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BAAO 2019, Теория — Задача 3. Stellar Mass Limits

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British Astronomy and Astrophysics Olympiad 2018-2019 — Astronomy & Astrophysics Competition Paper — Monday 21st January 2019 · 21 януари 2019 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 26th 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 August 2019; 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 IOAA this year will be held in Zanka, Hungary, from 2nd to 11th August 2019.

The best students taking this paper that are 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, (Saturday 13th April to Wednesday 17th April 2019). 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. A team of 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	$1.50 \times 10^{11}\ \mathrm{m}$
Semi-major axis of the Earth’s orbit		1 au
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	$m_{\mathrm{e}}$	$9.11 \times 10^{-31}\ \mathrm{kg}$

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

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 2018/19 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. 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

In the heart of every star, nuclear fusion is taking place. For most stars that involves hydrogen being turned into helium, a process that starts by bringing two protons close enough that the strong nuclear force can act upon them. The smallest stars are the ones that have a core that is only just hot enough for fusion to occur, whilst in the biggest ones the radiation pressure of the photons given out by the fusion reaction pushing on the stellar material can overcome the gravitational forces holding it together.

Left: artist's size comparison of Jupiter (gas giant), Saturn (gas giant), the orange star EBLM J0555-57Ab (low mass star) and TRAPPIST-1 (low mass star). Right: Hubble colour image of the Tarantula Nebula with a red arrow pointing to the bright star cluster containing R136a1. — Figure 5: Left: The lowest mass star we know of, EBLM J0555-57Ab, was found by von Boetticher et al. (2017) and is about the size of Saturn with a mass of $0.081\,M_{\odot}$ . Credit: Amanda Smith, University of Cambridge. Right: The highest mass star we know of, R136a1, is in the centre of the clump of stars on the right of this HST image of the Tarantula Nebula. Schneider et al. (2014) suggest it has a mass of $315\,M_{\odot}$ , which is above what stellar evolution models allow. Despite its large mass, other stars have far bigger radii. Credit: NASA & ESA.

For a spherical main sequence star made of a plasma (a fully ionized gas of electrons and nuclei) that is acting like an ideal gas, the temperature at the core can be approximately calculated as

T_{\mathrm{int}} \simeq \frac{GM\bar{\mu}}{k_{\mathrm{B}}R} \quad \text{where} \quad \bar{\mu} = \frac{m_{\mathrm{p}}}{2X + 3Y/4 + Z/2}\,.

In this equation, $M$ is the mass of the star, $R$ is its radius, $k_{\mathrm{B}}$ is the Boltzmann constant, and $\bar{\mu}$ is the mean mass of the plasma particles (i.e nuclei and electrons) with $m_{\mathrm{p}}$ the mass of a proton.

a. Given the Sun’s composition has hydrogen fraction, $X = 0.72$ , helium fraction $Y = 0.26$ and ‘metals’ (i.e. any element lithium and heavier) fraction $Z = 0.02$ , estimate the temperature at the centre of the Sun. [2 т.]

b. Classically, two protons need to have enough energy to overcome their electrostatic repulsion in order to fuse. Assuming the kinetic energy of each proton, $E_K$ , and the electric potential energy between two protons, $E_P$ , are

E_K = \frac{3}{2}k_{\mathrm{B}}T \quad \text{and} \quad E_P = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{b}\,,

where $e$ is the fundamental charge and $b$ is the distance between the protons, calculate the value of $T_{\mathrm{classical}}$ necessary to allow fusion to occur, given that at that point $b = 1$ fm (= $10^{-15}$ m). [You should find that it’s much larger than your answer to part a.] [4 т.]

Classically, the core of the Sun is not hot enough for fusion, and yet fusion is clearly happening. The key is that it is a fundamentally quantum process, and so protons are able to ‘quantum tunnel’ through the Coloumb barrier (see Figure 6), allowing fusion to occur at lower temperatures. In quantum mechanics, fusion will happen when $b = \lambda$ where $\lambda$ is the de Broglie wavelength of the proton, related to the momentum of the proton by $\lambda = h/p$ .

Red curve of a potential barrier against r, with a turquoise wave function: oscillating on the left (Wave function inside λ = h/p), decreasing exponentially inside the barrier (Exponential decrease inside barrier), and oscillating with small amplitude on the right (Wave function outside λ = h/p); a brace under the barrier is labelled Quantum tunnel. — Figure 6: A diagram showing the way a particle can pass through a classically impenetrable potential barrier due to its wave-like properties on the quantum scale. Credit: Brooks/Cole - Thomson Learning.

c. Given that the proton momentum is related to the average kinetic energy of a particle in the plasma by $E_K = p^2/2m_{\mathrm{p}}$ calculate the value of $\lambda$ and hence calculate $T_{\mathrm{quantum}}$ . [You should find that it’s below your answer to part a.] [6 т.]

In the smallest stars, electron degeneracy prevents them from compressing in radius and thus stops the core reaching $T_{\mathrm{int}} \gtrsim T_{\mathrm{quantum}}$ . At the limit of electron degeneracy, the number density of electrons $n_{\mathrm{e}} = 1/\lambda_{\mathrm{e}}^3$ where $\lambda_{\mathrm{e}}$ is the de Broglie wavelength of the electrons.

d. Assuming the star to be of uniform density at this limit with $\rho = m_{\mathrm{p}}n_{\mathrm{e}}$ and the electrons to be in thermal equilibrium with the plasma, show that the minimum mass of a star for which $T_{\mathrm{int}} = T_{\mathrm{quantum}}$ is $\sim 0.1\,M_{\odot}$ . [7 т.]

In the largest stars, radiation pressure pushes on the outer layers of the star stronger than gravity pulls them in. The brightest luminosity for a star is known as the Eddington luminosity, $L_{\mathrm{Edd}}$ . The acceleration due to radiation pressure can be calculated as

g_{\mathrm{rad}} = \frac{\kappa_{\mathrm{e}}I}{c} \quad \text{where} \quad \kappa_{\mathrm{e}} = \frac{\sigma_{\mathrm{T}}}{2m_{\mathrm{p}}}(1 + X)

and $\kappa_{\mathrm{e}}$ is the electron opacity of the stellar material, $\sigma_{\mathrm{T}}$ is the Thomson scattering cross-section for electrons ( $= 66.5\ \mathrm{fm}^2$ ), $X$ is the hydrogen fraction, and $I$ is the intensity of radiation (in W m $^{-2}$ ). Assuming main-sequence stars follow a mass-luminosity relation of $L \propto M^3$ , the maximum mass of a star can be found by considering one that is radiating at $L_{\mathrm{Edd}}$ .

e. By balancing the radiative acceleration with the gravitational acceleration at the surface of a star, derive a formula for $L_{\mathrm{Edd}}$ in terms of $M$ , and hence calculate the maximum mass of a star with a hydrogen fraction like the Sun. Give your answer in $M_{\odot}$ . [6 т.]

Отговори

Решение

Q3 – Stellar Mass Limits [25 marks]

$\bar{\mu} = \frac{m_p}{2X+3Y/4+Z/2} = \frac{1.67\times10^{-27}}{(2\times0.72)+(3\times0.26/4)+0.02/2} = 1.02\times10^{-27}\ \mathrm{kg}\ (= 0.608\,m_p)$ [1]

$T_{int} = \frac{GM_{\odot}\bar{\mu}}{k_BR_{\odot}} = \frac{6.67\times10^{-11}\times1.99\times10^{30}\times1.02\times10^{27}}{1.38\times10^{-23}\times6.96\times10^{8}} = 1.40\times10^{7}\ \mathrm{K}$ [1] [2]

Assuming both protons are heading straight towards each other,

$\text{Total } E_K = 2\times\frac{3}{2}k_BT_{classical}$ [1]

Equating with the potential energy at the point where they can fuse,

$\text{Total } E_K = E_P \therefore T_{classical} = \frac{1}{4\pi\epsilon_0}\frac{e^2}{b}\frac{1}{2\times\frac{3}{2}k_B}$ [1]

$= \frac{1}{4\pi\times8.85\times10^{-12}}\frac{(1.60\times10^{-19})^2}{10^{-15}}\frac{1}{3\times1.38\times10^{-23}}$ [1]

$= 5.56\times10^{9}\ \mathrm{K}$ [1] [4]

[Assuming one proton is stationary, resulting in $T_{\mathrm{classical}} = 1.11\times10^{10}$ K, gives 3 marks max]

$\lambda = \frac{h}{p} \therefore p = \frac{h}{\lambda}$ and $2\times E_K = 2\times\frac{p^2}{2m_p} = E_P = \frac{1}{4\pi\epsilon_0}\frac{e^2}{\lambda}$

$\therefore \frac{h^2}{\lambda^2m_p} = \frac{1}{4\pi\epsilon_0}\frac{e^2}{\lambda} \quad \therefore \quad \lambda = \frac{4\pi\epsilon_0h^2}{m_pe^2}$ [1]

$= \frac{4\pi\times8.85\times10^{-12}\times(6.63\times10^{-34})^2}{1.67\times10^{-27}\times(1.60\times10^{-19})^2}$ [1]

$= 1.14\times10^{-12}\ \mathrm{m}$ [1] [3]

[First mark is for eliminating p and equating with the expression for $E_{\mathrm{P}}$ . Allow full ecf for this calculation if assuming the stationary proton case, leading to $\lambda = 5.72\times10^{-13}$ m]

$\frac{3}{2}k_BT_{quantum} = \frac{h^2}{2\lambda^2m_p} \therefore T_{quantum} = \frac{h^2}{3\lambda^2m_pk_B}$ [1]

$= \frac{(6.63\times10^{-34})^2}{3\times(1.14\times10^{-12})^2\times1.67\times10^{-27}\times1.38\times10^{-23}}$ [1]

$= 4.86\times10^{6}\ \mathrm{K}$ [1] [3]

[First mark is for an expression for $T_{\mathrm{quantum}}$ in terms of known variables. In the stationary proton case you get $1.95\times10^{7}$ K, which is above the value in part a. and so only 2 marks max for this calculation]

[Despite our simplifying assumptions, this is close to the real value of the temperature needed in a star like to Sun to undergo hydrogen fusion, which is $\approx 3\times10^{6}$ K. Deuterium fusion is possible at lower temperatures since each molecule has more mass and hence more momentum at a given temperature, so some very small stars (called brown dwarfs) achieve the required temperature of $\approx 4.5\times10^{5}$ K. The ability to fuse for even a short time is how large gas giants and small stars are distinguished, and is investigated further in the next part of the question]

$M_{min} = \frac{4}{3}\pi R^3\rho = \frac{4}{3}\pi R^3m_pn_e = \frac{4}{3}\pi R^3\frac{m_p}{\lambda_e^3} \quad \therefore R = \lambda_e\left(\frac{M_{min}}{\frac{4}{3}\pi m_p}\right)^{1/3}$ [1]

$\frac{p_e^2}{2m_e} = \frac{3}{2}k_BT_{quantum} \therefore p_e = \sqrt{3m_ek_BT_q} \quad (= 1.35\times10^{-23}\ \mathrm{kg\,m\,s^{-1}})$ [1]

$\therefore \lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{3m_ek_BT_q}} \quad (= 4.90\times10^{-11}\ \mathrm{m})$ [1]

$T_{int} = \frac{GM_{min}\bar{\mu}}{k_BR} \therefore M_{min} = \frac{T_qk_BR}{G\bar{\mu}} = \frac{T_qk_B}{G\bar{\mu}}\frac{h}{\sqrt{3m_ek_BT_q}}\left(\frac{M_{min}}{\frac{4}{3}\pi m_p}\right)^{1/3}$

$\therefore M_{min}^{2/3} = \frac{\sqrt{T_qk_B}}{G\bar{\mu}}\frac{h}{\sqrt{3m_e}}\left(\frac{4}{3}\pi m_p\right)^{-1/3}$ [1]

$= \frac{\sqrt{4.86\times10^{6}\times1.38\times10^{-23}}}{6.67\times10^{-11}\times1.02\times10^{-27}}\frac{6.63\times10^{-34}}{\sqrt{3\times9.11\times10^{-31}}}\left(\frac{4}{3}\pi\times1.67\times10^{-27}\right)^{-1/3}$ [1]

$= 2.54\times10^{19}\ \mathrm{kg}^{2/3}$ [1]

$\therefore M_{min} = 1.28\times10^{29}\ \mathrm{kg} = 0.064\,M_{\odot}$ [Must be in $M_{\odot}$ ] [1] [7]

[If students assume the star is a fully ionised pure hydrogen plasma, such that $\bar{\mu} = 0.5\,m_p$ and leading to $M_{\mathrm{min}} = 0.086\ M_{\odot}$ give full marks (this was the intended answer for this question, but the value of $\bar{\mu}$ to choose was ambiguous so most students will use their value from part a.) If they use the value of $T_{\mathrm{quantum}}$ from the stationary proton case, getting $M_{\mathrm{min}} = 0.18\ M_{\odot}$ , give 6 marks max]

[The real lower limit for a star to join the main sequence is about 0.08 $M_{\odot}$ , or about 83 times the mass of Jupiter. In practice, the theoretical lower limit varies with the metallicity of the star – for a star like the Sun it is 0.075 $M_{\odot}$ , whilst for a pure hydrogen star it is 0.092 $M_{\odot}$ (most stars we can see have a metallicity that falls between these two)]

Balancing the radiative acceleration term with the gravitational one,

$g_{rad} = g_{grav} \therefore \frac{\kappa_eI}{c} = \frac{GM}{R^2}$ [1]

$I = \frac{L_{Edd}}{4\pi R^2} \quad \therefore L_{Edd} = \frac{4\pi GMc}{\kappa_e} \quad \left(= \frac{8\pi GMcm_p}{\sigma_T(1+X)}\right)$ [1] [2]

[ $L_{\mathrm{Edd}}$ can be given in terms of $\kappa$ or in terms of $\sigma_{\mathrm{T}}$ for the second mark]

$\kappa_e = \frac{\sigma_T}{2m_p}(1+X) = \frac{66.5\times10^{-30}}{2\times1.67\times10^{-27}}(1+0.72) = 0.03425\ \mathrm{m^2\,kg^{-1}}$ [1]

Using the given mass-luminosity relation,

$\frac{L_{Edd}}{L_{\odot}} = \left(\frac{M_{max}}{M_{\odot}}\right)^3 \therefore \frac{4\pi GM_{max}c}{L_{\odot}\kappa_e} = \frac{M_{max}^3}{M_{\odot}^3} \therefore M_{max} = \sqrt{\frac{4\pi GcM_{\odot}^3}{L_{\odot}\kappa_e}}$ [1]

$= \sqrt{\frac{4\pi\times6.67\times10^{-11}\times3.00\times10^{8}\times(1.99\times10^{30})^3}{3.85\times10^{26}\times0.03425}}$ [1]

$= 3.88\times10^{32}\ \mathrm{kg} = 195\,M_{\odot}$ [1] [4]

[The answer must be in $M_{\odot}$ for the final mark. Full marks can be awarded without $\kappa$ being calculated explicitly. Watch for incorrect conversions from $\mathrm{fm}^2$ to $\mathrm{m}^2$ ( $1\ \mathrm{fm}^2 = 10^{-30}\ \mathrm{m}^2$ ) with $\sigma_{\mathrm{T}}$ ]

[The real upper limit on stars is about 200 $M_{\odot}$ , so this simplified model has done rather well]

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Royal Grammar School, Guildford)

John Hayton (University of Cambridge)

Josh Brown (University of Cambridge)

Оригинал в Архива: BAAO_2019.pdf · официални решения: BAAO_2019_solutions.pdf