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
Phases of the Moon (29.53 days = 1 month) Classical Astronomy: it s about Time Observations: See stars, planets, Moon (comets, and meteors) Objects move on the sky WRT each other and the horizon Importantly there is a regularity to celestial motion NCP The sky is a giant clock steady and dependable The celestial poles = centers of rotation for the stars There is a north (NCP) and a south celestial pole (SCP) Projection of Earth s spin axis into the sky Sunrise / Sunset (24 hours = 1 day)
The seasons: Spring, Summer, Fall, Winter Caused by the tilt of Earth s axis and the orbital motion of the 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 For a given star these conditions take place just once per year
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 let us quantify this motion (that is, put a number against it)
Earth s spin axis Center of rotation on the Celestial Sphere
SCP 24 hour continuous star trails from the South Pole
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
Position after 40 minutes Position after 20 minutes NCP 5 o 5 o Star trail on photograph Start position Center of star rotation on the sky Astronomy is simple (honest): Look up Look up at nighttime Look up at nighttime with a clear sky Observe and quantify
Lab # 4 Star Map Of Ursa Major Astronomy begins with the act of measuring angles and time intervals Ya! Rock and roll. Roll and rock A handy guide to estimating angles on the sky
The Celestial Sphere North Celestial Pole (NCP) Fixed center of rotation for the stars Earth s spin axis Earth The celestial sphere upon which the positions of stars are mapped Celestial equator South Celestial Pole (SCP) Fixed center of rotation for the stars
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 is close to 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 All located close to the ecliptic Planets - astrology
The Celestial Sphere I say, I say, I say How does an astronomer cut his hair? The ecliptic is inclined by 23 ½ degrees to the celestial equator (NCP) E clips it Definition: The ecliptic corresponds to the projection of the Earth s orbit onto the celestial sphere (mapped out by Sun s annual (SCP) path through the constellations) Earth Ecliptic Celestial equator
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
The Celestial Sphere (NCP) Ecliptic path of Sun on the CS during one year Celestial equator Earth The Moon is always located within 5 degrees of the ecliptic within the Zodiacal band (SCP) Want to now bring the Moon into the picture
The Earth and Moon as seen by the MESSENGER spacecraft (now buried in an impact crater on Mercury)
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
Phases of the Moon The ancient astronomers knew that the time interval between repeat phases of the Moon (the synodic 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. First quarter Full Moon New Moon Earth Moon s orbit Earth - Sun reference line Sun
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
In the same time that the Moon is in motion about the Earth, the Earth is in motion about the Sun Earth s rate of motion about Sun = 360/365.25 = 0.986 degrees per day - So, the Earth, Sun direction changes relative to the initial Moon position Earth s orbit Way of to the right
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
Information given: 1) Moon s orbital period (back to the same place in the sky WRT the stars) = 27.322 days 2) 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 0.5 deg Moon s speed 360 27.322 deg /day 13.176 deg /day 13.176 24 0.549deg /hr
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 (1) Time = synodic period = 29.53 days (2) Moon s speed on the sky WRT stars = 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
Full-Moon to Full-Moon shift along ecliptic = 29.09 degrees Moon s eastward motion along the ecliptic Dec 2 nd Jan 30 th Dec 31 st Nov 2 nd Celestial equator Oct 4 th Ecliptic
Having distinguished, and measured, the sidereal and the synodic periods for the Moon, we now have the power to predict its location on the ecliptic in the celestial sphere (i.e. in which constellation) and its phase at any time into the future, or into the past We have the power Within, have you, the power
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 STS 232 and a not so good estimate of the relative distance to the Sun compared to the Moon Importantly, these steps begin to determine the actual scale of the inner solar system
Terminology on angle units 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
See your Astronomical Triangles handout: The BIG Result gives: d (km) (2 / 360) D(km) (degrees) diameter units distance measured Need to know one or other D d Planet of diameter d(km) Observer
The Sun has a diameter of d(km) = 1.39 x 10 6 km The Sun is at a distance of D(km) = 1.49 x 10 8 km From the Earth Question: what is the Sun s angular diameter? d (km) (2 / 360) D(km) (degrees)
Rearrange formula d (km) (2 / 360) D(km) (degrees) d ( degrees) (km) /[(2 / 360) D(km)] 6 1.39 10 ( degrees) 0.53 8 (0.01745) 1.49 10 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
The Distant Moon Given the Moon has an angular diameter of = 0.52 degrees and a physical diameter of d = 3475 km, what is the distance to the Moon Use the small triangle formula: distance D = (360/2 )[d(km) / (deg)] Putting in the numbers check with your calculator D = 383,000 kilometers This is about 60 Earth radii, or, The distance traveled by a light ray in a time of about 1.3 seconds
And now for a moving experience. With measures of distance and time, we can determine how fast something is moving A useful formula to remember Distance = Velocity x Time OOTETK distance And Captain Scarlet says, remember the triangle to velocity time
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 the Moon and back is about 2.563 seconds
Laser ranging from the Apache Point Observatory Earth Moon Time measurement of signal out + signal back TELESCOPES
Distance by light Speed of light c = 3 x 10 8 m/s OOTNTK A fundamental constant of physics - Change the speed of light and all physics and chemistry as we know it changes We know speed distance traveled time taken OOTETK For the Moon: Why ½? Distance = (time taken / 2) x speed of light = (2.563 / 2) x 3 x 10 8 = 3.8445 x 10 8 meters = 384,450 km
An expanding orbit The Moon s orbit is actually getting bigger with time use: velocity = speed of light = distance / time Experiment 0: the travel time is T 0 = 2D 0 / c Experiment 1 (1 year on): the travel time is T 1 = 2D 1 /c D 0 D 1 So D = c (T 1 T 0 ) / 2 D = expansion of orbit = D 1 D 0 RHS = measurable or known quantities
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)
Formula to use is D = c (T 1 T 0 ) / 2 Speed of light 3 x 10 8 m/s Time difference in signal travel times measured one year apart 0.25 x 10-9 Seconds Hence: D = 3 x 10 8 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
The Light Year distance by time The speed of light c = 3 x 10 8 m/s a fundamental constant 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 x10 8 x 365.25 x 24 x 60 x 60 (meters) = 9.46 x 10 15 meters I thought a light year meant only two classes
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 10 9 / 3 x 10 8 - 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