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

IOAA 2025, Теория — Задача 1. Daksha Mission

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Всички материали от IOAA 2025

IOAA 2025 — Theory Examination · 10 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

“Daksha” is a proposed Indian mission consisting of two satellites $S_1$ and $S_2$ orbiting the Earth in the same circular orbit of radius $r=7000\,\mathrm{km}$ but with $180^\circ$ phase difference. These satellites observe the universe in high energy domain (X-rays and $\gamma$ -rays). Each of the satellites of Daksha uses several flat, rectangular detectors.

To understand how to localize a source in the sky, we shall use a simplified model of the Daksha mission. Assume that $S_1$ has only two identical detectors $D_1$ and $D_2$ , each of area $A=0.50\,\mathrm{m^2}$ , attached on an opaque mount $M$ as shown in the figure below. The detectors lie symmetrically around the $y$ axis in planes perpendicular to the $x$ - $y$ plane and make an angle $\alpha=120^\circ$ with each other.

Complete original detector mount, both detectors, axes and incoming photon arrows.

(T01.1) When observing a distant source located in the $x$ - $y$ plane, detector $D_1$ records a power $P_1=2.70\times10^{-10}\,\mathrm{J\,s^{-1}}$ and detector $D_2$ records a power $P_2=4.70\times10^{-10}\,\mathrm{J\,s^{-1}}$ . Estimate the angle $\eta$ made by the position vector of the source with the positive $y$ -axis, with counter-clockwise angle from the positive $y$ axis being considered positive. [5 т.]

Consider a single pulse from a distant source (not necessarily in the $x$ - $y$ plane) recorded by both satellites ( $S_1$ and $S_2$ ) of Daksha. The times of the peaks of the pulses recorded by $S_1$ and $S_2$ are $t_1$ and $t_2$ , respectively.

(T01.2) If $t_1-t_2$ was measured to be $10.0\pm0.1\,\mathrm{ms}$ then determine the fraction, $f$ , of the celestial sphere where the source might lie. [5 т.]

Complete original Summary Answersheet T01, SA-1, FRONT: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Complete original Summary Answersheet T01, SA-2, BACK: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Решение

(T01.1)

Since the two detectors make an angle with each other, photons from the source will not be incident on the detectors at the same angle. Using this the source can be localized.

Let $\widehat n_1$ be the unit vector perpendicular to the plane of D₁ and $\widehat n_2$ be the unit vector perpendicular to the plane of D₂. We have,

\widehat n_1=\frac12\widehat x+\frac{\sqrt3}2\widehat y\qquad\text{(0.5)}

\widehat n_2=-\frac12\widehat x+\frac{\sqrt3}2\widehat y\qquad\text{(0.5)}

Let $\widehat s=s_x\widehat x+s_y\widehat y$ be the unit vector in the direction of the source and $F$ be the flux of the source. The effective area of capture for a detector will be

A_\mathrm{eff}=A(|\widehat s\cdot\widehat n|)\qquad\text{(0.5)}.

Hence, the power received is,

P=AF(|\widehat s\cdot\widehat n|)\qquad\text{(1.0)}

and therefore,

P_1=AF(|\widehat s\cdot\widehat n_1|),\qquad P_2=AF(|\widehat s\cdot\widehat n_2|).

Original detector/source-normal effective-area construction, full arrows and angle.

Hence,

\widehat s\cdot\widehat n_1=\frac12s_x+\frac{\sqrt3}2s_y=\frac{2.70\times10^{-10}}{AF},

\widehat s\cdot\widehat n_2=-\frac12s_x+\frac{\sqrt3}2s_y=\frac{4.70\times10^{-10}}{AF}.

Solving,

s_x=-\frac{2.00\times10^{-10}}{AF}\qquad\text{(0.5)},\qquad s_y=\frac{7.40\times10^{-10}}{\sqrt3 AF}\qquad\text{(0.5)}.

Hence, the angle the direction vector makes with respect to the positive $y$ -axis is

\eta=\tan^{-1}\left(-\frac{s_x}{s_y}\right)\qquad\text{(1.0)}.

Then, $\boxed{\eta=25.1^\circ=0.438\,\mathrm{rad}}$ (0.5).

Half credit lower limit	Full credit range	Half credit upper limit
–	24.9 to 25.3	–

The sign on $\eta$ carries 0.5 marks.

(T01.2)

The time delay occurs because the satellites S₁ and S₂ are located at different distances from the source. Light takes longer time to reach the satellite that is farther (here, say S₁). Let $\Delta t=t_1-t_2$ .

Complete original time-delay geometry, both detectors, source rays and all lengths/angles.

If the direction of the source makes an angle $\theta$ (see the figure above) with the line joining telescopes S₁ and S₂ we have the path difference ( $\Delta d$ ) given as,

\Delta d=c\Delta t\qquad\text{(0.5)}.

From the figure above the path difference can be written as,

\Delta d=2r\cos\theta\qquad\text{(1.0)}.

Therefore,

\cos\theta=\frac{c\Delta t}{2r}.

Since the measurement of $\Delta t$ has some uncertainty, that is $\Delta t\in[9.9,10.1]\,\mathrm{ms}$ , we get a range $[\theta_1,\theta_2]$ for the possible values of $\theta$ .

\theta_1=\cos^{-1}\left(\frac{c(10.1\,\mathrm{ms})}{2r}\right)=77.51^\circ\qquad\text{(0.5)},

\theta_2=\cos^{-1}\left(\frac{c(9.9\,\mathrm{ms})}{2r}\right)=77.76^\circ\qquad\text{(0.5)}.

Original full sky annulus geometry, sphere, both coloured circles and angular labels.

The source may lie anywhere in the annulus bounded by the red and blue circles on the unit sphere above, whose area is given by

2\pi(\cos\theta_1-\cos\theta_2)\qquad\text{(1.5)}.

Hence the fraction of sky where the source might lie is

f=\frac12(\cos\theta_1-\cos\theta_2)\qquad\text{(0.5)},

\boxed{f=2.13\times10^{-3}}\qquad\text{(0.5)}.

Half credit lower limit	Full credit range	Half credit upper limit
–	$2.12\times10^{-3}$ to $2.15\times10^{-3}$	–

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