Съдържание

Условие Отговори Решение

SAO 2021, Писмен тур — Задача 13. Sunset on a Plane

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Всички материали от SAO 2021

Singapore Astronomy Olympiad 2021 · 18 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:
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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

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  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.
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Section A

Each question is worth 2 marks.

Summary of answers:

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

Условие

Brendan is watching the sunset on board an airplane on Sept 23, 2021; from 11km above sea level.

We define the start of the sunset to be when the bottom of the disk of the Sun first touches the horizon and the end of the sunset to be when the top of the disk of the Sun detaches from the horizon.
The moment sunset starts, Brendan is at $\phi = 46.7655^{\circ} N$ and $\theta_L = 17.2480^{\circ} E$ , travelling at 250m/s due directly westwards.

How long will Sunset last from Brendan’s point of view?

You are to account for atmospheric refraction with a value of $32'$ .

Отговори

Решение

We first need to find out the angle of depression due to the height above sea-level. This is important as this will affect the setting angles of the Sun.

A circle (the Earth) with centre C; a vertical line from C through O on the surface up to the observer S at height h; a line from S tangent to the circle at F, the radius R from C to F, and the angle of depression ψ at S between the horizontal dotted line and SF.

Doing a little geometry, we find that the angle of depression, $\psi$ , is given by: M1

\psi = \cos^{-1}\left(\frac{R}{R + h}\right)

Where h is the height above ground and R is the radius of the Earth.

Next, we find the angular size of the Sun’s disk, $\psi_{Sun}$ . Simple trigonometry gives us the relation: M1

a_{Sun} = 2\tan^{-1}\left(\frac{R_{Sun}}{D_{Sun}}\right)

When the sun sets, its motion is not vertically downwards. Hence, we need to find the actual angle the sun has traversed over its arc during sunset. The following diagram will show the relation between the angle traversed, $\theta$ .

M2 (Correct Triangles) M1 (Refraction) M1 (Appt Horizon.) M1

The Sun (yellow disk) moving down along an inclined arc labelled “Sun’s Path” across a horizontal line labelled “Apparent Horizon”; the lower disk is below the line by α, and a curved arrow marks the angle θ traversed along the path.

Two horizontal lines labelled “Horizon” and “Apparent Horizon” and the Sun’s disk at two positions on the inclined “Sun’s Path”; vertical arrows mark the distances ψ + a/2 + α and ψ − a/2 below the horizon, and curved arrows mark the arcs x and θ along the path.

We use two triangles to get the two equations:

\frac{\sin x}{\sin 90^{\circ}} = \frac{\sin(\psi - \frac{a}{2})}{\sin\bar{\phi}}

\frac{\sin(x + \theta)}{\sin 90^{\circ}} = \frac{\sin(\psi + \frac{a}{2} + \alpha)}{\sin\bar{\phi}}

Hence,

\theta = \sin^{-1}\left(\frac{\sin(\psi + \frac{a}{2} + \alpha)}{\cos\phi}\right) - \sin^{-1}\left(\frac{\sin(\psi - \frac{a}{2})}{\cos\phi}\right)

We last turn to the speed at which the Sun will move below the horizon. Since the plane is travelling due west, the Sun will appear to move slower as the plane is “catching up with” the Sun.

For a stationary observer, the Sun will move with angular velocity equals to: M1

\omega_{Sun} = \omega_0

\omega_0 = \frac{2\pi}{T_{solar}}

T_{solar} = 24\ \mathit{hours}

M1 (Finding $\omega'_{Sun}$ )

For a moving observer, the new angular velocity of the Sun along its path is now:

\omega'_{Sun} = \omega_{Sun} - \frac{v_{plane}}{R\cos\phi}

M1 (Using $R\cos\phi$ )

With that, we can now find the duration of sunset: M1

t = \frac{\theta}{\omega'_{Sun}}

= \left[\sin^{-1}\left(\frac{\sin(\psi + \frac{a}{2} + \alpha)}{\cos\phi}\right) - \sin^{-1}\left(\frac{\sin(\psi - \frac{a}{2})}{\cos\phi}\right)\right]\left(\frac{1}{\omega_{Sun} - \frac{v_{plane}}{R\cos\phi}}\right)

Plugging in the values given in the question, we get the duration of the sunset is: A1

t = 3510s = 58^m30^s

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