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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)

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

(3)

• 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

(4)

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)

(5)

Earth’s spin axis

Center of rotation on the Celestial Sphere

(6)

24 hour continuous star trails from the South Pole

SCP

(7)

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

(8)

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

(9)

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

(10)

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

(11)

• 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

(12)

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…

(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)

Celestial equator NCP Ecliptic

(14)

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

(15)

The Earth and Moon as seen by the MESSENGER spacecraft (now buried in an impact crater on Mercury)

(16)

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

(17)

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

(18)

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

(19)

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

(20)

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

(21)

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

(22)

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

(23)

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

(24)

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

(25)

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

(26)

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

(27)

• 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)

(28)

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

(29)

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

( dD

 

53 . 10 0

49 . 1 )

01745 .

0 (

10 39

. ) 1

degrees

(

8

6

 

(30)

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

(31)

• 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

velocity time

And Captain Scarlet says, “remember the triangle to”

And now for a moving experience….

(32)

• 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

(33)

Laser ranging from the Apache Point Observatory

TELESCOPES Earth

Moon

Time measurement of signal out + signal back

(34)

Distance by light

• Speed of light c = 3 x 10

8

m/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 ?

(35)

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

(36)

• 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)

(37)

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

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

(38)

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

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

(39)

• 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

References

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