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Workshop on
Astronomy of Celestial Features
Bangladesh Olympiad on Astronomy and Astrophysics
Date: 12/21/2020
Speaker: Arman Hassan�Bronze Medalist , IOAA 2020
Prepared by: �Arman Hassan�Fahim Rajit Hossain (Bangladesh Team Leader, IOAA)
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Plan
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Celestial Features in the Sky
The key Celestial Features to notice in the sky are the Sun, the Moon, 5 Planets, stras , and occasional the rift of the Milky way.��From ancient times people have wondered towards the sky and eventually they started to incorporate the knowledge in their daily lives.
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Towards the Sky: Moon
The Moon changes shape, and appears in different parts of the sky, at different times of the day or night. But it’s not too hard to understand why it does this, and to know when and where to look for it.
�
প্রথমেই আমরা জেনে নিব, চাঁদের কলা বা Lunar Phase কীভাবে কাজ করে । আগেই জেনে নেওয়া উচিত যে, চাঁদ পৃথিবীর মহাকর্ষীয় ক্ষেত্রে এমন ভাবে আবদ্ধ যেন আমরা চাঁদের শুধু একটা দিকই দেখতে পাই (এটা কে জ্যোতির্বিজ্ঞানে আমরা বলি Tidally Locked অবস্থা ) । তাও চাঁদের কিছুটা গতির কারণে পুরা চন্দ্র পৃষ্ঠের ৫৭% আমরা পুর্নিমার চাঁদে দেখতে পায় । চাঁদের নিজস্ব আলো নেই তাই চাঁদকে দেখতে হলে চাঁদ দ্বারা সূর্যের আলো প্রতিফলিত কতটুকু পৃথিবীর দিকে আসছে তার ওপর নির্ভর করতে হয় । চাঁদের এই কলা বা চাঁদ-পৃথিবী-সূর্যের অবস্থান ভিত্তিতে চাঁদের এই পাশের কতখানি আমরা দেখেতে পাব এই নিয়ে আমাদের পুর্নিমা (Full Moon) বা অমাবস্যা (New Moon) নির্ধারন করা হয় ।
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Lunar Phases
চাঁদের কলা ��ছবির ডান দিক দিয়ে সূর্যের আলো আসছে । এবং চাঁদ ও পৃথিবীর অবস্থান অনুযায়ী আমরা বিভিন্ন সময় চাঁদের বিভিন্ন কলা দেখতে পায় ।
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Lunar Phases
চাঁদের কলাকে আমরা মূলত ২ টি পর্যায়ে ভাগ করি, কৃষ্ণপক্ষ এবং শুল্কপক্ষ। চাঁদ যখন সন্ধ্যার আকাশে পূর্ব দিকে থাকে তখন আমরা বলি পূর্ণিমার চাঁদ 🌕 । এভাবে পূর্ণিমার দিন (Full Moon) থেকে শুরু করে প্রতিদিন চাঁদ আগের দিনের চেয়ে প্রায় ৫০ মিনিট পরে উদয় হয় । আবার একটা সময় আসে যখন চাঁদকে আকাশে দেখা যায় না সেদিন আমরা বলি অমাবস্যা এর দিন (🌑 New Moon) আসলে সেই দিন আকাশে চাঁদ ঠিক পশ্চিমে থাকে কিন্তু যেমনটা চিত্রে দেখা যাচ্ছে সূর্য দ্বারা আলোকিত অংশ আমরা পৃথিবী থেকে দেখতে পায় না । আমরা চন্দ্রকলা গণনা শুরু করি অমাবস্যার দিন 🌑 থেকে । অমাবস্যার দিন এর পর থেকে চাঁদের আকার বৃদ্ধি পেতে থাকে । এই অবস্থাকে আমরা বলি শুল্ক পক্ষ (Waxing Phases) । আমাদের উত্তর গোলার্ধের জন্য শুল্ক পক্ষের আলোকিত দিকটা সবসময় ডান দিকে থাকে । এটির আকার দেখতে “D” এর মত (এরকম- 🌒→ 🌓→🌔)। আবার এভাবে পুর্নিমা 🌕 থেকে শুরু করে অমাবস্যা 🌑 পর্যন্ত চাঁদ ক্ষয় হতে শুরু করে তখন আমরা একে কৃষ্ণপক্ষ বলি (Waning Phases) । কৃষ্ণপক্ষের জন্য চাঁদের আলোকিত বক্রতল থাকে ডানদিকে যা দেখতে “C” এর মত ( এরকম -🌖→ 🌗 → 🌘) । এই নিয়ম কিন্তু শুধু উত্তর গোলার্ধের জন্য খাটবে । দক্ষিন গোলার্ধের জন্য ব্যাপারটা হয়ে যাবে উল্টো । আবার বিষুবরেখার কাছে বাঁকা চাঁদ এতটাই হেলে থাকে যেন মনে হয় আকাশে উজ্জ্বল নৌকা ভাসছে । প্রাচীন রোমরা যে শোয়া চাঁদকে ভুল “fallacious” (Lunar fallax) বলত তাতে বিস্ময়ের কিছু নেই । আমাদের মনে রাখতে হবে শুল্কপক্ষের চাঁদ উঠে অন্ধকার আকাশের পশ্চিম অংশে, কৃষ্ণপক্ষের চাঁদকে দেখা যায় ভোরের আকাশে পুব দিকে ।
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Lunar Phases- Few things to notice
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As we have seen, the time at which the Moon crosses the southern sky gets later every day, by about 50 minutes. So the time at which the Moon rises and sets will also vary from day to day. For example, when the Moon is heading northwards, it spends longer above the horizon each day. The time of moonrise gets later (because the Moon crosses the sky later), but it tries to get earlier too (because it is spending longer above the horizon). The net result can be that the time of moonrise gets only a few minutes later from one day to the next. Conversely, when the Moon is heading southwards, the time of moonrise can change by well over an hour from one day to the next.
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Problem 1: What phase is it?
Q1: At 6 A.M. you look up in the sky and see a moon with half its face bright and half dark near the Upper Meridian. What phase is it?
�A) first quarter
B) waxing gibbous
C) third quarter (last quarter)
D) half moon
Q2: You are an astronaut on the moon. You look up, and see the Earth in its full phase and on the meridian. What lunar phase do people on Earth observe? What if you saw a first quarter Earth? new Earth?
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Problem 1: Continued
Q1: Explanation
If no further information is given you can consider an idealized situation where both the sun and the moon above your horizon exactly for 12 hour, the sun rises in the east at 6 am, sets in the west at 6 pm and both the sun and moon takes exactly 6 hours to reach highest altitude. The moon was near upper meridian at 6 am. So it rose 6 hours ago at 12 pm.�Now, during new moon 🌑, the moon and sun are at exactly the same direction. So the moon and Sun will rise at the same time on 6am.�During full moon 🌕 , the Moon and the Sun are at exactly the opposite direction. So they Moon will rise 12 hours later than the Sun or vice-versa. So, at full moon 🌕 , the Moon will rise at 6 pm.�During 3rd quarter, the Moon is halfway between full moon 🌕 and new moon 🌑. This halfway position tells us that it will rise 6 hours later than full moon 🌕. So it will rise at 6 am exactly as the situation mentioned in the question. So the answer is third quarter.
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Problem 1: Continued
Q2:�
Lets us 1st examine 1 condition; when the people of earth observe waxing Gibbous moon. The configuration seems like this. If you are on the earth and you look at the moon you’ll see 75% of the moon illuminated as shown. The moon is farther from sun than earth��Now imagine you were suddenly teleported to moon. the earth is closer to sun so most of the dark side of earth will be facing you. So the projection of earth will seem to be 25% illuminated. And from the configuration, we can say that the earth will be on waning crescent phase.
try out the other phases on your own!�Drawing out the figure will make your life easier.
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Lunar Months and Synodic Period
পাশের ডায়াগ্রামটি দেখাচ্ছে সময়ে পুর্নিমার চাঁদ পিছের একটি তারার সাপেক্ষে একটি রেখায় আছে । এক Sidereal মাস পর চাঁদ পিছের সেই তারাটির সাথে আবারো একই রেখায় আছে সময়ে । কিন্তু চাঁদকে পুর্নিমায় আসতে আবার সময় লাগে যা কিনা ১ Synodic মাস পর ।
চাঁদের পৃথিবীকে একবার পরিভ্রমণ করতে লাগে ২৭.৩ দিন । একে জ্যোতির্বিজ্ঞানে আমরা বলি Sidereal Month । মানে একই আকাশপটের (যে তারামণ্ডলের ওপরে চাঁদ কে দেখা যায়) তার সাপেক্ষে একই অবস্থানে ফিরে আসতে । কিন্তু চাঁদের একটি কলা সাপেক্ষে একই কলাতে ফিরে আসতে লাগে মোটামোটি ২৯.৫৩ দিন যাকে আমরা বলি Synodic Month; এটি এসেছে গ্রীক শব্দ Synodos থেকে যার অর্থ Conjunction । পৃথিবী যদি স্থির থাকত তাহলে Synodic Month এবং Sidereal Month একই হত মানে ২৭.৩ দিন লাগত কিন্তু Synodic Month সম্পুর্ন হতে ২.২ দিন বেশী লাগে কারণ চাঁদ যে সময়ে পৃথিবীকে পরিভ্রমণ করে ঠিক সেসময় পৃথিবীও সূর্যকে একই দিকে আবর্তন করে । তাই চাঁদকে তার পরিভ্রমণ শেষ করে পৃথিবী এবং সূর্য সাপেক্ষে একই অবস্থানে গিয়ে একটি নির্দিষ্ট কলাতে আসতে একটু পিছিয়ে পড়তে হয় ।
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Synodic Period vs Sidereal Period
Earth
Sun
Planet
Situation 1: Earth’s Center | Surface | Moon
Situation 2: Sun | Planet | Earth
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Synodic Period vs Sidereal Period
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Synodic Period and Phase �[Try it yourself]
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Inclination of Lunar orbit
In one half of the sidereal month (27.32/2 =13.66) The Moon moves from the indicated position on the right to the position on the left. The declination changes from 28.58º to - 28.58.
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Inclination of Lunar orbit
The moon’s orbit is tilted by about 5° with respect to the ecliptic.
The orbit of the Moon is inclined at an angle of 5.145º to the ecliptic. The Moon has two nodes: the ascending node is where the Moon passes upward through the ecliptic, and the descending node is where it passes downward. The line of nodes is the line in the ecliptic that passes through the nodes. The angle between the zero longitude in the ecliptic and the line of nodes is the celestial declination of the Moon’s nodes. Owing to gravitational perturbations this angle makes a complete rotation in the ecliptic plane in 18.6 years
https://astronomy.stackexchange.com/questions/36512/why-is-the-moons-orbit-so-complicated/ �����
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Phases of Eclipse
Over time
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Phases of Eclipse
Over time
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Phases of Eclipse
Sun
Umbra Shadow Line
Ecliptic
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Summer Solstice = 21 June
Vernal Equinox - 21 March
Autumnal Equinox = 23 September
Winter Solstice = 21 December
North
North
North
Yearly Declination of Sun
Earth rotational axis and hence the equator is tilted at 23.5 degrees with respect to the ecliptic plane (the plane in which the earth moves round the sun) �
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As a result The angular distance of sun from the equator varies throughout the year with the highest being 23.5 on June 21 and lowest being -23.5 on December 21.�
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9.Motion of Sun
On June 21, the Sun rises in the most North-eastward direction possible, rises the highest in the sky, stays in sky for the longest, and sets in the North-westward direction possible��On September 23 and March 20, the sun rises exactly on east, the length of day and night is equal and sets exactly due west
On December 21, the Sun rises in the most South-eastward direction possible, rises the highest in the sky, stays in sky for the shortest, and sets in the South-westward direction possible.��
4 Special Day for Astronomers!
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Vernal and Autumnal Equinox
March 20 is vernal equinox and September 23 is autumnal equinox. The declination of su is 0 degree��The sun is directly above equator .Rays of sunlight hits equator perpendicularly. The day and night duration is equal.��
The result is autumn in southern and northern hemisphere
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Highest Altitude of Sun
We will try to find out the highest altitude / Upper culmination/ altitude of sun at upper meridian.�We assume the declination of sun to be
and the latitude of observer to be
Sun is very far from us so all the sunrays will seem parallel to us
[because S’O and SC are parallel]
so altitude of sun=
If , the formula will change to be�
This works for any object with a definite declination
�
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Summer solstice
The result is summer in Northern hemisphere and winter in southern hemisphere
June 21 is summer solstice. The Sun reaches maximum declination on this day��The sun is closer to Northern hemisphere.��Rays of sunlight hits Northern hemisphere almost perpendicularly and hits southern hemisphere obliquely������.
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Winter Solstice
December 21 is winter solstice. The Sun reaches minimum declination on this day��The sun is closer to southern hemisphere.��Rays of sunlight hits southern hemisphere almost perpendicularly and hits northern hemisphere obliquely��
The result is summer in southern hemisphere and winter in northern hemisphere
21 December
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Today the Jupiter and Saturn was so close together (6 arc minute) that they will basically look like e single point in the sky. this specific situation is called ‘Great conjuncion’.
�It is a very rare scenario which happens every 20 years (you can calculate it with what we learnt today)
�Last time we saw jupiter and saturn some this close was 400 years ago.
Longest Day + Winter Solstice + Great Conjunction = Astronomical Combo Pack
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Which latitudes will observe an 24 hour Day on 1st May?
Problem 2: 24 hour day
1st of all you would need the declination of sun. If the declination versus time graph is given you can easily find the declination of sun at a specific time. But you will not be given the graph most of the time. So you have to approach in another way.��The 4 most significant days are March 20, June 21, September 23, December 21. CLosest to May 21 is March 20 and June 21
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Problem 2 Continued
Number of days Between March 20 to June 21 is 92. Change of declination of sun in these 92 days is 23.5 degrees. Number of days between March 20 to 1st May is 42 days. Change of declination is this 42 days is (42/92*23.5)= 10.73 degrees. So declination of sun is 10.73 degrees in May 1st.�As the sun is really far from us, Sun rays will seem parallel to us. They hit the surface in different angles according to the latitude. However ray S’N is tangent to the earth at point N. Consider people living in the arc PN. They are now at the opposite side of the the face that is facing sun. So they were suppose to experience night now. But still they can see the sun. So we can say that they will see the sun the whole day.S’N is perpendicular at point N to earth. So CN is perpendicular to SN. S’W is parallel to SC.
In triangle NWC,
�
�So people living in latitudes 79.27 degrees or above will experience a 24 hour day. You can find the case of 24 hour night using tangent S”W
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Solar Day & Sidereal Day
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Sun Rays and Shadow
from the figure we see that shadow of anything depends highly on the altitude
The direction of the shadow will be exactly opposite to sun.�For example if the Azimuth of sun is 45 degree, the azimuth of shadow will be (45+180)=225 degrees
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Problem 3 : (IOAA)
In a typical Persian architecture, on the top of south side windows there is a structure called “Tabeshband” (shader), which controls sunlight in summer and winter. In summer when the Sun is high, Tabeshband prevents sunlight from entering rooms and keeps inside cooler. Modern studies Have verified that the Tabeshband saves about 20% of energy cost Figure 2.1 shows a vertical section of this design at latitude of 36.0° N with window and Tabeshband. Using the parameters given in the figure, calculate the maximum width of the Tabeshband (x), and maximum height of the window (h) in such a way that:
(a) No direct sunlight can enter to the room on the summer solstice at noon.
(b) The direct sunlight reaches the end of the room (indicated by the point A in the figure) on the winter solstice at noon.
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Problem 3: Continued
Declination of sun on Summer solstice= 23.5 degree. The altitude of Sun at 36 degree North on Summer solstice noon = 90-36+23.5 = 77.5 degree�Declination of sun on winter solstice= 23.5 degree. The altitude of Sun at 36 degree North on winter solstice noon = 90-36-23.5 = 30.5 degree�We don’t want sunlight to enter in in winter solstice. That can happen if sun-rays are in the situation of 1st figure and we want sunlight to reach at point a in winter solstice. That can happen if the sun-rays are in the situation of second figure.�
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Problem 3: Continued
In the second figure,
In the 1st figure,
��
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Concept of Celestial Sphere and �Basics Coordinate System
Celestial Sphere: An imaginary sphere of which the observer is the center and on which all celestial objects are considered to lie. In Bangla we call it “খ-গোলক” । The centre of the celestial sphere is always the observer.��In Astronomy in general the concept of Celestial Sphere and sky movement is essential.
Concepts that will be needed
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Small Circle
Great Circle
Drawing a Celestial Sphere with Alt �Azimuth and Equatorial coordinate System
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Z
O
Step 1
N
S
NCP
O
Z
Step 2��This will the base for all celestial sphere you will draw in the future�
S
Step 3
NCP
O
S
Z
= Altitude of NCP = Observer’s Latitude
Equator
Equator
Na
SCP
Horizon
is taken in an arbitrary position above horizon
Drawing a Celestial Sphere with Alt �Azimuth and Equatorial coordinate System
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NCP
Equator
Z
N
S
Alt
Azimuth
is taken in an arbitrary position above horizon
Equator
Z
NCP
S
N
= Declination
= Hour Angle
SCP
S
Alt-Azimuth System
Equatorial System
Horizontal perspective (2D)
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The north celestial pole lies at an altitude of Φ above the horizon, and the celestial equator is 90° away. At it's highest point, the celestial equator attains an altitude of 90-Φ, as shown above. The declination of the star is measured from the celestial equator, and so the transit altitude of our star is given by 90-Φ+δ. The star will be due south when it transits. What about a star in the direction given by the red arrow? For this star, the formula 90-Φ+δ yields an altitude of more than 90°, but as we have know altitude ranges from 0° to 90°. Instead, we must use the formula 180-(90-Φ+δ), as shown in the diagram above. This star will be due north when it transits.
Spherical triangle solution to all problems!
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Three points A, B and C on the surface of a sphere (and not all lying on a single great circle) define a unique small circle. A, B and C are the vertices of a spherical triangle, the sides of which are great circle arcs.
Theorems that will be needed
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Spherical cosine theorem:
or
The angle between the arcs is the angle between the planes of the circles to which these arcs belong
IMPORTANT: The cosine and sine theorems apply ONLY to arcs of large circles.
Spherical sine theorem:
Hemisphere Drawing for Solving Spherical Triangle
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We said earlier that the difficulty is knowing which spherical triangle to draw. In the figure we have solved that problem for you. See if you can work out what the angles and arc lengths labelled with question marks should be, then move on to next page to see if you're right.
Hemisphere Drawing for Solving Spherical Triangle
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For better understanding of the themes here please visit our article -��https://bdoaa.org/তারাচিত্র-এবং-আকাশ-চেনা/
Solving an IOAA level Problem
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Calculate the azimuth of the Sun at rising on Midsummer’s day at Stonehenge (latitude 50° 10′ N) at a time when the obliquity of the ecliptic was 23°48′.
Solution: On midsummer’s day the Sun reaches its greatest northerly declination, namely δ = +𝜖, where 𝜖 is the obliquity of the ecliptic. The required spherical triangle will be ZPX (zenith - pole - Sun), in which PS = 90° − 𝜖
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Thank You
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