IOAA 2024, Теория — Задача 1. T1. Sundial

IOAA 2024 — Theory Examination, English (Official) · 10 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=2.3143×1023 mR=2.3143\times10^{23}\ \mathrm m and later substitutes 3.857×1023 m3.857\times10^{23}\ \mathrm m, while retaining its boxed 3.9×1044 kg≃2×1014M⊙3.9\times10^{44}\ \mathrm{kg}\simeq2\times10^{14}M_\odot. The main double-sum potential expression also omits the factor 1/21/2 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 Vsky=12.9V_{sky}=12.9 mag but substitutes 12.812.8 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⁡(ϕ)-\sin(a)\cos(\phi) where the preceding cosine-law equation would give −sin⁡(a)sin⁡(ϕ)-\sin(a)\sin(\phi). It is retained as printed; the subsequent sunrise calculation sets a=0a=0.

T10: Solution page 25 prints the unit of ∣M⃗∣2−R⊕2=1.288×1017|\vec M|^2-R_\oplus^2=1.288\times10^{17} as m rather than m2\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∘W35.7^\circ W, while solution (e) uses 36.4∘W36.4^\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
  1. The theoretical competition will be 5 hours in duration and is marked out of a total of 300 points.
  2. 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.
  3. 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.
  4. There is a summary Answer Sheet with your student ID code for your final answers.
  5. 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).
  6. 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.
  7. 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).
  8. 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.
  9. A list of constants for this competition is given on the next page.
Fundamental Constants
ConstantValue
Speed of light in a vacuumc=2.998×108 m s−1c=2.998\times10^8\ \mathrm{m\,s^{-1}}
Planck constanth=6.626×10−34 J sh=6.626\times10^{-34}\ \mathrm{J\,s}
Boltzmann constantkB=1.381×10−23 J K−1k_B=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}
Stefan-Boltzmann constantσ=5.670×10−8 W m−2 K−4\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}
Elementary chargee=1.602×10−19 Ce=1.602\times10^{-19}\ \mathrm C
Universal Gravitational constantG=6.674×10−11 N m2 kg−2G=6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}
Universal gas constantR=8.315 J mol−1 K−1R=8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}
Avogadro constantNA=6.022×1023 mol−1N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}
Wien's displacement lawλmT=2.898×10−3 m K\lambda_mT=2.898\times10^{-3}\ \mathrm{m\,K}
Mass of the electronme=9.11×10−31 kgm_e=9.11\times10^{-31}\ \mathrm{kg}
Mass of the protonmp=1.67×10−27 kgm_p=1.67\times10^{-27}\ \mathrm{kg}
Mass of the neutronmn=1.67×10−27 kgm_n=1.67\times10^{-27}\ \mathrm{kg}
Astronomical Data and Solar Magnitudes
Astronomical DataValue
1 parsec1 pc=3.086×1016 m=206 265 AU=3.262 ly1\ \mathrm{pc}=3.086\times10^{16}\ \mathrm m=206\,265\ \mathrm{AU}=3.262\ \mathrm{ly}
1 Astronomical Unit (AU)1 AU=1.496×1011 m1\ \mathrm{AU}=1.496\times10^{11}\ \mathrm m
1 Jansky1 Jy=10−26 W m−2 Hz−11\ \mathrm{Jy}=10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}
Hubble constantH0=70 km s−1 Mpc−1H_0=70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}
Solar luminosityL⊙=3.826×1026 WL_\odot=3.826\times10^{26}\ \mathrm W
Angular diameter of the Sun (from Earth)32′32'
Effective temperature of the SunTeff=5778 KT_{\mathrm{eff}}=5778\ \mathrm K
Obliquity of the ecliptic23.5∘23.5^\circ
Inclination of the Moon's orbit (to ecliptic)5∘08′43′′5^\circ08'43''
Magnitude of full Moon−12.74-12.74
Coordinates of North Ecliptic Pole(18h00m00s,+66∘33′39′′)(18^{\mathrm h}00^{\mathrm m}00^{\mathrm s},+66^\circ33'39'')
Coordinates of North Galactic Pole(12h51m26s,+27∘07′42′′)(12^{\mathrm h}51^{\mathrm m}26^{\mathrm s},+27^\circ07'42'')
Sidereal day23h56m04s23^{\mathrm h}56^{\mathrm m}04^{\mathrm s}
Tropical year365.2422365.2422 solar days
Sidereal year365.2564365.2564 solar days

Solar Magnitudes

Apparent VisualAbsolute VisualApparent BolometricAbsolute Bolometric
−26.75-26.75+4.82+4.82−26.83-26.83+4.74+4.74
Solar System, Spherical Trigonometry and Calculus

Solar System Data

ObjectRadius (km)Mass (kg)Semi-major axis (AU)Orbital eccentricityAlbedo
Sun6955001.988×10301.988\times10^{30}n/an/an/a
Mercury24403.301×10233.301\times10^{23}0.3870.2060.088
Venus60524.867×10244.867\times10^{24}0.7230.0070.76
Earth63785.972×10245.972\times10^{24}1.0000.0167100.31
Moon17377.346×10227.346\times10^{22}3.844×105 km3.844\times10^5\ \mathrm{km}0.0549000.11
Mars33906.417×10236.417\times10^{23}1.5240.0930.25
Jupiter699111.898×10271.898\times10^{27}5.2030.0480.51
Saturn582325.683×10265.683\times10^{26}9.5370.0540.34
Uranus253628.681×10258.681\times10^{25}19.1890.0470.30
Neptune246221.024×10261.024\times10^{26}30.0700.0090.29

Spherical Trigonometry

Cosine formula:

cos⁡a=cos⁡bcos⁡c+sin⁡bsin⁡ccos⁡A\cos a=\cos b\cos c+\sin b\sin c\cos A

Sine formula:

sin⁡asin⁡A=sin⁡bsin⁡B=sin⁡csin⁡C\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}

Four parts formula:

cot⁡bsin⁡a=cos⁡acos⁡C+sin⁡Ccot⁡B\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:

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

Product Rule:

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}
∫xr dx=xr+1r+1+C(r≠−1)\int x^r\,dx=\frac{x^{r+1}}{r+1}+C\qquad(r\ne-1)
ddxln⁡(x)=1x,ddxtan⁡(x)=sec⁡2(x)\frac{d}{dx}\ln(x)=\frac1x,\qquad\frac{d}{dx}\tan(x)=\sec^2(x)

Условие

The following diagram represents a sundial, for which the triangle, called a gnomon, casts a shadow onto the surrounding surface, which has markings and numbers representing important information. It is known that this sundial is located either between the Tropic of Cancer and the Arctic Circle or between the Tropic of Capricorn and the Antarctic Circle.

Sundial with complete gnomon, hour markings, A/B/C curves, analemma, north and south labels.

In the image above, the angle between the dashed and solid lines for any given time is always equal to the longitude difference between the time shown by the sundial and the civil time (the time shown on your watch). For instance, the dashed line corresponding to 7h and the solid line corresponding to 7h form an angle equal to the longitude difference between the location of this sundial and the central meridian of the time zone.

Throughout the year, the shadow of the tip of the gnomon is always between curves A and C.

Read the following statements and indicate whether they are true or false. For each item, write a T on the answer sheet if you think the statement is true and an F if you think the statement is false. There is no need to explain your answers.

(a) This sundial will only function properly if it is located in the southern hemisphere.

(b) Curve A represents the trajectory of the shadow of the tip of the gnomon throughout the winter solstice (in the hemisphere the sundial is located in).

(c) Line B represents the trajectory of the tip of the gnomon's shadow throughout the equinoxes.

(d) The solid radial lines provide the mean local solar time.

(e) The analemma shape around the dashed line corresponding to 12h shows the position of the tip of the gnomon's shadow during the true solar noon at the central meridian of the time zone throughout the year.

Решение

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.

CaseMarking
Using absurd physical or mathematical concepts (despite correct answers)No points given
Giving correct answer but the method is unclearDeduct 50% of the marks for final answer
Minor mistakes in the calculations, e.g., wrong signs, numerical constantsDeduct 20% of the marks for that part, unless there is a physical incoherence
Too many or too few significant figures (n±1n\pm1 where nn is the expected number of significant figures)Deduct 50% of the marks for final answer
Correct answer but missing or wrong unitsNo points given for final answer
For numerical values, the answer is not within a 5% precision rangeNo 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 reasonableNo double penalization, full points given
Error resulting from an earlier part error, if the answer is physically unreasonableNo points given

First example: If due to an error in an earlier item, the student calculates the speed of a satellite as 4.0×104 m/s4.0\times10^4\ \mathrm{m/s} instead of 4.5×104 m/s4.5\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 4.5×1012 m/s4.5\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 LL at a distance rr. 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