Съдържание

Условие Part 1: Eclipse Geometry (10 т.)Part 2: Eclipse Dynamics (15 т.)Part 3: Eclipse Year (4 т.)Part 4: The Terminator (10 т.)Отговори Решение

SAO 2021, Писмен тур — Задача 14. Section C: Long Question

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Всички материали от SAO 2021

Singapore Astronomy Olympiad 2021 · 39 т.

Указания

The entire Olympiad lasts for 3.5 hours and is worth a total of 200 marks.
Fill in these details on the cover page (after the Contents page), the Summary Answer Sheet and each side of an SAO answer sheet:
- Year of competition
- Your participant code
- The page number – which should be continuous from 1 to N
- The part of the paper, and the question number
Cross out all workings or answers you do not wish to be evaluated.
You are allowed to leave the venue early upon submission of your answer script (see Instructions on Cover Page), but only after one hour after the start of the paper and not within 30 minutes to completion of the paper.
Please raise your hand to seek assistance from the invigilators (e.g. to visit the restroom, request for more answer sheets, enquire about an ambiguity, etc.)

Competition Rules and Regulation

Only the use of scientific calculators is permitted. No graphing or programmable calculators are allowed.
Disruptive behaviour, cheating, collusion to cheat or any integrity-related offences are grounds for immediate disqualification.
You may opt to retain the question paper and constants sheet for personal use. Return all unused answer sheets to the Organising Team.

References for Data Analysis Question

MS Star Details are derived from: Zombeck, Martin V. (1990). Handbook of Space Astronomy and Astrophysics (2nd ed.). Cambridge University Press
Additional MS Star Details are derived from: Vardavas, Ilias M.; et al. (2011), "Chapter 5. Incoming Solar Radiation" Radiation and Climate: Atmospheric Energy Budget from Satellite Remote Sensing.

Astronomy.SG would like to thank the organizing committee and the question setters. This document is typeset in LaTeX.

Cover Page

Fill in all required details below, before time is up.
Ensure that all materials (answer sheets, graphing paper, practical question paper) you intend to submit for grading have been attached. Amendments to any answer script after it has been collected will not be entertained.
Details on each side of answer sheet:
- Your participant code.
- Question number.
- Page number – continuous from 1 to N throughout your answer script.
- DO NOT WRITE YOUR NAME ON YOUR ANSWER SCRIPT.
Order of answer script:
- Stack 1: i. This cover page.
  ii. Summary Answer Sheet iii. Answer sheets from Part 1 to 3, in continuous order from page 1 to M.
- Stack 2: i. Answer sheets for Part 4, in continuous order from M to N.
  ii. Graph(s) from Part 4.
- Stack 3: i. Practical cover page.
  ii. Practical Component from Part 5.

1 Section A [12]

Write all your responses in the summary answer sheet provided.

Section B(1): Short Questions [36]

This section contains a total of 5 short-length questions worth 9 marks each.
You are to choose 4 to do.

Section B(2): Medium Questions [18]

This section contains a total of 2 medium-length questions worth 18 marks each.
You are to choose 1 to do.

Practical Cover Page

INSTRUCTIONS TO PARTICIPANTS:

Fill in all required details below, before time is up.
You are to detach this page and all subsequent pages as part of your response.
All detached pages are to be re-stapled and be part of your 3rd stack of your answer script.
All responses to this section of the paper are to be written on the question paper itself, in the spaces provided.
There will be a star chart provided as part of the question. Responses to the corresponding questions in that section are to be answered on the star chart itself.
All final responses shall be written legibly and unambiguously.
Details on each page:
- Your participant code.
- DO NOT WRITE YOUR NAME ON YOUR ANSWER SCRIPT.

Section A

Each question is worth 2 marks.

Summary of answers:

Q1	Q2	Q3	Q4	Q5	Q6
D	C	A	C	B	B

Условие

Write all responses on the answer sheets provided. The marks are stated in the brackets [] at the end of each sub-part.

Eclipses

On 26 December 2019, we observed an annular eclipse in Singapore. In this question, we concern ourselves with the dynamics of a solar eclipse occurring on the vernal equinox. The Earth-Sun distance on this day is $1.4897 \times 10^{11}m$ and the Earth-Moon distance is $3.5792 \times 10^8m$ .

Eclipse geometry: the Earth on the left and the yellow Sun on the right, with lines from the edges of the Sun past the Moon forming the shadow cone; the Moon's shadow touches the Earth, with the labels 'Umbra' and two 'Penumbra' labels.

The Earth seen along the Earth-Sun line: a circle labelled 'Earth' crossed by horizontal and vertical lines, a tilted dash-dotted line labelled 'Rotational Axis' through the centre, and a small grey disk (the shadow) at the centre. — View along Earth-Sun line

Part 1: Eclipse Geometry (10 т.)

(a) Determine the angular diameter of the moon and sun on the day of the eclipse. What type of solar eclipse is it? [4 т.]

(b) Calculate the radius of the umbra to 3 significant figures, considering a geometry like above. State any assumptions made. [6 т.]

Part 2: Eclipse Dynamics (15 т.)

(c) Determine the angular velocity of the moon and Sun during the eclipse, as viewed from the Earth. [6 т.]

(d) Calculate the longest duration of totality for an observer on the equator, given an eclipse geometry (at peak eclipse) as shown above. For simplicity, suppose that the moon moves along the ecliptic plane. [9 т.]

Part 3: Eclipse Year (4 т.)

(e) The eclipse year describes the approximate periodicity of conditions giving rise to an eclipse, taking into account the precession of the lunar nodes and the orbit of the moon. The draconic month (27.2122 days) is the period of time between two consecutive passages of the moon through the same orbital node, and the synodic month is 29.5306 days. Evaluate the length of the eclipse year, to 4 significant digits. You may neglect the eccentricity of the moon’s orbit. [4 т.]

Part 4: The Terminator (10 т.)

A grey world map in Mercator projection with a thick black terminator curve: it rises from the bottom left over the Americas, runs along the top over the North Atlantic and Europe, and falls through Asia to the bottom right; the region under the arch is labelled 'Night Time' and the regions on both sides 'Day Time'. — Terminator Line on a Mercator Projection

(f) On a Cartesian grid, with the Equator set as the x-axis and the Prime Meridian set as the y-axis, find the equation of the terminator line $^{1}$ on 21 December 2021, when it is local noon in Singapore. Singapore has a longitude of 103.82°E. In this question, you can treat the Sun as a point source. [10 т.]

$^{1}$ The terminator is a line that divides the daylit side and the dark night side of a planetary body.

Отговори

Решение

Part 1: Eclipse Geometry

Small angle approximation will also be accepted ( $\theta_M = \frac{R_m}{d_E}$ ) M1 Expression for $\theta$

\theta_M = 2\tan^{-1}\left(\frac{R_M}{d_E}\right) = 0.556^{\circ}

\theta_S = 2\tan^{-1}\left(\frac{R_S}{d_S}\right) = 0.536^{\circ}

(accepted range $0.535^{\circ} - 0.536^{\circ}$ )

M2 (correct geometry) M1 (similar triangles)

\frac{x}{R_{sun}} = \frac{x - d_{sun} + d_{moon}}{R_{moon}} \to x = \frac{(d_{sun} - d_{moon})}{(1 - \frac{R_{moon}}{R_{sun}})} = 1.4898 \times 10^{11}m

M2 (correct manipulation)

r_{umbra} = (x - d_{sun} + R_{earth}) \times \frac{R_{sun}}{x}

= 95.397\mathrm{km}

= 95.4\mathrm{km}

M1 A1

A more correct diagram would in fact be the following, but we note that the similar triangles approach yields the same answer either way; as such, both diagrams (or results of) would be accepted and given full credit

The Sun (large circle, left, with a dotted vertical diameter and the label R_sun) and the Earth (circle, right); two lines from the top and bottom of the Sun converge past the grey Moon (label R_moon, dotted vertical diameter) to a point beyond the Earth’s surface; a blue double arrow below spans the distance x from the Sun’s centre to the apex of the cone.

The more correct version of the same diagram: the lines are tangent to the Sun and to the Moon, with right-angle marks at the tangent points; labels R_sun and R_moon, and the blue double arrow x below.

Part 2: Eclipse Dynamics

We may invoke a form of kepler’s second law:

\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt} = \frac{\pi ab}{T}

\frac{1}{2}r^2\frac{d\theta}{dt} = \frac{\pi a^2\sqrt{1 - e^2}}{T}

\left(\frac{d\theta}{dt}\right)_{moon} = 3.065 \times 10^{-6}\mathrm{rad/s}

\left(\frac{d\theta}{dt}\right)_{Sun} = 2.008 \times 10^{-7}\mathrm{rad/s}

Essentially, what is important here is to calculate the velocity of the moon’s shadow on the surface of the Earth. The previous parts of the question were meant to guide the readers to the following formulation:

v_{E,t} = d_{sun} \times \left(\frac{d\theta}{dt}\right)_{sun} = 2.991 \times 10^4m/s

v_{M,t} = d_M \times \left(\frac{d\theta}{dt}\right)_{moon} = 1097m/s

v_{shadow} = (v_{E,t} - v_{M,t})\frac{d_E}{d_E - d_M} - v_{E,t} = -1027m/s

Another solution that students may provide could be the following, which is a valid approximation – however, it is important to be careful when considering the direction of the shadow’s velocity.

v_{shadow} = \left(-\left(\frac{d\theta}{dt}\right)_{sun} + \left(\frac{d\theta}{dt}\right)_{moon}\right) \times d_{moon} = 1025m/s

It is also important to take into the account the Earth’s rotational velocity at the surface $v_{ES} = \omega_E R_E = 463m/s$

Note that this velocity is at angle of $23^{\circ}26'$ to the velocity of the shadow (the ecliptic makes that angle with the equator), and their relative velocities summed through use of the cosine rule M1

v = \sqrt{v_S^2 + v_{ES}^2 - 2v_Sv_{ES}\cos 23^{\circ}26'} = 630\mathrm{m/s}

T = \frac{2r}{v} = 303s

(accepted range: $301 - 305$ s)

M1 M1 A1

The Sun (circle, left), the grey Moon and the Earth (circle, right) with the shadow cone between them; arrows mark v_E − v_M at the Moon, v_E at the Earth and v_ES at the Earth’s surface; a large arc (the Earth’s orbit) and a dash-dotted circle (the Moon’s orbit); blue double arrows below mark the distances d_E − d_M and d_E.

The Earth (circle labelled “Earth”) with a tilted dash-dotted line labelled “Rotational Axis”, a line across the disk and the grey shadow at the centre with two small blue arrows showing the directions of the velocities.

Part 3: Eclipse Year

The eclipse year is in fact the beat period/synodic period of the draconic month and the synodic month – the conditions repeat when the moon is at the same position with respect to the nodes again and along the lunar cycle M2

T_{syn} = \frac{1}{27.2122} - \frac{1}{29.5306} = 346.6\ \mathrm{days}

M1 (Sign) A1

Part 4: The Terminator

Given the set-up in the question, we are hinted to plot a latitude vs longitude graph, where the line in the graph symbolises the points on Earth that experience Sunset/Sunrise. Hence, we should turn our attention to finding an expression that gives us every pair of latitude and longitude points that experience sunset/sunrise.

To do that, we need to employ more celestial geometry. We will want to find out how many hours after noon will sunset/sunrise occur, given a specific latitude. M1

(Figure) M1

From the figure, we can find a relation between the Hour Angle (HA), latitude $\phi$ and the declination of the sun $\delta$ . Remember that $\delta$ is negative, and $\phi$ is negative for the southern hemisphere. M1

We can conjure up a spherical cosine law to get the relation.

\cos 90^{\circ} = \cos\bar{\delta}\cos\bar{\phi} + \sin\bar{\delta}\sin\bar{\phi}\cos HA

We can simplify it to:

0 = \sin\delta\sin\phi + \cos\delta\cos\phi\cos HA

Hence, we get:

HA = \cos^{-1}(-\tan\phi\tan\delta)

Looking at this, we may get stuck, but we need to remind ourselves the importance of the hour angle expression we have gotten. The $HA$ in question represents the number of hours that will have pass before sunset occurs. M1 Putting a spin to that by bringing in the idea of time zones, it tells us which place that is an $\frac{HA}{15}$ number of hours ahead of Greenwich will be experiencing sunset. M1

So, for example at the Equator, $HA = 90^{\circ}$ which translates to 6 hours, and so a place that is 6 hours ahead of Greenwich at the Equator will be experiencing Sunset! This is basically how time zones work.

Using time zones, we can then realise that the $HA$ we calculated is just the longitude coordinate we are looking for. Coincidentally, we have found the expression, we just need to put $\phi$ in terms of $HA$ .

\phi = \tan^{-1}\left[-\frac{\cos HA}{\tan\delta}\right]

Finally, we need to translate the graph to the right, since the expression above is for local noon at Greenwich, but we require it when it is local noon in Singapore. Luckily, this tweak is a simple one. M1

\phi = \tan^{-1}\left[-\frac{\cos(HA - \theta)}{\tan\delta}\right]

Where $\theta = 103.82^{\circ}$ , the longitude of Singapore. A1

When plotted, we see that the graph looks neat:

Graph of the terminator: Latitude (vertical axis, −60 to 60) against Longitud[e] (horizontal axis, −3 to 3); a blue sinusoid-like curve starts near 30 at −3.1, falls through 0 near −2.9 to a minimum of about −66 near −1.3, rises through about −30 at 0 and 0 near 0.25 to a maximum of about 66 near 1.8, and falls to about 30 at 3.1.

Оригинал в Архива: SAO 2021 Question Paper.pdf · официални решения: SAO 2021 Compiled Solutions.pdf