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IOAA 2024, Теория — Задача 7. T7. Castaway

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Всички материали от IOAA 2024

IOAA 2024 — Theory Examination, English (Official) · 20 т.

Внимание: Бележка към темата

The printed instructions refer to blank Detailed Worksheets and a summary Answer Sheet. Those forms are absent from the available 2024 archive packet. The complete question booklet, solution booklet and separate marking scheme are included. Printed source inconsistencies are identified in the relevant tasks and retained without silently changing the source.

T2: The solution first gives $R=2.3143\times10^{23}\ \mathrm m$ and later substitutes $3.857\times10^{23}\ \mathrm m$ , while retaining its boxed $3.9\times10^{44}\ \mathrm{kg}\simeq2\times10^{14}M_\odot$ . The main double-sum potential expression also omits the factor $1/2$ used for unique pairs in the appendix. These printed expressions are retained; the marking scheme allows valid alternative estimates.

T6: The question names Palomar 4; the solution's repeated question names Palomar 6. The solution gives $V_{sky}=12.9$ mag but substitutes $12.8$ in the following surface-brightness line. The native question and printed calculation are retained.

T7: The printed expansion on solution page 13 has $-\sin(a)\cos(\phi)$ where the preceding cosine-law equation would give $-\sin(a)\sin(\phi)$ . It is retained as printed; the subsequent sunrise calculation sets $a=0$ .

T10: Solution page 25 prints the unit of $|\vec M|^2-R_\oplus^2=1.288\times10^{17}$ as m rather than $\mathrm m^2$ . In (j), the final vector prints 77.59 m/s after previously using 77.76 m/s; the magnitude uses 77.76 m/s. These printed values are retained.

T11: The question gives Maceió's longitude as $35.7^\circ W$ , while solution (e) uses $36.4^\circ W$ . The solution's printed margin marks total 11 points in (g), versus 12 points in the question and separate marking scheme; its (j) heading and marks total 7 points, versus 6 in the question and separate marking scheme. All three sources are retained with their original point values.

Common solar-system table: The native solar-system table labels the semi-major-axis column AU, while explicitly giving the Moon's entry in km; its printed unit is retained.

Указания

The theoretical competition will be 5 hours in duration and is marked out of a total of 300 points.
There are Detailed Worksheets for carrying out detailed work / rough work. On each of the Detailed Worksheets, please fill in Student Code, Question No., Page no. and total number of pages.
Start each problem on a new page of the Detailed Worksheets. Please write only on the printed side of the sheet. Do not use the reverse side. If you have written something on any sheet which you do not want to be marked, cross it out.
There is a summary Answer Sheet with your student ID code for your final answers.
Please remember that the graders may not understand your language. As far as possible, write your solutions only using mathematical expressions and numbers. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your exam desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, need more Detailed Worksheets, etc.), please put up your hand to signal the invigilator.
The beginning and end of the competition will be indicated by a long sound signal. Additionally, there will be a short sound signal fifteen minutes before the end of the competition (before the final long sound signal).
Wait at your table until your envelope is collected. Once all envelopes are collected, your student guide will escort you out of the competition room.
A list of constants for this competition is given on the next page.

Fundamental Constants

Constant	Value
Speed of light in a vacuum	$c=2.998\times10^8\ \mathrm{m\,s^{-1}}$
Planck constant	$h=6.626\times10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e=1.602\times10^{-19}\ \mathrm C$
Universal Gravitational constant	$G=6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Universal gas constant	$R=8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}$
Wien's displacement law	$\lambda_mT=2.898\times10^{-3}\ \mathrm{m\,K}$
Mass of the electron	$m_e=9.11\times10^{-31}\ \mathrm{kg}$
Mass of the proton	$m_p=1.67\times10^{-27}\ \mathrm{kg}$
Mass of the neutron	$m_n=1.67\times10^{-27}\ \mathrm{kg}$

Astronomical Data and Solar Magnitudes

Astronomical Data	Value
1 parsec	$1\ \mathrm{pc}=3.086\times10^{16}\ \mathrm m=206\,265\ \mathrm{AU}=3.262\ \mathrm{ly}$
1 Astronomical Unit (AU)	$1\ \mathrm{AU}=1.496\times10^{11}\ \mathrm m$
1 Jansky	$1\ \mathrm{Jy}=10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
Hubble constant	$H_0=70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Solar luminosity	$L_\odot=3.826\times10^{26}\ \mathrm W$
Angular diameter of the Sun (from Earth)	$32'$
Effective temperature of the Sun	$T_{\mathrm{eff}}=5778\ \mathrm K$
Obliquity of the ecliptic	$23.5^\circ$
Inclination of the Moon's orbit (to ecliptic)	$5^\circ08'43''$
Magnitude of full Moon	$-12.74$
Coordinates of North Ecliptic Pole	$(18^{\mathrm h}00^{\mathrm m}00^{\mathrm s},+66^\circ33'39'')$
Coordinates of North Galactic Pole	$(12^{\mathrm h}51^{\mathrm m}26^{\mathrm s},+27^\circ07'42'')$
Sidereal day	$23^{\mathrm h}56^{\mathrm m}04^{\mathrm s}$
Tropical year	$365.2422$ solar days
Sidereal year	$365.2564$ solar days

Solar Magnitudes

Apparent Visual	Absolute Visual	Apparent Bolometric	Absolute Bolometric
$-26.75$	$+4.82$	$-26.83$	$+4.74$

Solar System, Spherical Trigonometry and Calculus

Solar System Data

Object	Radius (km)	Mass (kg)	Semi-major axis (AU)	Orbital eccentricity	Albedo
Sun	695500	$1.988\times10^{30}$	n/a	n/a	n/a
Mercury	2440	$3.301\times10^{23}$	0.387	0.206	0.088
Venus	6052	$4.867\times10^{24}$	0.723	0.007	0.76
Earth	6378	$5.972\times10^{24}$	1.000	0.016710	0.31
Moon	1737	$7.346\times10^{22}$	$3.844\times10^5\ \mathrm{km}$	0.054900	0.11
Mars	3390	$6.417\times10^{23}$	1.524	0.093	0.25
Jupiter	69911	$1.898\times10^{27}$	5.203	0.048	0.51
Saturn	58232	$5.683\times10^{26}$	9.537	0.054	0.34
Uranus	25362	$8.681\times10^{25}$	19.189	0.047	0.30
Neptune	24622	$1.024\times10^{26}$	30.070	0.009	0.29

Spherical Trigonometry

Cosine formula:

\cos a=\cos b\cos c+\sin b\sin c\cos A

Sine formula:

\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}

Four parts formula:

\cot b\sin a=\cos a\cos C+\sin C\cot B

Native common spherical-trigonometry reference sphere and triangle A, B, C with sides a, b, c.

Calculus

Chain Rule:

\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

Product Rule:

\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}

\int x^r\,dx=\frac{x^{r+1}}{r+1}+C\qquad(r\ne-1)

\frac{d}{dx}\ln(x)=\frac1x,\qquad\frac{d}{dx}\tan(x)=\sec^2(x)

Условие

After surviving a shipwreck and reaching a small island in the southern hemisphere, a sailor had to estimate the island's latitude using the Sun.

However, due to poor eyesight, the sailor couldn't see the night stars very well, so his best option was to rely on the Sun. He had no information about the date, but he realized the days were longer than the nights.

(a) The sailor noticed that on his first day on the island, the angle between the positions of sunrise and sunset on the horizon was $120^{\circ}$ . With this piece of information, determine the range of possible values for the latitude of the island. Neglect the daily motion of the Sun across the ecliptic. [7 т.]

(b) The angle between the positions of sunrise and sunset kept increasing daily. After 40 days, this angle was equal to $163^{\circ}$ . Estimate the latitude of the island. You may neglect the eccentricity of the Earth's orbit. [13 т.]

Решение

Solution:

The first step is to find an expression for the azimuth of an object during sunrise:

Spherical triangle with Zenith and Celestial South Pole, altitude, azimuth, latitude and declination labels.

Using the law of cosines:

\cos(90^{\circ} + \delta) = \cos(90^{\circ} - a)\cos(90^{\circ} + \phi) + \sin(90^{\circ} - a)\sin(90^{\circ} + \phi)\cos(180^{\circ} - A)

\therefore -\sin(\delta) = -\sin(a)\cos(\phi) - \cos(a)\cos(\phi)\cos(A)

Since the altitude $a = 0^{\circ}$ at the sunrise:

\cos(\phi) = \frac{\sin(\delta)}{\cos(A)}

[2.0]

Note that due to symmetry over the local meridian, the azimuth of sunrise corresponds to $180^{\circ}$ minus half of the angle given in the question statement (the $180^{\circ}$ term comes from the fact that the observer is on the southern hemisphere).
The scenarios that lead to the minimum and maximum values of latitude correspond to the extreme values for the declination of the Sun in the southern hemisphere summer ( $-23.5^{\circ}$ and $0^{\circ}$ ).
If the declination of the Sun is $-23.5^{\circ}$ , the latitude has to be

\cos(\phi) = \frac{\sin(-23.5^{\circ})}{\cos\left(180^{\circ} - \frac{120^{\circ}}{2}\right)}

\cos(\phi) = -2\sin(-23.5^{\circ})

\phi = -37.1^{\circ}

[2.0]

Only the negative solution of this equation is relevant in this case since the observer is in the southern hemisphere.
If the declination is the Sun is $0^{\circ}$ , simply plugging in this value to the formula will result in a latitude of $-90^{\circ}$ . Since celestial objects do not rise or set at the poles, this approach would not be conceptually correct. However, at a point infinitesimally close to the pole, the Sun would still mathematically rise and set, and it would be possible to achieve the difference in azimuth from the problem statement with a declination very close to $0^{\circ}$ . Thus, although a latitude of $-90^{\circ}$ is impossible, any latitude infinitesimally close to it would still mathematically be possible.
Therefore, the range of possible latitudes for the island is the following:

\boxed{-90^{\circ} < \phi \leqslant -37.1^{\circ}}

[3.0]

(b) The angle between positions of the sunrise and the sunset kept increasing daily. After 40 days, this angle was equal to $163^{\circ}$ . Estimate the latitude of the island. (13 points)

Solution:

Since the latitude is constant, the following expression must be true:

\cos(\phi) = \frac{\sin(\delta_0)}{\cos(A_0)} = \frac{\sin(\delta_{40})}{\cos(A_{40})}

[2.0]

It is also possible to apply the law of sines to the following triangle to derive another expression for the declination of the Sun on each day:

Celestial equator, ecliptic and Vernal Point with two solar declinations, two orbital angles and obliquity. [1.0]

\frac{\sin(\theta_0)}{\sin(90^{\circ})} = \frac{\sin(-\delta_0)}{\sin(\epsilon)}

\sin(\delta_0) = -\sin(\theta_0)\sin(\epsilon)

Analogously:

\frac{\sin(\theta_{40})}{\sin(90^{\circ})} = \frac{\sin(-\delta_{40})}{\sin(\epsilon)}

\sin(\delta_{40}) = -\sin(\theta_{40})\sin(\epsilon)

[1.0]

Combining the three previous expressions:

\sin(\delta_0) = \frac{\sin(\delta_{40})\cos(A_0)}{\cos(A_{40})}

-\sin(\theta_0)\sin(\epsilon) = \frac{-\sin(\theta_{40})\sin(\epsilon)\cos(A_0)}{\cos(A_{40})}

\sin(\theta_0) = \frac{\sin(\theta_{40})\cos(A_0)}{\cos(A_{40})}

\sin(\theta_0) = k\ \sin(\theta_{40})

[2.0]

Where $k = \frac{\cos(A_0)}{\cos(A_{40})} = \frac{\cos(120^{\circ})}{\cos(98.5^{\circ})} = 3.38$ .
In order to replace one variable and solve for $\theta_0$ , it is possible to consider that the angular velocity of the Sun on the Ecliptic is approximately constant due to the Earth's very low eccentricity. In that case, the following expression must be true:

\theta_{40} = \theta_0 - \frac{40}{T_{year}} \times 360^{\circ}

\theta_{40} = \theta_0 - \beta

[2.0]

Where $\beta = \frac{40}{T_{year}} \times 360^{\circ} = 39.4^{\circ}$ . Combining the expressions and solving for $\theta_0$ :

\sin(\theta_0) = k\ \sin(\theta_0 - \beta)

\sin(\theta_0) = k\ \left(\sin(\theta_0)\cos(\beta) - \sin(\beta)\cos(\theta_0)\right)

\sin(\theta_0)(k\cos(\beta) - 1) = k\sin(\beta)\cos(\theta_0)

\therefore \tan(\theta_0) = \frac{k\sin(\beta)}{k\cos(\beta) - 1} = \frac{3.38\sin(39.4^{\circ})}{3.38\cos(39.4^{\circ}) - 1}

\theta_0 = 53.1^{\circ}

[3.0]

Solving for $\delta_0$ :

\sin(\delta_0) = -\sin(\theta_0)\sin(\epsilon)

\delta_0 = -18.6^{\circ}

[1.0]

Solving for the latitude of the island:

\cos(\phi) = \frac{\sin(\delta_0)}{\cos(A_0)} = \frac{\sin(-18.6^{\circ})}{\cos\left(180^{\circ} - \frac{120^{\circ}}{2}\right)}

\boxed{\phi = -50.4^{\circ}}

[1.0]

Note that only the negative solution to the equation above is relevant since it is known that the island is in the southern hemisphere.

Separate Marking Scheme — T7

(a)

2 points: deriving an expression for the latitude as a function of the declination and the azimuth. Note that the students don’t necessarily need to derive an expression for the azimuth during the sunrise and the sunset. They should also get full marks on this criterion if they simply write down the expression for the azimuth and proceed from there.
2 points: determining the correct lower bound for the latitude of the island. Students who identify that the lower bound corresponds to the Southernmost declination of the Sun but fail to determine the latitude will receive 1 point.
3 points: determining the correct upper bound for the latitude of the island. Students who identify that the upper bound corresponds to a $0^\circ$ declination but fail to determine the latitude will receive 1.5 point.

(b)

2 points: noticing that the ratio between the sine of the declination and the cosine of the azimuth remains constant for a given latitude.
1 point: correctly drawing or describing the geometry of the situation.
1 point: finding an expression for the declination on day 40 as a function of $\theta$ on day 40 and the obliquity of the ecliptic.
2 points: finding an expression for $\theta$ on day 0 as a function of $\theta$ on day 40.
2 points: finding a second expression that relates $\theta$ on day 0 and $\theta$ on day 40.
3 points: combining the equations to find the value of $\theta_0$ .
1 point: finding the value of $\delta_0$
1 point: correct final answer.

General Rules for Marking

These are general guidelines for the grading of all problems. All of the rules listed below refer to deductions on specific items in the marking scheme.

Case	Marking
Using absurd physical or mathematical concepts (despite correct answers)	No points given
Giving correct answer but the method is unclear	Deduct 50% of the marks for final answer
Minor mistakes in the calculations, e.g., wrong signs, numerical constants	Deduct 20% of the marks for that part, unless there is a physical incoherence
Too many or too few significant figures ( $n\pm1$ where $n$ is the expected number of significant figures)	Deduct 50% of the marks for final answer
Correct answer but missing or wrong units	No points given for final answer
For numerical values, the answer is not within a 5% precision range	No points given for final answer, unless differently stated in the problem or marking scheme
Error resulting from an earlier part error, for which the student already lost marks, if the answer is physically reasonable	No double penalization, full points given
Error resulting from an earlier part error, if the answer is physically unreasonable	No points given

First example: If due to an error in an earlier item, the student calculates the speed of a satellite as $4.0\times10^4\ \mathrm{m/s}$ instead of $4.5\times10^4\ \mathrm{m/s}$ , they will only lose marks for the earlier item. However, if, for the same reason, a student calculates the velocity of the satellite as $4.5\times10^{12}\ \mathrm{m/s}$ , they should realize this is wrong (higher than the speed of light) and thus should lose some marks for this item as well.

Second example: Consider a question that requires a student to determine the temperature of a planet that orbits a star with luminosity $L$ at a distance $r$ . To answer correctly, the student must calculate the star's radiant flux, the power incident on the planet, the power radiated by the planet, and then, using the condition of thermal equilibrium, the planet's temperature. If the student only calculates the radiant flux and arrives at the correct final answer without clearly explaining the other steps, they will receive marks only for the flux calculation, but not for the other steps, neither for the final answer.

Оригинал в Архива: theory_examination_304b6eb66a8940d79fccc404cf1ba9ab_preview.pdf · официални решения: theory solutions.pdf · допълнителни данни: theory marking scheme.pdf