• “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
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
• 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)
Phases of the Moon (29.53 days = 1 month) NCP
• 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
24 hour continuous star trails from the South Pole
SCP
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
For example: during a 20 minute exposure time, the Earth will spin through an angle = 15 o/hr x (20/60) = 5 degrees
Hence the star trails will swept out 5 degree arcs about the NCP
Converting 20 minutes to hour fraction
Star trails due to Earth’s rotation
5o NCP
Center of star rotation
on the sky
Start position Position after
20 minutes
Star trail
on photograph
5o
Position after 40 minutes
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
A handy guide to estimating angles on the sky
Ya! Rock and roll.
Roll and rock
Astronomy begins with the act of measuring angles and time intervals
The Celestial Sphere
Earth
North Celestial Pole (NCP)
South Celestial Pole (SCP)
Celestial equator Earth’s spin axis
The celestial sphere upon which the positions of stars are mapped
Fixed center of
rotation for the stars
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
• Planets - astrology
All located close to the ecliptic
The Celestial Sphere
(NCP)
(SCP)
Celestial equator Ecliptic
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 annual path through the constellations)
Earth
I say, I say, I say…
How does an astronomer cut his hair?
E clips it…
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)
Celestial equator NCP Ecliptic
The Celestial Sphere
(SCP) Celestial equator
Ecliptic – path of Sun on the CS during one year (NCP)
Earth
Want to now bring the Moon into the picture The Moon is always
located within 5 degrees of the ecliptic – within the Zodiacal band
The Earth and Moon as seen by the MESSENGER spacecraft (now buried in an impact crater on Mercury)
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.
Humm…..”The Moon she is an arrant thief and her
pale light she snatches from the Sun” …. (wrote Shakespeare)
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.
Earth
Sun New Moon
Full Moon
First quarter
Earth - Sun reference line Moon’s orbit
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
0.5 deg
We know:
Moon completes 360 deg
path on the sky in 27.322 day Moon’s speed deg/day
322 . 27
360
/day deg
176 .
13
/hr deg 549 . 24 0
176 .
13
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
Dec 2nd
Dec 31st
Oct 4th Nov 2nd Jan 30th
Ecliptic
Celestial equator Moon’s eastward motion along the ecliptic
We have the power
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
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
– and a not so good estimate of the relative distance to the Sun compared to the Moon
STS 232
Importantly, these steps begin to determine the actual scale of the inner solar system
Angular units and conversions 360 degrees in a circle
60 arc minutes in 1 degree
60 arc seconds in 1 arc minute
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”
Terminology on angle units
So, 60 x 60 = 3600 arc seconds in 1 degree
• See your Astronomical Triangles handout:
– The BIG Result gives:
) degrees (
(km) )
360 /
2 (
(km) D
d
measured distance
diameter ‘units’
Need to know one or other
Observer
D d
Planet of diameter d(km)
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 ?
) degrees (
(km) )
360 /
2 (
(km) D
d
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
Rearrange formula
) degrees (
(km) )
360 /
2 (
(km) D
d
(km)]
) 360 /
2 ( /[
(km) )
degrees
( d D
53 . 10 0
49 . 1 )
01745 .
0 (
10 39
. ) 1
degrees
(
86
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
• With measures of distance and time, we can determine how fast something is moving
• A useful formula to remember
Distance = Velocity x Time
OOTETKdistance
velocity time
And Captain Scarlet says, “remember the triangle to”
And now for a moving experience….
• The distance between the Earth and Moon
– Measured by sending a burst of laser light to a reflector and timing the light travel time
Laser Ranging the Moon’s orbit
Light travel time to the Moon and back is about 2.563 seconds
Laser ranging from the Apache Point Observatory
TELESCOPES Earth
Moon
Time measurement of signal out + signal back
Distance by light
• Speed of light c = 3 x 10
8m/s
• A fundamental constant of physics - Change the speed of light and all physics and chemistry as we know it changes
OOTNTK
We know
taken time
traveled distance
speed OOTETK
For the Moon
:Distance = (time taken / 2) x speed of light
= (2.563 / 2) x 3 x 108 = 3.8445 x 108 meters = 384,450 km
Why ½ ?
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
• So D = c (T
1– T
0) / 2
D0
D1
D = expansion of orbit = D1 – D0
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 108 m/s
Time difference in signal travel times measured one year apart
0.25 x 10-9 Seconds
Hence: D = 3 x 10
8x (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
1 light year = speed of light x number of seconds in a year
= 3 x10
8x 365.25 x 24 x 60 x 60 (meters)
= 9.46 x 10
15meters
The Light Year – distance by time
Definition:
The distance traveled by a light ray in one year
I thought a light year meant only two classes
The speed of light c = 3 x 108 m/s – a fundamental constant
• 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