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

IOAA 2025, Теория — Задача 6. Galactic Rotation

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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

The rotation curve of our Galaxy is determined using line-of-sight velocity measurements of neutral hydrogen (HI) clouds along various Galactic longitudes, observed through the 21 cm HI line.

Consider an HI cloud with Galactic longitude $l$ , located at a distance $R$ from the Galactic Centre (GC) and a distance $D$ from the Sun. Consider Sun to be at a distance $R_0=8.5\,\mathrm{kpc}$ from the GC. Assume that both the Sun and the HI cloud are in circular orbits around the GC in the Galactic plane, with angular velocities $\Omega_0$ and $\Omega$ , and rotational velocities $V_0$ and $V$ , respectively.

The line-of-sight velocity ( $V_r$ ) and transverse velocity ( $V_t$ ) components of the cloud, as observed from the Sun, can be expressed as

V_r=(\Omega-\Omega_0)R_0\sin l

V_t=(\Omega-\Omega_0)R_0\cos l-\Omega D

Seen from the North Galactic Pole, the Galactic rotation is clockwise. Throughout this problem, we shall take line-of-sight velocity to be positive when receding and clouds will be treated as point objects.

(T06.1) In the graph provided on the Summary Answersheet, sketch $V_r$ as a function of $D$ from $D=0$ to $D=2R_0$ for two lines of sight defined by (i) $l=45^\circ$ and (ii) $l=135^\circ$ . Label each of your lines/curves with the value of $l$ . [5 т.]

(T06.2) The graph below shows the average radial (solid, red curve) and transverse (dashed, blue curve) velocity components of stars at a distance of 100 pc from the Sun, plotted as a function of Galactic longitude.

Original red radial and blue dashed tangential velocity curves, all longitude axes and labels.

Using the graph, estimate the Sun’s orbital period ( $P$ ) around the GC in mega-years (Myr). [3 т.]

(T06.3) Jan Oort noted that in the solar neighbourhood ( $D\ll R_0$ ), the difference in angular velocities ( $\Omega-\Omega_0$ ) will be small, and hence, derived the following first order approximation for the line-of-sight and the transverse velocity components:

V_r=AD\sin2l

V_t=AD\cos2l+BD

where $A$ and $B$ are known as Oort’s constants.

Let us consider two cases:

(I) the actual observed rotation curve of the Galaxy, and

(II) the rotation curve is for a hypothetical scenario where the Galaxy is devoid of dark matter and the whole mass of the Galaxy is assumed to be concentrated at its centre.

(T06.3a) Derive expressions for the radial gradient of the rotational velocity at the location of the Sun, $\left.\dfrac{dV}{dR}\right|_{R=R_0}$ for the two cases. [2 т.]

(T06.3b) Express $A$ and $B$ in terms of $V_0$ , $R_0$ , and the radial gradient of rotational velocity at the location of the Sun, $\left.\dfrac{dV}{dR}\right|_{R=R_0}$ . [8 т.]

(T06.3c) The ratio ( $A/B$ ) of Oort’s constants for the two given cases, (I) and (II), are defined as $F_\mathrm I$ and $F_\mathrm{II}$ , respectively. Determine $F_\mathrm I$ and $F_\mathrm{II}$ . [2 т.]

Complete original Summary Answersheet T06, SA-11, FRONT: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Complete original Summary Answersheet T06, SA-12, BACK: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Решение

(T06.1)

Complete original galactic longitude/orbit geometry, Galactic Centre, Sun and all rays/radii.

The concept of differential rotation of the Galaxy is used here.

For $l=45^\circ$ :

From Sun to the GC, the quantity $(\Omega-\Omega_0)$ increases, which implies $V_r$ increases.
As shown in the figure, the line of sight intersects circular orbits that are at varying distances from the GC.
The line of sight is closest to the GC when $D=R_0\cos l=6.01\,\mathrm{kpc}$ . $V_r$ reaches the maximum value here.
Beyond this, $(\Omega-\Omega_0)$ decreases implying $V_r$ decreases.
$V_r=0$ when the line of sight intersects the Sun’s orbit. Here, $D=2R_0\cos l=12.02\,\mathrm{kpc}$ .
Beyond the Sun’s orbit, $\Omega<\Omega_0$ . $V_r$ is negative and its magnitude increases with increasing $D$ .

$l=135^\circ$ :

The line of sight lies outside the Sun’s orbit. Here, $V_r$ is negative, there is no peak, and its magnitude increases with increasing $D$ .

The plot of the line-of-sight velocity as a function of the distance from the Sun for two values of $l$ is as follows:

Complete original handwritten radial-velocity curves for both longitudes, axes, zero crossings and full grid.

For $l=45^\circ$ :

	Marks
Starts at (0,0)	0.5
Has one and only one positive maximum	0.5
Maximum is at $D=6\,\mathrm{kpc}$	0.5
Has a zero crossing at $D=12\,\mathrm{kpc}$	1.0
Is negative throughout beyond zero crossing	0.5
Is smooth throughout	0.5

For $l=135^\circ$ :

	Marks
Starts at (0,0)	0.5
Is negative throughout without any maxima/minima	0.5
Is smooth throughout	0.5

(T06.2)

Consider $l=90$ ,

V_r=0\ \text{implying}\ \Omega=\Omega_0\qquad\text{(0.5)},

V_t=-\Omega_0D=-2.75\,\mathrm{km\,s^{-1}}\qquad\text{(0.5)},

\Omega_0=8.91\times10^{-16}\,\mathrm{rad\,s^{-1}}\qquad\text{(1.0)},

P=\frac{2\pi}{\Omega_0}=\boxed{224\,\mathrm{Myr}}\qquad\text{(1.0)}.

	Marks
Expression of $V_r$ for a chosen $l$	0.5
Expression of $V_t$ for a chosen $l$	0.5
Calculation of $\Omega_0$ for a chosen $l$ (acceptable range $8.8$ to $9.1\times10^{-16}\,\mathrm{rad\,s^{-1}}$ )	1.0
Calculation of $P$ for a chosen $l$	1.0

Half credit lower limit	Full credit range	Half credit upper limit
	$218\,\mathrm{Myr}$ to $228\,\mathrm{Myr}$

(T06.3a)

(I) The observed rotation curve is flat near the Sun.

V\ \text{is constant}\ \to\ \boxed{\frac{dV}{dR}\bigg|_{R=R_0}=0}\qquad\text{(1.0)}.

(II) If the whole mass of the Galaxy is assumed to be concentrated at the centre, then we consider Keplerian motion.

V=\sqrt{\frac{GM}R}\ \to\ \boxed{\frac{dV}{dR}\bigg|_{R=R_0}=-\frac12\frac{V_0}{R_0}}\qquad\text{(1.0)}.

(T06.3b)

In the solar neighbourhood, $(\Omega-\Omega_0)$ will be very small. Using the approximation

f(x)\simeq f(x_0)+\frac{df}{dx}\bigg|_{x=x_0}(x-x_0),\quad\text{for}\ x\approx x_0,

given in the Data Sheet,

\Omega-\Omega_0=\frac{d\Omega}{dR}\bigg|_{R=R_0}(R-R_0)\qquad\text{(2.0)}

=\left[\left(\frac1R\frac{dV}{dR}\right)\bigg|_{R=R_0}-\left(\frac1{R^2}V\right)\bigg|_{R=R_0}\right](R-R_0)

=\frac1{R_0^2}\left[R_0\frac{dV}{dR}\bigg|_{R=R_0}-V_0\right](R-R_0).

Thus the line-of-sight velocity expression becomes

V_r=\frac1{R_0^2}\left[R_0\frac{dV}{dR}\bigg|_{R=R_0}-V_0\right](R-R_0)R_0\sin l.

Further, in the solar neighbourhood, $D\ll R_0$ . So we can approximate

R-R_0\approx-D\cos l\qquad\text{(1.0)}.

Thus, the line-of-sight velocity can be expressed as,

V_r=\frac1{R_0^2}\left[R_0\frac{dV}{dR}\bigg|_{R=R_0}-V_0\right](-D\cos l)R_0\sin l

=\frac12\left[\frac{V_0}{R_0}-\frac{dV}{dR}\bigg|_{R=R_0}\right]D\sin2l.

Comparing with $V_r=AD\sin2l$ , we get

\boxed{A=\frac12\left[\frac{V_0}{R_0}-\frac{dV}{dR}\bigg|_{R=R_0}\right]}\qquad\text{(2.0)}.

Also, in the solar neighbourhood, $\Omega D\approx\Omega_0D$ . Using this approximation and $(R-R_0)\approx-D\cos l$ , the tangential velocity is (1.0)

V_t=(\Omega-\Omega_0)R_0\cos l-\Omega D

=\frac1{R_0^2}\left[R_0\frac{dV}{dR}\bigg|_{R=R_0}-V_0\right](-D\cos l)R_0\cos l-\Omega_0D

=\left[\frac{V_0}{R_0}-\frac{dV}{dR}\bigg|_{R=R_0}\right]D\cos^2l-\Omega_0D

=\frac12\left[\frac{V_0}{R_0}-\frac{dV}{dR}\bigg|_{R=R_0}\right]D(1+\cos2l)-\frac{V_0}{R_0}D

=AD\cos2l+\frac12\left[\frac{V_0}{R_0}-\frac{dV}{dR}\bigg|_{R=R_0}\right]D-\frac{V_0}{R_0}D

=AD\cos2l-\frac12\left[\frac{V_0}{R_0}+\frac{dV}{dR}\bigg|_{R=R_0}\right]D.

Comparing with $V_t=AD\cos2l+BD$ , we get

\boxed{B=-\frac12\left[\frac{V_0}{R_0}+\frac{dV}{dR}\bigg|_{R=R_0}\right]}\qquad\text{(2.0)}.

Main credit points:

	Marks
$(\Omega-\Omega_0)$ first order approximation in terms of Taylor series	2.0
$(R-R_0)\approx-D\cos l$ approximation	1.0
$\Omega D\approx\Omega_0D$ approximation	1.0
Working through to arrive at correct expression for $A$	2.0
Working through to arrive at correct expression for $B$	2.0

(T06.3c)

For a flat rotation curve $(dV/dR=0)$ , $\boxed{F_\mathrm{I}=-1}$ (1.0).

For Keplerian rotation $(dV/dR=-\frac12 V/R)$ , $\boxed{F_\mathrm{II}=-3}$ (1.0).

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