Съдържание

Earth Moon Sun Jupiter Saturn Physical constants Условие Решение

IOAA 2017, Теория — Задача 12. Shipwreck

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Всички материали от IOAA 2017

IOAA 2017 — Theoretical Examination · 40 т.

Original Examination Instructions

The theoretical examination lasts for 5 hours and is worth a total of 300 marks.
Dedicated IOAA Summary Answer Sheets are provided for writing your answers. Enter the final answers into the appropriate boxes in the corresponding Summary Answer Sheet. On each Summary Answer Sheet, please fill in
- Student Code (Country Code and 1 digit)
There are Answer Sheets for carrying out detailed work/rough work. On each Answer Sheet, please fill in
- Student Code (Country Code and 1 digit)
- Question no.
- Page no. and total number of pages.
Start each problem on a separate Answer Sheet. 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 evaluated, cross it out.
Use as many mathematical expressions as you think may help the evaluator to better understand your solutions. The evaluator may not understand your language. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your work desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, need more Answer Sheets, etc.), please draw the attention of the proctor using the Help card.
The beginning and end of the examination will be indicated by a long sound signal. Additionally, there will be a buzzer sound, fifteen minutes before the end of the examination (before the final sound signal).
At the end of the examination you must stop writing immediately. Sort and put your sheets in separate stacks, a) Stack 1: Summary Answer Sheets, Answer Sheets of part 1
b) Stack 2: Summary Answer Sheets, Answer Sheets of part 2
c) Stack 3: Summary Answer Sheets, Answer Sheets of part 3
d) Stack 4: question papers and paper sheets you do not want to be graded.
Wait at your table until your envelope is collected. Once all envelopes are collected, your student guide will escort you out of the examination room.
A list of constants and a table of the mark distribution for this exam are given on the next two pages.

Original List of Constants

Earth

Quantity	Value
Mass ( $M_\oplus$ )	$5.98\times10^{24}$ kg
Radius ( $R_\oplus$ )	$6.38\times10^6$ m
Acceleration of gravity ( $g$ )	$9.8 \mathrm{m\cdot s^{-1}}$
Obliquity of Ecliptic	$23^\circ27'$
Length of Tropical Year	365.2422 mean solar days
Length of Sidereal Year	365.2564 mean solar days
Albedo	0.39

Moon

Quantity	Value
Mass (M with subscript ☾)	$7.35\times10^{22}$ kg
Radius (R with subscript ☾)	$1.74\times10^6$ m
Mean distance from Earth	$3.84\times10^8$ m
Orbital inclination with the Ecliptic	$5.14^\circ$
Albedo	0.14
Apparent magnitude (mean full moon)	−12.74

Sun

Quantity	Value
Mass ( $M_\odot$ )	$1.99\times10^{30}$ kg
Radius ( $R_\odot$ )	$6.96\times10^8$ m
Luminosity ( $L_\odot$ )	$3.83\times10^{26}$ W
Absolute Magnitude ( $\mathcal M_\odot$ )	4.80 mag
Angular diameter	0.5 degrees
Rotational velocity in the Galaxy	$220 \mathrm{km\,s^{-1}}$
Distance from Galactic centre	8.5 kpc

Jupiter

Quantity	Value
Mass	$1.89\times10^{27}$ kg
Orbital semi-major axis	5.20 au
Orbital period	11.86 year

Saturn

Quantity	Value
Mass	$5.68\times10^{26}$ kg
Orbital semi-major axis	9.58 au
Orbital period	29.45 year

Physical constants

Quantity	Value
1 au	$1.50\times10^{11}$ m
1 pc	206,265 au
Gravitational constant ( $G$ )	$6.67\times10^{-11} \mathrm{N\cdot m^2\cdot kg^{-2}}$
Planck constant ( $h$ )	$6.62\times10^{-34} \mathrm{J\cdot s}$
Boltzmann constant ( $k_B$ )	$1.38\times10^{-23} \mathrm{J\cdot K^{-1}}$
Stefan-Boltzmann constant ( $\sigma$ )	$5.67\times10^{-8} \mathrm{W\cdot m^{-2}\cdot K^{-4}}$
Hubble constant ( $H_0$ )	$67.8 \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Speed of light in vacuum ( $c$ )	$299,792,458 \mathrm{m\cdot s^{-1}}$
Proton mass	$938.27 \mathrm{MeV\cdot c^{-2}}$
Deuterium mass	$1875.60 \mathrm{MeV\cdot c^{-2}}$
Neutron mass	$939.56 \mathrm{MeV\cdot c^{-2}}$
Helium-3 mass	$2808.30 \mathrm{MeV\cdot c^{-2}}$
Helium-4 mass	$3727.40 \mathrm{MeV\cdot c^{-2}}$

Original Mark Distribution

Problem number	Marks
T1	10
T2	10
T3	10
T4	10
T5	10
T6	15
T7	20
T8	20
T9	20
T10	25
T11	50
T12	40
T13	60
Total	300

Условие

You are shipwrecked on an island. Fortunately, you are still wearing a watch that is set to Bangkok time, and you also have a compass, an atlas and a calculator. You are initially unconscious, but wake up to find that it has recently become dark. Unfortunately, it is cloudy. An hour or so later you see Orion through a gap in the clouds. You estimate that the star “Rigel” is about $52.5^\circ$ above the horizon and with your compass you find that it has an astronomical azimuth of $109^\circ$ . Your watch says it is currently 01:00 on the 21st November 2017. You happen to remember from your astronomy class that Greenwich Sidereal Time (GST) at 00h UT 1st January 2017 is about 6h 43min and that the R.A. and Dec of Rigel are 5h 15min and $-8^\circ11'$ , respectively. Bangkok is in the UT+7 time zone.

a) Find the Local Hour Angle (LHA) of Rigel. [10 т.]

b) Find the current Greenwich sidereal time (GST). [10 т.]

c) Find the longitude of the island. [5 т.]

d) Find, to the nearest arcminute, the Latitude of the island. [15 т.]

Решение

a) Solution

Draw a correct celestial sphere. [4.0]

Complete original Shipwreck celestial sphere, both pole arrows, horizon/cardinal directions, Zenith/Rigel, all arcs and angular labels.

\frac{\sin\mathrm{LHA}^*}{\sin z}=\frac{\sin A}{\sin(90^\circ-\delta)}.

[2.0]

\mathrm{LHA}^*=\sin^{-1}\left(\frac{\sin A\sin z}{\sin(90^\circ-\delta)}\right)=\sin^{-1}\left(\frac{\sin109^\circ\sin37.5^\circ}{\sin98.18^\circ}\right).

[2.0]

\mathrm{LHA}^*=35.56^\circ=2\mathrm h\ 22\ \mathrm{min},

\mathrm{LHA}=24\mathrm h-2\mathrm h\ 22\ \mathrm{min}=21\mathrm h\ 38\ \mathrm{min}.

[2.0]

b) Solution

Bangkok time of 01:00 21st November is Universal Time (UT) 18:00 20th November.

Day number + Time = 323 day 18 h. [1.0]

GST = GST of 1st Jan + The angle that Υ moves away from 1st Jan. [2.0]

\mathrm{GST}=6\mathrm h\ 43\mathrm{min}+\left(323\frac{18}{24}\ \mathrm{days}\right)\left(\frac1{365.2422}\right)\times24\frac{\mathrm h}{\mathrm{day}}+\left[18\mathrm h\times\left(\frac{24}{23.9344}\right)\right].

[5.0]

GST = 6h 43min + 21h 16min + 18h 3min

GST = 46h 2min

GST = 21h 58min. [2.0]

Any other correct alternative methods that provide the correct answer also acceptable.

c) Solution

Complete original longitude/sidereal-time circles, meridians, Rigel, Aries and all GST/LST/RA/LHA/L arrows/labels.

24\mathrm h-\mathrm{LHA}=\mathrm{RA}-\mathrm{LST},

\mathrm{LST}=\mathrm{RA}-24\mathrm h+\mathrm{LHA},

L=\mathrm{RA}+\mathrm{LHA}-\mathrm{GST},

L=5\mathrm h\ 15\mathrm{min}+21\mathrm h\ 38\ \mathrm{min}-21\mathrm h\ 58\mathrm{min}=4\mathrm h\ 55\mathrm{min}.

[3.0]

L=73.75^\circ\mathrm E.

[2.0]

d) Solution

\cos(90^\circ-\delta)=\cos z\cos(90^\circ-\phi)+\sin z\sin(90^\circ-\phi)\cos A,

\sin\delta=\sin(h)\sin\phi+\cos(h)\cos A\cos\phi.

[5.0]

\alpha=\sin\delta,\quad\beta=\sin(h),\quad\gamma=\cos(h)\cos A,\quad x=\sin\phi,

\alpha=\beta x+\gamma\sqrt{1-x^2},

(\beta^2+\gamma^2)x^2-2\alpha\beta x+(\alpha^2-\gamma^2)=0,

[1.0]

x=\frac{\alpha\beta\pm\gamma\sqrt{\gamma^2+\beta^2-\alpha^2}}{\beta^2+\gamma^2},

[1.0]

\sin\phi=\frac{\sin\delta\sin(h)\pm\cos(h)\cos A\sqrt{(\cos(h)\cos A)^2+\sin^2(h)-\sin^2\delta}}{\sin^2(h)+(\cos(h)\cos A)^2}.

[1.0]

\phi=-24.05^\circ,\ 4.01^\circ.

[2.0]

For $\phi=-24.05^\circ\to A=71.0^\circ$ . [2.0]

For $\phi=4.01^\circ\to A=109^\circ$ . [2.0]

The Latitude of the island is $4^\circ1'\mathrm N$ . [1.0]

Alternative Solutions — b)

−LMT (Local Mean Time) → UT. [1.0]

Calculate sidereal day using given tropical year

\gamma=\frac{\text{Sidereal time}}{\text{Solar time}}=\frac{24^h\times\left(1+\frac1{T_{\mathrm{tropic}}}\right)}{24^h}=1+\frac1{T_{\mathrm{tropic}}}=1.002737909.

[3.0]

On the 21st November 2017, GST can be calculated from GST on 1st January 2017

Day number + Time = 323 d 18 h. [2.0]

\mathrm{GST}=\mathrm{GST}_0+\Delta\theta=\mathrm{GST}_0+\gamma\square\Delta T.

[2.0]

\Delta\theta=\left(323^d\frac{18}{24}\right)\square1.002737909=324^d.6363981=15^h16^m24^s.8,

\mathrm{GST}=\mathrm{GST}_0+15^h16^m24^s.8=6^h43^m+15^h16^m24^s.8=21^h59^m24^s.8.

[2.0]

Alternative Solutions — d)

From Cosine law:

\sin\delta=(\cos z)\square\sin\phi+(\sin z\cos A)\square\cos\phi,

\cos z=(\sin\delta)\square\sin\phi+(\cos\delta\cos\mathrm{LHA}^*)\square\cos\phi.

[5.0]

Correct calculation steps that can lead to the final answers. [5.0]

Final answers

\sin\phi=0.0698,\quad\phi=4.00^\circ

\cos\phi=0.9976,\quad\phi=4.00^\circ.

[5.0]

Source discrepancies retained: the main GST key prints 46h 2min followed by 21h 58min; the alternative gives 21h 59min 24.8s. Main latitude is 4°1′N and alternative latitude is 4.00°. The alternative page displays missing-glyph boxes between some factors; those boxes remain $\square$ .

Бележки за изданието

The original question and key contain several differing expressions. The Galactic-arm question uses a flat curve beyond 5 kpc and its worked key uses 2 kpc. The HI survey key prints differing flux-threshold exponents and an unusual Hubble-constant unit. The Bethlehem key has differing multiplication and remainder values. The Galactic Outflow key heading says T2 while its full prompt matches T9. GOTO assigns a different pair of marks to parts a) and d) in the question and key. The Shipwreck key prints differing sidereal-time values in its main and alternative methods. Visible missing-font glyph squares and the Exomoon key’s grading dependency are preserved. The original constants table prints the acceleration unit as m·s^-1. These printed forms remain literal.

Оригинал в Архива: Theory.pdf · официални решения: Theory_Solution.pdf