Take the mass of a proton to be $1.67 \times 10^{-27}$ kg.] [2 т.]

a. (ii) Show that $\dot{M}$ due to nuclear fusion is greater than that from the solar wind. [2 т.]

a. (iii) Estimate how much the Earth-Sun distance and the Earth’s orbital period will have changed after the Sun has lost mass (via both routes) for one year. Assume the orbit remains circular throughout and ignore gravitational effects from all other bodies. [5 т.]

In practice, the mass loss rate can vary quite considerably during a star’s lifetime, particularly once it has left the main sequence when the stellar wind can become much more substantial. Wolf-Rayet stars are massive stars near the end of their lives, presumed to be the in the stage just before a supernova, and are losing substantial amounts of mass due to very fast stellar winds. This deposits considerable energy into the surrounding interstellar medium (ISM) and can sweep up material into a thin bubble around the star, visible as a type of planetary nebula.

Colour photograph of the nebula NGC 2359 (Thor’s Helmet): a pale blue bubble-shaped shell in the centre with red-pink filaments extending up to the left and right like the wings of a helmet, on a background of many stars. — Figure 1: The nebula NGC 2359 around the Wolf-Rayet star WR7. The nebula is known as Thor’s Helmet due to its resemblance to the helmet worn by the character from the Marvel Comics series. Credit: Star Shadows Remote Observatory and PROMPT/UNC.

b. (i) b. If every photon leaving a star is able to transfer all of its momentum (given by $p_{\mathrm{photon}} = \frac{E_{\mathrm{photon}}}{c}$ ) to drive a stellar wind (known as the single scattering limit):

(i) Show that the maximum mass loss rate for a star can be written in terms of the luminosity of the star, $L$ , the (terminal) velocity of the stellar wind far from the star, $v_{\infty}$ , and the speed of light, $c$ , as

\dot{M}_{\mathrm{max}} = \frac{L}{v_{\infty} c}.

[3 т.]

b. (ii) Hence, derive an expression for the maximum kinetic energy deposited per second by the stellar wind as a function of the luminosity of the star. Comment on your answer. [2 т.]

c. (i) c. The Wolf-Rayet star WR7 is in the constellation of Canis Major and its strong winds are responsible for the nebula known as Thor’s Helmet (see Figure 1). The star has a mass of $16\,M_{\odot}$ , a radius of $1.41\,R_{\odot}$ , and a surface temperature of $112\,000$ K, with a measured $v_{\infty}$ of $1545\ \mathrm{km\,s^{-1}}$ .

(i) Predict the mass loss rate of WR7 based upon the properties of the star, assuming the single scattering limit. Give your answer in units of $M_{\odot}\ \mathrm{yr^{-1}}$ . [3 т.]

c. (ii) The nebula is 5 arcmins in diameter ( $1^{\circ} = 60$ arcmin) and 4.8 kpc away, and at its edge is a bright thin shell of swept up material expanding at a rate of $30\ \mathrm{km\,s^{-1}}$ . The age of such a nebula, $t$ , is related to the current values of radius, $R$ , and expansion speed, $v$ , by $t = 0.55R/v$ . Using this model, determine the age of the nebula. [3 т.]

c. (iii) If the expansion is purely driven by the direct impact of the stellar winds, then the radius at time $t$ , $R(t)$ , can be related to $\dot{M}$ with the formula

R(t) = \left(\frac{3\dot{M} v_{\infty} t^2}{2\pi n_0 m_{\mathrm{H}}}\right)^{1/4}

where $n_0$ is the ambient hydrogen number density in the nebula. If $n_0 = 16\ \mathrm{cm^{-3}}$ and $m_{\mathrm{H}} = 1.67 \times 10^{-27}$ kg, calculate the observed mass loss rate based upon the properties of the nebula. Compare it with the predicted one from earlier and comment on your answer. [3 т.]

c. (iv) Using your new value for $\dot{M}$ , calculate the total mass expelled from the star and hence the total kinetic energy the stellar wind has so far deposited into the ISM during this stage of the star’s life. [2 т.]

Отговори

Решение

Q1 - Stellar Mass Loss [Total = 25]

a. (i)

$\dot{M}_{\mathrm{wind}} =$ mass flux $\times$ surface area of sphere 1 au in radius

$= (3.0 \times 10^{12} \times 1.67 \times 10^{-27}) \times (4\pi \times (1.50 \times 10^{11})^2)$ [1]

$= 1.42 \times 10^{9}\ \mathrm{kg\,s^{-1}}$

$= 2.24 \times 10^{-14}\ M_{\odot}\ \mathrm{yr^{-1}}$ (so $\sim 10^{-14}\ M_{\odot}\ \mathrm{yr^{-1}}$ ) [1] [2]

a. (ii)

$L = \dot{M}_{\mathrm{fusion}} c^2 \quad \therefore \quad \dot{M}_{\mathrm{fusion}} = L_{\odot}/c^2 = \frac{3.85 \times 10^{26}}{(3.00 \times 10^{8})^2}$ [1]

$= 4.28 \times 10^{9}\ \mathrm{kg\,s^{-1}}$ [1] [2]

$(= 6.78 \times 10^{-14}\ M_{\odot}\ \mathrm{yr^{-1}})$ [ $so > \dot{M}_{\mathrm{wind}}$ ]

a. (iii)

$\dot{M}_{\mathrm{tot}} = \dot{M}_{\mathrm{wind}} + \dot{M}_{\mathrm{fusion}} = 9.02 \times 10^{-14}\ M_{\odot}\ \mathrm{yr^{-1}}$ [1]

In solar units, Kepler's 3rd Law is $M = a^3 / T^2$ , and since the change is mass is very small compared with $M_{\odot}$ we can make the approximation $\frac{\Delta a}{a} = \frac{\Delta T}{T} = \frac{\Delta M}{M}$ [2]

So: $\Delta a = \frac{\Delta M}{M_{\odot}} a = 9.02 \times 10^{-14} \times 1.50 \times 10^{11} = 1.35 \times 10^{-2}\ \mathrm{m}$ [1]

$\Delta T = \frac{\Delta M}{M_{\odot}} T = 9.02 \times 10^{-14} \times (365.25 \times 24 \times 3600)$

$= 2.85 \times 10^{-6}\ \mathrm{s}$ [1] [5]

[Allow angular momentum conservation arguments too, as well as ΔT being 3 times larger]

b. (i)

$p_{\mathrm{max}} = m v_{\infty} \quad \therefore \quad \frac{p_{\mathrm{max}}}{t} = \dot{M}_{\mathrm{max}} v_{\infty}$ [1]

In the single scattering limit: $p_{\mathrm{max}} = p_{\mathrm{photon}} = \frac{E}{c} \quad \therefore \frac{E}{ct} = \dot{M}_{\mathrm{max}} v_{\infty}$ [1]

(since L = E/t) $\quad \therefore \frac{L}{c} = \dot{M}_{\mathrm{max}} v_{\infty}$ [1] [3]

$\left(\therefore \dot{M}_{\mathrm{max}} = \frac{L}{v_{\infty} c}\right)$

b. (ii)

$\dot{E}_{\mathrm{K,max}} = \frac{1}{2} \dot{M}_{max} v_{\infty}^2 = \frac{1}{2} \frac{L}{v_{\infty} c} v_{\infty}^2 = \frac{1}{2} \frac{v_{\infty}}{c} L$ [1] [1]

Since $v_{\infty} \ll c$ , this is $\ll L$ (so photons carry away much more energy) [1] [1]

[Also allow that as $v_{\infty} \to c$ , $E_{\mathrm{K,max}} \to$ ½L]

c. (i)

$L = 4\pi R^2 \sigma T^4$

$= 4\pi \times (1.41 \times 6.96 \times 10^{8})^2 \times 5.67 \times 10^{-8} \times (112000)^4$ [1]

$= 1.08 \times 10^{32}\ \mathrm{W} \quad (= 2.80 \times 10^{5}\ L_{\odot})$ [1]

$\dot{M}_{\mathrm{max}} = \frac{L}{v_{\infty} c} = \frac{1.08 \times 10^{32}}{1545 \times 10^{3} \times 3.00 \times 10^{8}} = 2.33 \times 10^{17}\ \mathrm{kg\,s^{-1}}$

$= 3.69 \times 10^{-6}\ M_{\odot}\ \mathrm{yr^{-1}}$ [1] [3]

[Award 2 max if they forget to convert the velocity into m s $^{-1}$ ]

c. (ii)

Radius = (2.5 arcmin in radians) × (distance in metres) (we're given the diameter)

$= \left(\frac{2.5}{60} \times \frac{\pi}{180}\right) \times (4.8 \times 10^{3} \times 3.09 \times 10^{16})$ [1]

$= 1.08 \times 10^{17}\ \mathrm{m} \quad (= 3.49\ \mathrm{pc})$ [1]

$t = \frac{0.55R}{v} = \frac{0.55 \times 1.08 \times 10^{17}}{30 \times 10^{3}} = 1.98 \times 10^{12}\ \mathrm{s} \quad (\approx 63000\ \mathrm{years})$ [1] [3]

[Award 2 max if they forget to convert the velocity into m s $^{-1}$ ]

c. (iii)

$R(t) = \left(\frac{3\dot{M} v_{\infty} t^2}{2\pi n_0 m_{\mathrm{H}}}\right)^{\frac{1}{4}} \therefore \dot{M} = \frac{2\pi n_0 m_{\mathrm{H}} R(t)^4}{3 v_{\infty} t^2}$

$= \frac{2\pi \times 16 \times 10^{6} \times 1.67 \times 10^{-27} \times (1.08 \times 10^{17})^4}{3 \times 1545 \times 10^{3} \times (1.98 \times 10^{12})^2}$ [1]

$= 1.25 \times 10^{18}\ \mathrm{kg\,s^{-1}} \quad (= 1.99 \times 10^{-5}\ M_{\odot}\ \mathrm{yr^{-1}})$ [1] [2]

This is (about 5.4 times) larger than the previously predicted value, suggesting the single scattering limit is not a good approximation in this situation [1] [1]

[In practice, multiple scattering occurs due to the Doppler Effect in the winds]

c. (iv)

Total mass, $M_{\mathrm{tot}} = \dot{M} t = 1.25 \times 10^{18} \times 1.98 \times 10^{12}$

$= 2.48 \times 10^{30}\ \mathrm{kg} \quad (= 1.25\ M_{\odot})$ [1] [1]

Total $E_{\mathrm{K,tot}} = \frac{1}{2} M_{\mathrm{tot}} v_{\infty}^2 = \frac{1}{2} \times 2.48 \times 10^{30} \times (1545 \times 10^{3})^2$

$= 2.96 \times 10^{42}\ \mathrm{J}$ [1] [1]

[This means Wolf Rayet stars are depositing a very large amount of kinetic energy into the ISM in only a few tens of thousands of years (in this case about 2.5% of the total energy the Sun will radiate in its lifetime!)]

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Royal Grammar School, Guildford)

Dr Emile Doran (The Langley Academy)

Sandor Kruk (University of Oxford)

Оригинал в Архива: BAAO_2018.pdf · официални решения: BAAO_2018_solutions.pdf