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Douglas Adams The Hitchhikers Guide to the Galaxy

“There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened.”. Douglas Adams

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Douglas Adams The Hitchhikers Guide to the Galaxy

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  1. “There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened.” Douglas Adams The Hitchhikers Guide to the Galaxy

  2. The very basics of astronomy Look up Look up at night Ask questions HORIZON

  3. Classical Astronomy: it’s about Time • Observations: • See stars, planets, Moon (comets, and meteors) • Objects move on the sky and with respect to each other • Importantly there is a regularity to celestial motion • The sky is a ‘giant clock’ – steady and dependable • The celestial poles = centers of rotation for the stars • Sunrise / Sunset (24 hours = 1 day) • Phases of the Moon (29.5 days = 1 month)

  4. The seasons: Spring, Summer, Fall, Winter • Caused by the tilt of Earth’s axis (see later) and orbital motion of Earth around the Sun • Time to orbit Sun = 1 year = 365.25 days • Measured by Helical rise/set times of bright stars and prominent constellation • The star rises in the East just before the Sun • Star sets in the west just as Sun rises in the East

  5. NCP Star Trail Image The stars ‘move’ in circles upon the sky centered on the north celestial pole NCP (in the northern hemisphere) - due to Earth’s rotation – see 21st Century fig. 2.11

  6. Star Trail Math We know: Earth rotates on its axis once every 24 hours Hence: Earth’s spin rate = 360 degrees / 24 hours = 15 o/hr We can now calculate the angle swept out by a star in a long-exposure photograph Star trails due to Earth’s rotation For example: during a 20 minute exposure time, the Earth will spin through an angle = 15 o/hr x (20/60) = 5 degrees Converting 20 minutes to hour fraction Hence the star trails will swept out 5 degree arcs about the NCP

  7. 5o Position after 40 minutes Position after 20 minutes Star trail on photograph 5o NCP Center of star rotation on the sky Start position

  8. Lab # 4 – Star Map of the Ursa Major A handy guide to estimating angles on the sky

  9. The Celestial Sphere North Celestial Pole (NCP) Earth’s spin axis The celestial sphere upon which the positions of stars are mapped Earth Celestial equator South Celestial Pole (SCP) See 21st Century: fig. 2.6

  10. The utility of the celestial sphere - This is how the heavens appear to us from Earth • Provides a backdrop (reference system) of stars and prominent star groups = constellations • North Star = Polaris = NCP = a fixed sky point • Provides navigators with a north direction (and indicates their latitude on Earth’s surface) • A reference system for mapping the locations of • The Sun - seasons • The Moon - calendar • Planets - astrology All located on (close to) the ecliptic

  11. The Celestial Sphere (NCP) The ecliptic is inclined by 23 ½ degrees to the celestial equator Definition: The ecliptic corresponds to the projection of the Earth’s orbit onto the celestial sphere (mapped out by Sun’s path through the constellations) Ecliptic Earth Celestial equator The Moon is always within 5 degrees of the ecliptic, and the visible planets within 7 deg. (SCP)

  12. Zodiacal band (the circle of animals) is a 10 degree wide strip, centered on the ecliptic. The position of the Sun in the Zodiac indicates the seasons (Winter, Spring, Summer, Fall). The Moon along with all of the planets visible to the naked eye are found to lie within its girth The zodiac is divided into 12, 30-degree wide divisions (or signs)

  13. CS is still the bases for constructing star maps to this very day  First lab Planisphere = the projection of the CS onto a flat disc (analog computer for star and constellation locations) NCP Ecliptic Celestial equator

  14. Humm…..”The Moon she is an arrant thief and her pale light she snatches from the Sun” …. (wrote Shakespeare) Phases of the Moon Illuminated portion varies between zero (New Moon) and 100% (Full Moon) Time to repeat New Moon to New Moon is 29.530 days The Moon’s motion is actually very complex, and we see more than half of its surface area during one phase cycle. Its apparent size also varies during each phase cycle. See Lab # 2

  15. First quarter Full Moon New Moon Earth Sun Phases of the Moon • The ancient astronomers knew that the time interval between repeat phases of the Moon (thesynodic cycle) is 29.530 days, and that the phase variation was due to the motion of the Moon about the Earth as measured from the line joining the Earth and the Sun. Earth - Sun reference line Moon’s orbit

  16. See 21st Century Fig. 2.23 Moon phase repeats on the Synodic period of 29.530 days

  17. Definitions - lunar • Sidereal (or orbital) period • Time for Moon to get back to the same position in the sky (with respect to the stars) = 27.322 days • Synodic period: • Time to repeat alignment of Earth, Moon and Sun (repeat of phase illumination) = 29.530 days Synodic period  Sidereal period because of Earth’s motion about Sun See 21st Century – figure 2.24

  18. Moon in same position WRT the stars Moon’s orbit Very distant “fixed” stars used to time sidereal period Earth’s orbit SUN

  19. The Moon’s Motion on the Sky • If the Moon’s angular diameter on the sky is 0.5 degrees, how long does it take to move its own diameter on the sky WRT the stars? 0.5 degree across as seen from the Earth

  20. Information given: • Moon’s orbital period (back to the same place in the sky WRT the stars) = 27.322 days • Moon’s diameter is 0.5 degree on the sky Question: How long will it take the Moon to move its own diameter (0.5 deg) on the sky We know: Moon completes 360 deg path on the sky in 27.322 day Moon’s speed 0.5 deg

  21. So, final result Moon moves its own diameter in time = 0.5 deg / 0.549 (deg/hr) = 0.91 hrs = 54.6 min Another question How many degrees does the Moon move on the sky (with respect to the stars) between each full Moon to full Moon cycle? • We know • Time = synodic period = 29.53 days • Moon’s speed on the sky = 13.176 deg/day Hence Over full Moon cycle the moon travels 29.53 x 13.176 = 389.09 deg. So: Motion along the ecliptic = 389.09 – 360 = 29.09 deg

  22. Full-Moon to Full-Moon shift along ecliptic = 29.09 degrees Moon’s eastward motion along the ecliptic 29.09 deg Dec 2nd Nov 2nd Dec 31st Jan 30th Oct 4th Celestial equator Ecliptic

  23. What is the same Moon phase + same position on the sky repeat time ? Moon moves 29.09o per full phase cycle – need to find number of phase cycles so that Moon accumulates exactly 360o motion on the sky – it is then back in the same spot that it started from N = 235

  24. The Metonic Cycle First described by Meton of Athens in 432 BC - it is the time for the Moon to show the same phase in the same position (WRT the stars) on the Sky We have just seen that the number of phase cycles before the Moon is back in the same position on the sky with the same phase is N = number of Moon phases = 235 We know that each Moon phase cycle lasts 29.530 days Hence: Metonic cycle = 235 x 29.530 = 6939.55 days  19 years

  25. STS 232 It’s amazing what you can do • With just basic tools for measuring angles on the sky the ancient astronomers deduced highly accurate values for: • the size of the Earth • the distance to the Moon • and a not so good estimate of the relative distance to the Sun compared to the Moon These steps begin to determine the actual scale of the inner solar system

  26. Johannes Muller (1436 – 1476) Regiomontanus “Royal Mountain” “ No one can bypass the science of triangles and reach a satisfying knowledge of the stars…A new student should neither be frightened nor despair. Good things are worthy of their difficulties…..” On triangles of every kind (1464)

  27. Aristarchus – circa 220 BC The relative distance to the Sun Moon’s disk is 50% illuminated at the 1st and 3rd quarter phases Hence: Earth – Moon – Sun angle must be 90 degrees ¼ phase moon Galileo Spacecraft image December 1990

  28. Earth Sun First quarter Moon phase 90 degrees in here Dmoon Dsun Measure this angle (A)

  29. A bit of trig gets our result • Procedure – measure angle A, then Aristarchus found A = 87o Hence: DSun = DMoon / cos (A) So, Aristarchus found: DSun = DMoon / cos (87) = 19.1 Dmoon That is, the Sun is 19.1 times further from Earth than the Moon This result is actually 20 times too small! It is very difficult to measure angle A accurately

  30. From last class • Metonic cycle: repeat Moon phase + location in sky time = 235 luantions or about 19 years • Use geometry to begin measuring the physical scale of the solar system • Measure an angle in a right-angle triangle • Recall handout – sine, cosine and tangent • Aristarchus – observes Moon at ¼ phase to determine that Sun is 19.1 times further away from Earth than the Moon • Good start (but no cigar) - Sun is actually 400 times further away than the Moon (modern result)

  31. Cross-staff measures angle a a Ouch, my eye, my eye!!! What’s the angular size? Angles on the sky  star maps  locations of planets  distances to planets Angular size  physical size (if distance is known)

  32. Terminology – see Trig handout Angular units and conversions 360 degrees in a circle 60 arc minutes in 1 degree 60 arc seconds in 1 arc minute So, 60 x 60 = 3600 arc seconds in 1 degree 1 arc minute - angular size of a quarter at 82.5-meters 1 arc second - angular size of a quarter at 4.95-kilometers distance - hence the expression “small change” 

  33. Need to know one or other • See your Astronomical Triangles handout: • The BIG Result gives: diameter measured ‘units’ distance Planet of diameter d(km) d D a Observer

  34. The Sun has a diameter of d(km) = 1.39 x 106 – km The Sun is at a distance of D(km) = 1.49 x 108 – km From the Earth Question: what is the Sun’s angular diameter ?

  35. Rearrange formula The Moon is 400 times smaller than the Sun but 400 times closer to us – hence it has the same angular diameter as the Sun ~ ½ a degree This is why the total eclipse track is very small on the Earth’s surface – ASTRO: figure 2.7

  36. Moon’s shadow only just reaches Earth’s surface Same angular size from Earth

  37. Question – why do we not see two eclipses per Moon phase cycle? Answer - Moon’s orbit inclined by 5.2 degrees to the ecliptic See 21 Century fig. 2.33

  38. The line connecting the two points at which the Moon’s orbit cuts through the ecliptic is called the line of nodes For a solar eclipse to take place, the Moon must be in its New Moon phase, and located at one of the nodes Definition – the Draconic month = 27.212221 days the time for the Moon to complete one orbit about the ascending node Moon moves from below to above ecliptic The Draconic month is slightly smaller than the sidereal period of the Moon because the line of nodes is slowly rotating

  39. Looking up the numbers (no need to know them) Sidereal period – Draconic month = 27.321662 – 27.212221 (days) = 0.109441 day = 2.627 hours Moon’s line of nodes rotates about the Earth by 19.3548 degrees per year (or about 1 degree every 19 days) Hence: takes 360 / 19.3548 = 18.6 years for the line of nodes to complete one rotation about the sky This is the basis of the Saros Cycle for the repeat times between similar (lunar or solar) eclipses

  40. Laser Ranging the Moon’s orbit • The distance between the Earth and Moon • Measured by sending a burst of laser light to a reflector and timing the light travel time Light travel time to Moon and back is about 2.563 seconds

  41. Distance by light See ASTRO Page 56 & Table A.2 p 326  OOTNTK • Speed of light c = 3 x 108 m/s Relationship between speed, time and distance  OOTETK For the Moon: Distance = (time taken / 2) x speed of light = (2.563 / 2) x 3 x 108 = 3.8445 x 108 meters Why ½ ?

  42. An expanding orbit • The Moon’s orbit is actually getting bigger with time • use: velocity = speed of light = distance / time • 1st observation, the travel time is T1 = 2D1 / c • 2nd observation 1 year later, travel time is T2 = 2D2/c • So D = c (T2 – T1) / 2 D1 D2 D = expansion of orbit = D2 – D1 RHS = measurable or known quantities

  43. Results • Timing measurements indicate • From one year to the next, light signals take 0.25 nano (0.25 x 10-9) seconds longer to return • Hence, the Moon’s orbit is expanding by 3.75 centimeters per year Lunokhod 1 (1970)

  44. Formula to use is D = c (T2 – T1) / 2 Speed of light 3 x 108 m/s Time difference in signal travel times measured one year apart 0.25 x 10-9 Seconds Hence: D = 3 x 108 x (0.25 x 10-9) / 2 = 0.0375 meters or, D = 3.75 centimeters (per year) Trivial Pursuits: A nail biting result – the Moon is moving away from the Earth at about the same rate as fingernails grow ~ 3mm per month

  45. The Light Year See ASTRO Pages 5 & 6 Definition: The distance traveled by a light ray in one year 1 light year = speed of light x number of seconds in a year = 3 x108 x 365.25 x 24 x 60 x 60 (meters) = 9.46 x 1015 meters I thought a lite year meant only two classes

  46. The Sun again…. • Distance to the Sun from Earth is about 150 million kilometers (150 billion meters) Hence: Light travel time from Sun to Earth = 150 x 109 / 3 x 108 - seconds = 500 seconds = 8.3 - minutes The Sun is 8.3 light minutes away from Earth The nearest star to the Sun (Proxima Centauri) is 4.243 light years away Implication: space is very big and mostly empty

  47. The fundamental unit of measure 1 AU • Definition: The Astronomical Unit (AU) is the average distance between the Earth and the Sun 1 AU = 1.496 x 108 km  five star definition Reason: this is a distance that can be directly (in the modern era) measured by radar

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