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BAAO 2018, Теория — Задача 3. Great American Eclipse

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British Astronomy and Astrophysics Olympiad 2017-2018 — Astronomy & Astrophysics Competition Paper — Monday 22nd January 2018 · 22 януари 2018 г. · 25 т.

Указания

This question paper must not be taken out of the exam room

Time: 3 hours plus 15 minutes reading time (no writing permitted). Approx 45 minutes per question.

Questions: All four questions should be attempted.

Marks: The questions carry similar marks.

Solutions: Answers and calculations are to be written on loose paper or in examination booklets. Students should ensure their name and school is clearly written on all answer sheets and pages are numbered. A standard formula booklet with standard physical constants should be supplied.

Instructions: To accommodate students sitting the paper at different times, please do not discuss any aspect of the paper on the internet until 8 am Saturday 27th January.

Clarity: Solutions must be written legibly, in black pen (the papers are photocopied), and working down the page. Scribble will not be marked and overall clarity is an important aspect of this exam paper.

Eligibility: The International Olympiad will be held during September 2018; all sixth form students are eligible to participate, even if they will be attending university from October.

Training Dates and the International Astronomy and Astrophysics Olympiad (IOAA)

The team will be selected from sixth form students taking this paper and Y12 students taking the AS Challenge in March. The best students eligible to represent the UK at the IOAA will be invited to attend the Training Camp to be held in the Physics Department at the University of Oxford, (Monday 9th April to Thursday 12th April 2018). Astronomy material will be covered; problem solving skills and observational skills (telescope and naked eye observations) will be developed. At the Training Camp a data analysis exam and a short theory paper will be sat. Five students (plus one reserve) will be selected for further training. From May there will be mentoring by email to cover some topics and problems, followed by additional training camps in the summer.

Important Constants

Constant	Symbol	Value
Speed of light	$c$	$3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$
Earth’s rotation period	1 day	24 hours
Earth’s orbital period	1 year	365.25 days
parsec	pc	$3.09 \times 10^{16}\ \mathrm{m}$
Astronomical Unit	au	$1.50 \times 10^{11}\ \mathrm{m}$
Radius of the Earth	$R_{\oplus}$	$6.37 \times 10^{6}\ \mathrm{m}$
Semi-major axis of the Earth’s orbit		1 au
Radius of the Sun	$R_{\odot}$	$6.96 \times 10^{8}\ \mathrm{m}$
Mass of the Sun	$M_{\odot}$	$1.99 \times 10^{30}\ \mathrm{kg}$
Mass of the Earth	$M_{\oplus}$	$5.97 \times 10^{24}\ \mathrm{kg}$
Luminosity of the Sun	$L_{\odot}$	$3.85 \times 10^{26}\ \mathrm{W}$
Stephan-Boltzmann constant	$\sigma$	$5.67 \times 10^{-8}\ \mathrm{J\,m^{-2}\,K^{-4}}$
Gravitational constant	$G$	$6.67 \times 10^{-11}\ \mathrm{m^{3}\,kg^{-1}\,s^{-2}}$

You might find the diagram of an elliptical orbit below useful in solving some of the questions (the diagram is shown with Question 3).

Elements of an elliptic orbit:

$a = \mathbf{OA}\ (= \mathbf{OP})$	semi-major axis
$b = \mathbf{OB}\ (= \mathbf{OC})$	semi-minor axis
$e = \sqrt{1 - \frac{b^2}{a^2}}$	eccentricity
F	focus
P	periapsis (point nearest to F)
A	apoapsis (point furthest from F)

Keplers Third Law: For an elliptical orbit, the square of the period, $T$ , of orbit of an object about the focus is proportional to the cube of the semi-major axis, $a$ (the average of the minimum and maximum distances from the Sun). The constant of proportionality is $4\pi^2/GM$ , where $M$ is the mass of the central object and $G$ is the universal gravitational constant.

Magnitudes: The apparent magnitudes of two objects, $m_1$ and $m_0$ , are related to their apparent brightnesses, $b_1$ and $b_0$ , via the formula:

\frac{b_1}{b_0} = 10^{-0.4(m_1 - m_0)}

Note for markers

BAAO 2017/18 Solutions and Marking Guidelines

Note for markers:

Answers to two or three significant figures are generally acceptable. The solution may give more in order to make the calculation clear.
There are multiple ways to solve some of the questions; please accept all good solutions that arrive at the correct answer. If a candidate gets the final (numerical) answer then allow them all the marks for that part of the question (as indicated in red), so long as there are no unphysical / nonsensical steps or assumptions made.

Условие

On 21st August 2017 the continental United States experienced a total solar eclipse. Dubbed the ‘Great American Eclipse’, it was estimated to be one of the most watched eclipses in history.

Globe map titled “Total Solar Eclipse of 2017 Aug 21” centred on North America, with compass marks N, E, S, W. A narrow blue band (path of totality) crosses the United States from the north-west to the south-east; an arrow marks “Greatest Eclipse” in the middle of the USA. Green lines are labelled with times 17:00 UT to 20:00 UT, blue curves with eclipse magnitudes 0.20, 0.40, 0.60, 0.80; pink limit curves are marked P1–P4, and an arrow marks the “Sub Solar” point. — Figure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA’s GSFC.

The path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse (“GE”; where the path was widest since the axis of the cone of the Moon’s shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question:

The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15'48.7''$ and $16'03.4''$ , respectively, where the notation $xx'yy.y''$ corresponds to $xx$ arcminutes and $yy.y$ arcseconds (60 arcminutes = 1 degree, and 60 arcseconds = 1 arcminute)
The latitude and longitude of the location of GE are $36^{\circ}58.0'$ N and $87^{\circ}40.3'$ W, respectively
Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\odot} = 695\,700$ km, $R_{\oplus} = 6371$ km and $R_{\mathrm{Moon}} = 1737$ km, and a day to be 24 hours
Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149\,600\,000$ km and $384\,400$ km, respectively
As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction

For an ellipse with semi-major axis $a$ it can be shown that the velocity $v$ , at a distance $r$ from mass $M$ , can be written as:

v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)

Diagram of an elliptical orbit (from the constants page): ellipse with dashed major axis from P (left) through focus F (blue dot) and centre O to A (right), labelled a between O and A; dashed minor axis from B (top) to C (bottom), labelled b between O and B.

a. Calculate the width of the path of totality at GE (in km). You may use the approximation that the Moon is directly overhead so that the shadow is circular, and small enough that the curvature of the Earth can be neglected. [6 т.]

b. Show that at this point the shadow is moving across the Earth’s surface at approximately $0.68\ \mathrm{km\,s^{-1}}$ . You may use the simplifying approximation that for short time intervals it can be considered as travelling with a constant latitude, and that the apparent movement of the Sun due to the Earth’s orbit can be neglected. [5 т.]

c. Hence calculate the duration of the eclipse. Give your answer to the nearest 0.1 s. [1 т.]

The point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration (“GD”) was at co-ordinates of $37^{\circ}35'$ N latitude and $89^{\circ}07'$ W longitude, reached about 4 minutes before GE, and where totality lasted 0.1 s longer than the value calculated in part c.

Road map of southern Illinois, western Kentucky and surroundings (towns such as Carbondale, Marion, Cape Girardeau, Paducah, Evansville, Hopkinsville, Clarksville). Two parallel purple lines labelled “Path of totality limits” run from upper left to lower right; a red line between them is labelled “Path of maximum totality”. A purple pin “GD” lies on the red line near Carbondale, a black dot “CP” marks the closest point to Carbondale, and a green pin “GE” lies on the red line near Hopkinsville. — Figure 4: The route of the Moon’s shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse - and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow’s journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak & Google Maps.

d. (i) d. The town of Carbondale, Illinois, is the closest big town to the point of GD, with co-ordinates $37^{\circ}44'$ N latitude and $89^{\circ}13'$ W longitude. Assuming the path of maximum totality can be treated as linear as it passes through the region around GD and GE:

(i) Calculate the co-ordinates of the closest point (“CP”) to Carbondale on the path of maximum totality. [5 т.]

d. (ii) Calculate the distance (in km) between Carbondale and CP. [2 т.]

d. (iii) Hence calculate (to the nearest 0.1 s) how much shorter totality was for residents in Carbondale compared with CP. Take the duration at CP to be the same as GD, the width of the path to be 0.5 km less than for GE (so the Moon’s shadow is elliptical), and the speed of the Moon’s shadow to have only been affected by the change in latitude. [6 т.]

Отговори

Решение

Q3 - Great American Eclipse [Total = 25]

Suitable diagram [1]

Geometry of the umbra: a large circle (Sun) with radius R_Sun on the left, a small circle (Moon) with radius R_M and a medium circle (Earth) with radius R_E on the right. Two tangent lines from the Sun’s limbs past the Moon converge to a point beyond the Earth, making angle α with the axis; the red segment s_umbra marks the shadow width at the Earth. Distances d_Sun (Sun to Earth centre), d_Moon (Moon to Earth centre) and d_umbra (Moon to cone apex) are marked with arrows.

From the apparent angular radii we can calculate the distance to the Moon and the Sun:

Moon: $16'03.4'' = 4.67 \times 10^{-3}\ \mathrm{rad} \quad \therefore d_{\mathrm{Moon}} = \frac{R_{\mathrm{Moon}}}{\theta} = \frac{1737}{4.67 \times 10^{-3}} = 371\,900\ \mathrm{km}$ [1]

Sun: $15'48.7'' = 4.60 \times 10^{-3}\ \mathrm{rad} \quad \therefore d_{\mathrm{Sun}} = \frac{R_{\mathrm{Sun}}}{\theta} = \frac{695700}{4.60 \times 10^{-3}} = 151\,258\,000\ \mathrm{km}$ [1]

Angle of shadow: $\alpha = \sin^{-1}\left(\frac{R_{\mathrm{Sun}} - R_{\mathrm{Moon}}}{d_{\mathrm{Sun}} - d_{\mathrm{Moon}}}\right) = 4.60 \times 10^{-3}\ \mathrm{rad} \quad (= 0.2635^{\circ})$ [1]

Shadow cone length: $d_{\mathrm{umbra}} = \frac{R_{\mathrm{Moon}}}{\sin\alpha} = 377\,670\ \mathrm{km}$ [1]

Full width of shadow: $s_{\mathrm{umbra}} = 2\left(d_{\mathrm{umbra}} - (d_{\mathrm{Moon}} - R_{\mathrm{Earth}})\right)\tan\alpha = 111.7\ \mathrm{km}$ [1] [6]

[Allow s $_{\mathrm{umbra}}$ = 55.9 km if it is clear that the student recognises that's the radius of the circular shadow, otherwise they lose one mark. Allow use of small angle approximations throughout or other slight variations in the geometry resulting in a width within a few km of the correct answer]

At GE, the latitude is $\theta = 36^{\circ}58' = 36.967^{\circ}$

$v_{\mathrm{rot}} = \frac{2\pi R_E \cos\theta}{T} = \frac{2\pi \times 6371 \times \cos 36.967}{24 \times 60 \times 60}$ [1]

$= 0.370\ \mathrm{km\,s^{-1}}$ [1]

$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right) \quad \therefore v_{\mathrm{Moon}} = \sqrt{GM_E\left(\frac{2}{d_{\mathrm{Moon}}} - \frac{1}{a}\right)}$

$= \sqrt{6.67 \times 10^{-11} \times 5.97 \times 10^{24}\left(\frac{2}{371900 \times 10^{3}} - \frac{1}{384400 \times 10^{3}}\right)}$ [1]

$= 1.051\ \mathrm{km\,s^{-1}}$ [1]

Hence: $v_{\mathrm{rel}} = v_{\mathrm{Moon}} - v_{\mathrm{rot}} = 1.051 - 0.370 = 0.681\ \mathrm{km\,s^{-1}}$ [1] [5]

[Since this is a 'show that' question, ensure the student has shown evidence that they have calculated the correct relative velocity (rather than simply quoted it) for the final mark]

$t_{\mathrm{eclipse}} = \frac{s_{\mathrm{umbra}}}{v_{\mathrm{rel}}} = \frac{111.74}{0.681} = 164.0\ \mathrm{s} \quad (= 2\ \mathrm{mins}\ 44.0\ \mathrm{secs})$ [1] [1]

[Answer must be to the nearest 0.1 s for the mark]

Despite many of the simplifying assumptions made, this is within a few seconds of the measured value of 2 mins 41.6 secs or the official predicted value of 2 mins 40.3 secs

d. (i)

Converting the co-ordinates of GE and GD:

GE: $x_1 = 87.67^{\circ}, y_1 = 36.97^{\circ}$ $\quad$ GD: $x_2 = 89.12^{\circ}, y_2 = 37.58^{\circ}$

Equation of a straight line is $y = mx + c$ , so:

$m = \frac{x_2 - x_1}{y_2 - y_1} = 0.427$ [1]

$c = y_1 - m x_1 = -0.448$ [1]

[Allow reversed signs of m and c so long as internally consistent]

Equation of the line passing through Cardondale (with co-ordinates x $_{\mathrm{C}}$ , y $_{\mathrm{C}}$ ) and perpendicular to the initial line at CP (with co-ordinates x $_{\mathrm{CP}}$ , y $_{\mathrm{CP}}$ ) must satisfy

$m x_{CP} + c = -\frac{1}{m}(x_{CP} - x_C) + y_C$ [1]

$\therefore x_{CP} = \frac{x_C + m y_C - mc}{m^2 + 1} = 89.26^{\circ} \quad (= 89^{\circ}15.3')$ [1]

$y_{CP} = m x_{CP} + c = 37.64^{\circ} \quad (= 37^{\circ}38.6')$ [1] [5]

d. (ii)

We can calculate the angular distance, utilising Pythagoras' theorem since they are very close together:

$\theta_{\mathrm{dist}} = \sqrt{(x_{CP} - x_C)^2 + (y_{CP} - y_C)^2} = 0.0987^{\circ} = 0.0017\ \mathrm{rad}$ [1]

Using the small angle approximation (and ignoring the surface curvature on this scale):

$d = \theta_{\mathrm{dist}} \times R_E = 0.0017 \times 6371 = 11.0\ \mathrm{km}$ [1] [2]

d. (iii)

At CP: $t_{\mathrm{eclipse,CP}} = 164.1\ \mathrm{s} \quad s_{\mathrm{umbra,CP}} = s_{\mathrm{umbra}} - 0.5 = 111.24\ \mathrm{km}$

$v_{\mathrm{Moon}} = 1.051\ \mathrm{km\,s^{-1}}$ (same as before)

$v_{\mathrm{rot,CP}} = \frac{2\pi R_E \cos(y_{CP})}{T} = 0.367\ \mathrm{km\,s^{-1}} \quad \therefore v_{\mathrm{rel,CP}} = 0.684\ \mathrm{km\,s^{-1}}$ [1]

Eclipse shadow is now elliptical with:

Minor axis $= s_{\mathrm{umbra,CP}} = 111.24\ \mathrm{km}$

Major axis $= t_{\mathrm{eclipse,CP}} \times v_{\mathrm{rel,CP}} = 112.36\ \mathrm{km}$ [1]

For an ellipse, $x = a\cos\theta$ and $y = b\sin\theta$

Ellipse with dashed horizontal and vertical axes through the centre; semi-axes labelled a (horizontal) and b (vertical); a radius line drawn at angle θ to the horizontal axis meets the ellipse at a point whose projections are labelled x (on the horizontal axis) and y (on the vertical axis).

Hence if $y = 11.0\ \mathrm{km}$ and $b = \frac{111.24}{2} \quad \Rightarrow \theta = 11.38^{\circ} \quad \therefore x = 55.07\ \mathrm{km}$ [2]

[First mark is for an appropriate diagram and use of the equation of an ellipse]

So the length of the shadow over Carbondale $= 2x = 110.1\ \mathrm{km}$ [1]

$\therefore t_{\mathrm{eclipse,C}} = \frac{110.1}{0.684} = 160.9\ \mathrm{s}$

$\therefore \Delta t = 164.1 - 160.9 = 3.2\ \mathrm{s}$ [1] [6]

[Answer must be to the nearest 0.1 s for the mark]

Carbondale is lucky enough that it will also be on the path of totality of the next eclipse to cross the mainland of the USA on 8th April 2024!

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Royal Grammar School, Guildford)

Dr Emile Doran (The Langley Academy)

Sandor Kruk (University of Oxford)

Оригинал в Архива: BAAO_2018.pdf · официални решения: BAAO_2018_solutions.pdf