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Universal constants Astronomical data Calculus related formulas Условие Решение General Marking Principles

IOAA 2025, Теория — Задача 7. Neutron Star Binary

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Всички материали от IOAA 2025

IOAA 2025 — Theory Examination · 20 т.

Theory Examination Instructions

You are given an envelope marked B. Please verify the Contestant Code printed on it. Inside you will find:

This Instruction Sheet and/or This Instruction Sheet in your native language (if applicable)
Data Sheet and/or Data Sheet in your native language (if applicable)
A set of Working Sheets
One envelope marked S (please verify the Contestant Code marked on it).

You are allowed to open the envelope S only when the whistle is blown to mark the start of the examination.

Inside the envelope S you will find the following:

Question Paper and/or Question Paper in your native language (if applicable)
A set of Summary Answersheets.

Total duration of this examination is 5 hours.

This examination consists of the following questions:

Questions	Marks each	Subtotal
T01 and T02	10	$2\times10=20$ marks
T03 and T04	15	$2\times15=30$ marks
T05–T08	20	$4\times20=80$ marks
T09 and T10	35	$2\times35=70$ marks
T11 and T12	50	$2\times50=100$ marks
Total		300 marks

Writing your answers:

Summary Answersheet:

For each question a separate Summary Answersheet has been provided. Write your contestant code on each page of the Summary Answersheet.
Final answer(s) for each question / each part of a question must be written in the corresponding box in the Summary Answersheet.

Working Sheets:

Your detailed work must be shown on the Working Sheet.
Write your Contestant Code on both sides of each Working Sheet.
Start a new question on a fresh Working Sheet. Write the question number in the box provided at the top.
Both sides of the Working Sheet should be used. Write only inside the boxed area.
For rough work that you do not want to be evaluated, clearly strike that part out.
For each question write the page number in the box provided at the top of the Working Sheet, starting with 1 (each sheet has 2 pages). For example, if you use only 1 Working Sheet for T01, the question number and page number should be “T01, 1” and “T01, 2” on the FRONT and BACK of that Working Sheet, respectively. Next, if you use 2 Working Sheets for T02, the question number and page number should be “T02, 1” and “T02, 2” for the first Working Sheet, and “T02, 3” and “T02, 4” on the second Working Sheet.
You may ask for extra Working Sheets, if required.
Note that failure to mark the Working Sheets, as instructed above, may result in your answers not being evaluated.

At the end of the examination:

Check that you have correctly written your Contestant Code and page numbers on all used sheets.
Arrange your sheets in the following way: Summary Answersheet of each question (starting with T01) followed by Working Sheet(s) used for that question.
Similarly stack the Summary Answersheet and Working Sheet(s) used for each question, in sequence.
Place the bunch of all your used sheets arranged in the above manner inside the smaller envelope S. All Summary Answersheets must be placed inside the envelope S irrespective of whether you have used them or not.
Only the contents of the smaller envelope S will be evaluated. Any sheet that is not placed inside this envelope will not be evaluated.
Place the smaller envelope S inside the bigger envelope B.
All other sheets, namely, Instruction Sheet(s), Question Paper(s), Data Sheet(s) and unused Working Sheet(s) should be placed directly inside the bigger envelope B.
Tie the thread seal on the envelope B.
Hand over the envelope B to the examiner. Make sure that no sheet is left on the table, except the green and the red card.

General Instructions:

If you need a visit to the toilet during the examination, please raise your hand holding the green card kept on your table. You will be escorted by a volunteer.
If you need any other help during the examination, please raise your hand holding the red card kept on your table.
Use only black or blue pen for writing. For drawing, you may use a pencil.
Some marks may be deducted if the final answer is given without detailed solution.
Some marks may be deducted if the final answers have improper number of significant digits, no units or wrong units.
The necessary values of fundamental and astronomical constants should be taken from the Data Sheet provided to you.

Original IOAA 2025 Data Sheet

Universal constants

Quantity	Symbol	Value
Avogadro constant	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Charge of electron	$e$	$1.602\times10^{-19}\ \mathrm C$
Planck constant	$h$	$6.626\times10^{-34}\ \mathrm{J\,s}$
Speed of light in vacuum	$c$	$2.998\times10^8\ \mathrm{m\,s^{-1}}$
Universal gravitational constant	$G$	$6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Universal gas constant	$R$	$8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Wien’s displacement constant	$b$	$2.898\times10^{-3}\ \mathrm{m\,K}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}\ \mathrm{m^{-3}\,kg^{-1}\,s^4\,A^2}$
Permeability of free space	$\mu_0$	$1.257\times10^{-6}\ \mathrm{N\,A^{-2}}$
Mass of electron	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
Mass of proton	$m_p$	$1.673\times10^{-27}\ \mathrm{kg}=938.272\ \mathrm{MeV}/c^2$
Mass of neutron	$m_n$	$1.675\times10^{-27}\ \mathrm{kg}=939.565\ \mathrm{MeV}/c^2$
Mass of deuteron	$m_D$	$3.344\times10^{-27}\ \mathrm{kg}=1875.613\ \mathrm{MeV}/c^2$
Mass of He nucleus	$m_{\mathrm{He}}$	$6.645\times10^{-27}\ \mathrm{kg}=3727.181\ \mathrm{MeV}/c^2$

Astronomical data

Quantity	Symbol	Value
Mass of Sun	$M_\odot$	$1.988\times10^{30}\ \mathrm{kg}$
Radius of Sun	$R_\odot$	$6.957\times10^8\ \mathrm m$
Luminosity of Sun	$L_\odot$	$3.828\times10^{26}\ \mathrm W$
Effective temperature of Sun	$T_{\mathrm{eff},\odot}$	$5772\ \mathrm K$
Apparent magnitude of Sun (in V-band)	$m_{V,\odot}$	$-26.74$
Absolute magnitude of Sun (in V-band)	$M_{V,\odot}$	$+4.82$
Apparent bolometric magnitude of Sun	$m_{\mathrm{bol},\odot}$	$-26.83$
Absolute bolometric magnitude of Sun	$M_{\mathrm{bol},\odot}$	$+4.74$
Solar constant (above atmosphere of Earth)	$S_\odot$	$1361\ \mathrm{W\,m^{-2}}$
Apparent angular diameter of Sun (from Earth)	$\theta_\odot$	$\approx32'$
Mass of Earth	$M_\oplus$	$5.972\times10^{24}\ \mathrm{kg}$
Radius of Earth	$R_\oplus$	$6.378\times10^6\ \mathrm m$
Axial tilt of Earth	$\epsilon$	$23^\circ26'$
Inclination of the lunar orbit w.r.t. the ecliptic		$5^\circ8'43''$
Mass of Jupiter	$M_J$	$1.898\times10^{27}\ \mathrm{kg}$
Radius of Jupiter	$R_J$	$6.991\times10^7\ \mathrm m$
1 Astronomical Unit	1 au	$1.496\times10^{11}\ \mathrm m$
1 parsec	1 pc	$3.086\times10^{16}\ \mathrm m$
1 light-year	1 ly	$9.461\times10^{15}\ \mathrm m$
1 jansky	1 Jy	$10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
1 tropical year		$365.2422$ solar days $=3.156\times10^7\ \mathrm s$ ; $365\ \mathrm d\ 5\ \mathrm h\ 48\ \mathrm{min}\ 46\ \mathrm s$
1 sidereal year		$365.2564$ solar days $=3.156\times10^7\ \mathrm s$ ; $365\ \mathrm d\ 6\ \mathrm h\ 9\ \mathrm{min}\ 13\ \mathrm s$
Rate of precession of Vernal Equinox		$1^\circ$ per $71.6$ years

Calculus related formulas

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ .
$\frac{d}{dx}x^n=nx^{n-1}$ .
$\frac{d}{dx}\sin kx=k\cos kx$ .
$\frac{d}{dx}\cos kx=-k\sin kx$ .
$\frac{d}{dx}\tan kx=k\sec^2kx$ .
$\int x^n\,dx=\frac{x^{n+1}}{n+1}+\text{constant}$ ; for $n\ne-1$ .
$f(x)\simeq f(x_0)+\left.\frac{df}{dx}\right|_{x=x_0}(x-x_0)$ ; for $x\approx x_0$ .

Условие

In a binary system involving a compact star, where the companion star does not overflow its Roche lobe, the primary source of accretion for the compact star is the stellar wind from the companion star. This wind-fed accretion is especially significant in systems that include an early-type star (such as an O or B star, indicated henceforth as an OB star), alongside a compact object like a neutron star (NS) in a close orbit.

Consider such a NS-OB star binary system where a neutron star of mass $M_{\mathrm{NS}}=2.0\,M_\odot$ and radius $R_{\mathrm{NS}}=11\,\mathrm{km}$ is orbiting in a circular orbit of radius $a$ around the centre of the OB star with velocity $v_{\mathrm{orb}}=1.5\times10^5\,\mathrm{m\,s^{-1}}$ (see figure below). Throughout this problem the mass loss from the OB star is assumed to be spherically symmetric and its rate is $\dot M_{\mathrm{OB}}=1.0\times10^{-4}\,M_\odot\,\mathrm{yr^{-1}}$ .

Complete original OB/neutron-star wind geometry, orbital and wind arrows and separation.

(T07.1) The accretion radius, $R_{\mathrm{acc}}$ , is defined as the maximum distance from the NS at which the stellar wind can be captured by the NS. If the stellar wind speed at the orbital distance of the NS is $v_{\mathrm w}=3.0\times10^6\,\mathrm{m\,s^{-1}}$ , find $R_{\mathrm{acc}}$ for the above system in km using standard escape velocity calculations. [3 т.]

(T07.2) Assuming that all captured material is accreted by the NS, estimate the mass accretion rate, $\dot M_{\mathrm{acc}}$ , from the stellar wind onto the NS in units of $M_\odot\,\mathrm{yr^{-1}}$ if $a=0.5\,\mathrm{au}$ . Neglect the effects of radiation pressure and finite cooling time of the accreting gas. [3 т.]

(T07.3) Now consider the situation where the stellar wind speed at the orbital distance $a$ (near the NS) becomes comparable with the orbital speed of the NS. The mass accretion rate from the stellar wind onto the NS in this case would be given by an expression of the form $\dot M_{\mathrm{acc}}=\dot M_{\mathrm{OB}}f(\tan\beta,q)$ , where $q=M_{\mathrm{NS}}/M_{\mathrm{OB}}$ is the mass ratio of the binary and $\beta$ is the angle in the frame of the NS between the wind velocity direction and radial direction away from the OB star. Obtain the expression for $f(\tan\beta,q)$ assuming $M_{\mathrm{OB}}\gg M_{\mathrm{NS}}$ . [6 т.]

(T07.4) Consider that the fully ionized material accretes radially and is hindered by the strong magnetic field $\vec B$ of the NS. The pressure exerted by the magnetic field is given by $\dfrac{B^2}{2\mu_0}$ . We shall assume that the NS has a dipole magnetic field whose magnitude in the equatorial plane varies with the distance $r$ from the NS for $r\gg R_{\mathrm{NS}}$ as

B(r)=B_0\left(\frac{R_{\mathrm{NS}}}{r}\right)^3

where $B_0$ is the magnetic field at the equator of the NS. Assume that the axis of the magnetic dipole aligns with the rotation axis of the NS.

(T07.4a) Obtain the magnetic pressure, $P_{\mathrm{eq,mag}}$ , in the equatorial plane in terms of $B_0$ , $R_{\mathrm{NS}}$ , $r$ , and other suitable constants. [1 т.]

(T07.4b) The maximum distance where the accretion flow is stopped by the magnetic field at the equatorial plane is called the magnetospheric radius $R_m$ . This flow of matter will exert a pressure due to the relative motion between incoming stellar wind and the NS. Obtain an approximate expression for the critical magnetic field $B_{0,c}$ for which $R_m$ coincides with $R_{\mathrm{acc}}$ and calculate its value in Tesla. Magnetic effects are neglected for $r>R_m$ and consider $v_{\mathrm w}\gg v_{\mathrm{orb}}$ . [7 т.]

Complete original Summary Answersheet T07, SA-13, FRONT: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Complete original Summary Answersheet T07, SA-14, BACK: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Решение

(T07.1)

The accretion radius can be estimated by equating the gravitational potential energy of the stellar wind to its kinetic energy

\frac12\rho_wv_\mathrm{rel}^2=\frac{GM_\mathrm{NS}\rho_w}{R_\mathrm{acc}}\qquad\text{(0.5)},

\implies R_\mathrm{acc}=\frac{2GM_\mathrm{NS}}{v_\mathrm{rel}^2}\qquad\text{(0.5)},

where, $\rho_w$ is the wind density and $v_\mathrm{rel}$ is the relative velocity of the wind. For the case considered above,

v_\mathrm{rel}^2=v_w^2+v_\mathrm{orb}^2.

Here, $v_w\gg v_\mathrm{orb}$ , therefore, $v_\mathrm{rel}^2\approx v_w^2=9\times10^{12}\,\mathrm{m^2\,s^{-2}}$ (1.0). which gives,

R_\mathrm{acc}=\frac{2\times6.674\times10^{-11}\times2.0\times1.988\times10^{30}}{9\times10^{12}}\,\mathrm{m},

\boxed{R_\mathrm{acc}=5.9\times10^4\,\mathrm{km}}\qquad\text{(1.0)}.

(T07.2)

In the case of spherically symmetric mass loss from the companion and $v_w\gg v_\mathrm{orb}$ , the companion mass loss is

\dot M_\mathrm{OB}=4\pi a^2\rho_wv_w\qquad\text{(1.0)},

where $a=0.5\,\mathrm{au}=7.480\times10^{10}\,\mathrm{m}$ is the orbital separation. Mass accretion rate on the compact object is

\dot M_\mathrm{acc}\approx\pi R_\mathrm{acc}^2\rho_wv_\mathrm{rel}\qquad\text{(1.0)}.

Also here, since $v_w\gg v_\mathrm{orb}$ , $v_\mathrm{rel}\approx v_w$ . This implies,

\dot M_\mathrm{acc}\approx\left(\frac{\dot M_\mathrm{OB}}4\right)\left(\frac{R_\mathrm{acc}}a\right)^2\approx9.74\times10^{11}\,\mathrm{kg\,s^{-1}},

\implies\boxed{\dot M_\mathrm{acc}\approx1.6\times10^{-11}M_\odot\,\mathrm{yr^{-1}}}\qquad\text{(1.0)}.

If the student does not use the approximation for speed, then he will get $9.75\times10^{11}\,\mathrm{kg\,s^{-1}}$ , but the final answer is same for required significant digits. No marks should be deducted if this is the case.

(T07.3)

The orbital speed of the NS in the frame corresponding to the center of OB star (assuming the star is stationary) is given by

v_\mathrm{orb}^2=\frac{GM_\mathrm{OB}}a\qquad\text{(1.0)}.

Using vector algebra we get $\tan\beta=v_\mathrm{orb}/v_w$ (1.0). Now, the ratio of the mass accretion rate to that of mass loss rate from the star

\frac{\dot M_\mathrm{acc}}{\dot M_\mathrm{OB}}\sim\frac{\pi R_\mathrm{acc}^2\rho_wv_\mathrm{rel}}{4\pi a^2\rho_wv_w}

=\frac14\left(\frac{R_\mathrm{acc}}a\right)^2\sqrt{\frac{v_\mathrm{orb}^2+v_w^2}{v_w^2}}=\frac14\left(\frac{R_\mathrm{acc}}a\right)^2\sqrt{1+\tan^2\beta}\qquad\text{(1.0)}.

To find the expression of the ratio of the two radii, $R_\mathrm{acc}$ and orbital separation $a$ ,

\frac{R_\mathrm{acc}}a=\frac{2GM_\mathrm{NS}}{v_\mathrm{rel}^2}\frac1a=\frac{2GM_\mathrm{OB}q}{v_\mathrm{rel}^2}\frac1a=\frac{2v_\mathrm{orb}^2}{v_\mathrm{rel}^2}q\qquad\text{(1.0)}

=\left(\frac{2\tan^2\beta}{1+\tan^2\beta}\right)q\qquad\text{(1.0)}.

On substituting and simplifying, we can express,

\implies\boxed{f(\tan\beta,q)=q^2\left(\frac{\tan^4\beta}{(1+\tan^2\beta)^{3/2}}\right)}\qquad\text{(1.0)}.

(T07.4a)

The magnetic pressure $P_\mathrm{mag}=B^2/2\mu_0$ increases rapidly towards the NS magnetic poles.

\implies\boxed{P_{\mathrm{eq,mag}}=\frac{B_0^2R_\mathrm{NS}^6}{2\mu_0r^6}}\qquad\text{(1.0)}.

(T07.4b)

The critical magnetic field $B_{0,c}$ can be obtained by equating pressure due to incoming wind with the $P_{\mathrm{eq,mag}}$ at $r=R_\mathrm{acc}$ (1.0). The pressure due to incoming wind at $r=R_\mathrm{acc}$ is given by

\rho_wv_\mathrm{rel}^2\sim\rho_wv_w^2\qquad\text{(1.0)}.

If the student writes the pressure due to incoming wind as dynamic pressure, i.e., $\frac12\rho_wv_\mathrm{rel}^2\sim\frac12\rho_wv_w^2$ , 0.5 credit will be given.

From part (T07.2), one can obtain the expression for $\rho_w$ as follows:

\rho_w=\frac{\dot M_\mathrm{acc}}{\pi R_\mathrm{acc}^2v_w}=\frac{\dot M_\mathrm{OB}}{4\pi a^2v_w}\qquad\text{(1.0)}.

If the student writes the density of wind as $\dot M_\mathrm{acc}/(4\pi R_\mathrm{acc}^2v_w)$ , 0.5 credit will be given.

Now equating the pressures due to wind and magnetic field, we have:

\frac{B_{0,c}^2R_\mathrm{NS}^6}{2\mu_0R_\mathrm{acc}^6}=\frac{\dot M_\mathrm{acc}}{\pi R_\mathrm{acc}^2}\left(\frac{2GM_\mathrm{NS}}{R_\mathrm{acc}}\right)^{1/2}\qquad\text{(1.0)},

which gives the value of the critical magnetic field as

\implies\boxed{B_{0,c}=\left(\frac{2\mu_0\dot M_\mathrm{acc}\sqrt{2GM_\mathrm{NS}}}{\pi R_\mathrm{NS}^6}\right)^{1/2}R_\mathrm{acc}^{7/4}}\qquad\text{(1.0)}.

No credit will be given if the expression has wrong exponent, or is missing a constant quantity.

Now putting the values we get $\implies\boxed{B_{0,c}\approx4.0\times10^9\,\mathrm{T}}$ (2.0).

General Marking Principles

STRICT adherence to given marking scheme.
No double penalty – deduction only at the original place of error; no further penalty in consequent steps, if they are otherwise correct – if final answer is physically unviable or dimensionally incorrect, deduct for final answer also.
Significant figures: one more or one less than appropriate no. of significant figures allowed with full credit.
Units / dimensions:
- In cases where units are not mentioned in the Summary Answersheet, 0.5 deduction for missing units.
- For missing units on the same kind of quantities in a given part, deduct only once.
- For wrong dimensions in final answer because of some error in earlier steps, no credit for final answer, irrespective of no-double penalty principle.
If expression is asked in terms of x, y, …: deduction of final part credit if expression is not in terms of appropriate quantities.
Cumulative part marking: all previous part credits to be added if final expression / number is correct (unless otherwise specified).
Final answer not in Summary Answersheet, but present inside: if final answer is present on Working Sheet, but not transferred on Summary Answersheet, it is to be credited.

Бележки за изданието

The printed question, key and blank answer sheets contain differing expressions and units. The Balmer key uses different rounded colour excesses; the quasar scale changes its angular unit; the neutron-star key prints differing mass-rate units/values. The black-hole-shadow question prints an unsquared radial speed in its comparison equation, while the repeated key uses a square. Atmospheric Seeing asks for a diameter but its key describes a radius. The nucleosynthesis key changes strict inequalities, has a differently grouped decay exponent, and includes an extra time-arrow field also present on its answer sheet. The stellar-profile key omits a factor in one line that appears in the next. The Hawking key prints a separation step differing from its integral and a horizon volume differing from its prose radius. Its answer sheet labels a temperature field with “s”. These native forms are preserved, including every original blank answer page.

Оригинал в Архива: Theory-Questions.pdf · официални решения: Theory-Solution.pdf · допълнителни данни: Theory-Summary-Answersheet.pdf · допълнителни данни: Theory-Instructions.pdf · допълнителни данни: Data_Sheet.pdf · допълнителни данни: IOAA2025_General_Marking_Scheme.pdf