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BAAO 2016, Теория — Задача 2. Supermoons

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British Astronomy and Astrophysics Olympiad 2015-2016 — Astronomy & Astrophysics Competition Paper — Monday 18th January 2016 · 18 януари 2016 г.

Внимание: Бележка към темата

The official solutions cover Questions 1–3 only (Question 3 as a handwritten scan); there are no official solutions for Questions 4 and 5.

Указания

Time: 3 hours (approximately 35 minutes per question).

Questions: All five 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 may 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 23rd 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 December 2016; all A Level students are eligible to participate, even if they will be attending university in December.

Training Dates and the International Astronomy and Astrophysics Olympiad

Following this round the best students eligible to represent the UK at the International Olympiad in Astronomy and Astrophysics (IOAA) will be invited to attend the Training Camp to be held in the Physics Department at the University of Oxford, (Monday 4th April –Thursday 7th April 2016). Astronomy material will be covered; problem solving skills and observational skills (telescope and naked eye observations) will be developed. At the Training Camp a practical exam and a short theory paper will be sat. Five will be selected for further training. From May there will be mentoring by email to cover some topics and problems, followed by a training camp at the beginning of July and a weekend training camp in autumn.

The IOAA this year will be held in Bhubaneswar, India, from 9th to 19th December 2016.

Important constants

Speed of light in free space	$c$	$3.00 \times 10^{8}\ \mathrm{m\ s^{-1}}$
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	$1.49 \times 10^{11}\ \mathrm{m}$
Radius of the Earth	$R_{\mathrm{E}}$	$6.37 \times 10^{6}\ \mathrm{m}$
Radius of the Earth’s orbit	$r_{\mathrm{E}}$	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	$M_{E}$	$5.97 \times 10^{24}\ \mathrm{kg}$
Luminosity of the Sun	$L_{\odot}$	$3.85 \times 10^{26}\ \mathrm{W}$
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: (see the ellipse figure attached to Qu 2)

Elements of an elliptic orbit:

a = OA (=PO) semi-major axis
b = OB (=CO) 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)

Kepler’s 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.

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)}

Условие

Two photographs of the full Moon side by side on a black background: the larger one labelled “Perigee Moon May 5, 2012” and the smaller one labelled “Apogee Moon November 28, 2012”; credit “Pete Lardizabal”. — Figure 2 Supermoons at Perigee and Apogee. Image credit: John Gaughan/Pete Lardizabal/ WJLA.

A “supermoon” is a new or full moon that occurs with the Moon at or near its closest approach to Earth in a given orbit (perigee). The media commonly associates supermoons with extreme brightness and size, sometimes implying that the Moon itself will become larger and have an impact on human behaviour, but just how different is a supermoon compared to the ‘normal’ Moon we see each month?

Lunar Data:

Synodic Period	= 29.530589 days (time between same phases e.g. full moon to full moon)
Anomalistic Period	= 27.554550 days (time between perigees i.e. perigee to perigee)
Semi-major axis	= $3.844 \times 10^{5}\ \mathrm{km}$
Orbit eccentricity	= 0.0549
Radius of the Moon	= 1738.1 km
Mass of the Moon	= $7.342 \times 10^{22}\ \mathrm{kg}$

In this question, we will only consider a full moon that is at perigee to be a supermoon.

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

a. Calculate how many days separate two supermoons.

b. Show that the difference in distance between the apogee and perigee is $4.22\ \times\ 10^{4}\ \mathrm{km}$ . (The data given in this question allows the mean orbital parameters to be calculated. Note that perturbations in the lunar orbit mean that the perigee and apogee continually change over the course of the year.)

c. Determine the difference in the angular diameter of a supermoon and a full moon observed at apogee. Thus, determine the percentage difference in the brightness of a supermoon and a full moon observed at apogee. (Ignore the effects of the Moon’s orbital tilt with respect to the Earth.)

d. What change in magnitude does this brightness difference correspond to?

e. Suggest why it can be difficult to detect any differences in the brightness of supermoon compared to a ‘normal’ full moon when observing with the naked eye?

f. Calculate the gravitational field of the supermoon at the Earth. What fractional mass increase would a Moon at apogee need in order to create the same gravitational field?

Отговори

Решение

The same pair of full-Moon photographs as Figure 2 of the paper: the larger “Perigee Moon May 5, 2012” and the smaller “Apogee Moon November 28, 2012”, credit “Pete Lardizabal”. — Figure 1: Image credit: John Gaughan / Pete Lardizabal / WJLA

Supermoons

The term“supermoon” was coined by astrologer Richard Nolle in 1979. He defined a supermoon as a new or full moon that occurs with the Moon at or near (within 90% of) its closest approach to Earth in a given orbit (perigee). The value of 90% was arbitrarily chosen, and other definitions are often used.

The media commonly associates supermoons with extreme brightness and size, sometimes implying that the Moon itself will become larger and have an impact on human behaviour, but just how different is a supermoon compared to the ‘normal’ Moon we see each month?

Lunar Data:

Synodic Period $= 29.530589$ days - time between same phases e.g. full moon to full moon, new moon to new moon.
Anomalistic Period $= 27.554550$ days - time between perigees e.g. perigee to perigee.
Semimajor axis ( $a$ ) $= 3.844 \times 10^{5}\ \mathrm{km}$
Orbit eccentricity ( $e$ ) $= 0.0549$
Radius of the Moon ( $R_{\text{☽}}$ ) $= 1738.1\ \mathrm{km}$
Mass of the Moon ( $M_{\text{☽}}$ ) $= 7.342 \times 10^{22}\ \mathrm{kg}$

In this question, we will only consider a full moon that is at perigee to be a supermoon.

a) Calculate how many days separate a supermoon.

b). Show that the mean difference in the distance between the apogee and perigee is $4.22 \times 10^{4}\ \mathrm{km}$ . (The data given in this question allows the mean orbital parameters to be calculated. Note that in reality, perturbations in the lunar orbit mean that the perigee and apogee continually change over the course of the year).

c). Determine the difference in the angular diameter of a supermoon and a full moon observed at apogee. Thus, determine the percentage difference in the brightness of a supermoon and a full moon observed at apogee. (Ignore the effects of the Moon’s orbital tilt with respect to the Earth).

d). What change in magnitude does this brightness difference correspond to?

e). Suggest why it can be difficult to detect any differences in the brightness of a supermoon when observing with the naked eye?

f). Calculate the gravitational force of the supermoon on the Earth. What mass increase would a Moon at apogee need, to create the same gravitational force?

Solutions

a). 27.554550/(29.530589 − 27.554550) = 13.9443 synodic periods between each supermoon,

so 13.9443 × 29.530589 = 411.78 days between each supermoon.

b). From equation of ellipse we have that: $R_a = a(1 + e)$ and $R_p = a(1 - e)$ , where $R_a$ is the lunar apogee, $R_p$ is the lunar perigee, $a$ is the semi-major axis and $e$ is the eccentricity.

Therefore the difference in apogee and perigee is: $R_a - R_p = a(1+e) - a(1-e) = 2ae = 4.22 \times 10^{4}\ \mathrm{km}$ .

c). Angular diameter of a full moon at apogee: $R_{\text{☽}}/R_a = 4.286 \times 10^{-3}$

Angular diameter of a full moon at perigee: $R_{\text{☽}}/R_p = 4.784 \times 10^{-3}$

So the Moon appears $4.784 \times 10^{-3}/4.286 \times 10^{-3} = 1.116$ times or 11.6% larger at perigee.

Since light will follow an inverse square law, the brightness will decrease according to the distance $^2$ . Hence, in terms of brightness, a Moon appears $1.11^2 = 1.246$ times or 25% brighter at perigee.

d). Difference in magnitude = $\log_{10}$ (difference in brightness)/0.4

$= \log_{10}(1.246)/0.4 = 0.239$ magnitudes.

e). Typically, a difference of 0.239 magnitudes should be detectable with the naked eye. However, several months separate these two extremes of the brightest and dimmest full moons. In addition, the Moon is passing through different phases (e.g. crescent, quarter, gibbous) as it reaches full Moon, all of which have different brightnesses. Therefore, it is unlikely that a naked eye observer will notice, and furthermore remember, the difference in brightness between a supermoon and a full moon at apogee.

f). Gravitational force of Moon at perigee: $F_{grav,p} = \frac{G\ \mathrm{M}_{\odot} M_{\text{☽}}}{R_p^2}$ ,

and gravitational force of Moon at apogee: $F_{grav,a} = \frac{G\ \mathrm{M}_{\odot} M_{\text{☽}}}{R_a^2}$

or write as: $F_{grav,a} = \frac{G\ \mathrm{M}_{\odot} M_{\text{☽}} + m}{(R_p + r)^2}$ .

But from part b), know that $r = 4.22 \times 10^{4}\ \mathrm{km}$ .

If $F_{grav,p} = F_{grav,a}$ , the increase in lunar mass, $M_{\text{☽}} + m = 1.246 M_{\text{☽}}$ , or $m = 0.246 M_{\text{☽}}$ . (since gravity also follows the inverse square law, this is analogous to the answer in part c).

Оригинал в Архива: BAAO_2016.pdf · официални решения: BAAO_2016_solutions.pdf