The Phases of the moonNew moon First Quarter Full moon Evening Sky

The Phases of the moonFull moon Third Quarter New moon Morning Sky

Eclipses of the Sun and Moon

The Cause of Eclipses: Shadows

• The Umbra & Penumbra

Two types of shadows

•A solar eclipse occurs when the Moon comes between the Earth and Sun. The Earth enters the Moon’s shadow.

•Solar eclipses only occur at New Moon.Solar Eclipses

Anatomy of a Solar Eclipse

Observers in the penumbra shadow will see a partial eclipse, Those in the Umbra shadow will see a total eclipse.

Totality:• During a total Solar Eclipse there are a number

of phenomena typically observed:– The sky darkens enough so that we can often

see bright stars in the sky.– Animals become quiet– The Sun’s corona (and prominences if present)

are observed– The diamond ring phenomena can occur.– Shadow fringes can be seen moving across the

ground.

The Diamond Ring

Eclipse: Solar From Space

Eclipses• In principal there should be an eclipse each new

and full moon if the earth-moon-sun system was properly aligned, but the Moon’s orbital plane is inclined about 5° with respect to the Ecliptic.

The Moon passes through the plane of the Earth’s orbit at two points on opposite sides called nodes.

Eclipses and Nodes

• To predict when an eclipse is likely to occur, we need to know where the line of nodes is in the sky.

• Eclipses can occur when the line of nodes is pointing toward the Sun.

• This happens twice a year, and lasts for ~ 1 month.

• These two months are called the “Eclipse Seasons”

Solar Eclipse Occurs by Node

Some times the Moon rides above the Sun, sometimes below.

Is the time of new moon within + or - 28 hours of node?

Is it within +or – 20 hrs of node ?

No eclipse

Partial Central

Is it within + or – 8 days of apogee ?

Annular

Solar Eclipse

Total

YES NO

NO YES

YES NO

Movement of the Nodes

• If the Moon’s orbit was fixed in the sky with Earth’s then the Eclipse season would always happen at the same time of year.

• But the orbital nodes precess with a period of roughly 18.6 years.

• This causes the Eclipse season to occur about 3 weeks earlier/year

Partial or Total?• Our location within the Moon’s shadow

determines whether we see a total or partial solar eclipse.

• The Moon’s umbra makes a circle generally about 170 miles in diameter on the surface of the Earth and the Moon’s orbital motion causes that shadow to sweep rapidly along the surface of the earth, and totality usually only lasts a few minutes.

Annular Eclipses

•Because the Moon and Sun are not a constant distance from the Earth, their angular size changes.

Annular Eclipses

• When the Moon’s angular size is too small

to completely cover

the disk of the Sun,

we observe an

Annular Eclipse.

Partial Solar Eclipse : The Moon moves in front of the sun

Lunar Eclipses• The most common eclipse seen on

Earth is a Lunar Eclipse

• Lunar eclipses occur at Full Moon when the Moon enters the Earth’s shadow.

Description of a lunar eclipse

• As the Moon enters the Earth’s penumbra, the disk shows only a small amount of change.

• When the Moon enters the Earth’s Umbra, the Lunar disk will appear to get smaller. Before the disk is completely dark it will become slightly redder,due to the scattering of light from the Earth’s atmosphere.

• When the Moon enters the Earth’s Umbra completely, the eclipse is said to have reached “totality”.

• An eclipse can last up to an hour and a half or even longer.

Lunar Eclipse:

Is the time of new moon within + or - 28 hours of node?

Is it within +or – 20 hrs of node ?

No eclipse

Penumbral Umbral

Lunar Eclipse

YES

NO YES

NO

Total Lunar Eclipse

Partial Eclipses

• If the Moon does not completely enter the Earth’s Umbra, then we say that eclipse is a partial eclipse.

• A penumbral eclipse occurs when the Moon only enters the Earth’s penumbra, they are not very impressive, and can be hard to observe

AU (astronomical unit)One AU is the average distance from which the Earth orbits the Sun. The AU is most commonly used for the distances of objects with in our solar system. The Earth is 1.0 au from the sun, and Neptune is a distance of 30.06 au from the Sun.

•AVERAGE EARTH-SUN DISTANCE

•15O x 106 KILOMETERS

•93 x 106 MILES

Distances in    Light years ly        The distance light travels in one year.         6 trillion miles = 1016 meters.         We are 8.3 light minutes away form the Sun.

Pluto is about 13 light hours.

The nearest star is 4.2 light years away

Sirius is 8.6 ly away

The Andromeda Galaxy is 2.4 million ly away

Measuring Distance• How can you measure the distance to

something? Direct methods, e.g. a tape measure. Not good

for things in the sky. Sonar or radar: send out a signal with a know

velocity and measure the time it takes for the reflected signal. Works for only relatively nearby objects (e.g. the Moon, Mercury, Venus Mars & certain asteroids).

Triangulation: the use of parallax.

Parsec: where a star shifts by 1 arcsec over a 1/2 year

Baseline (Earth’s orbit)

Dis

tan

ce

to S

tar

Parallax(Angle)

Parallax ~ 1Distance

““ParsecParsec” ” is short foris short for parparallaxallax arcarcsecsecondondBaseline

Parallax ~ Distance1

Calculating distance using Stellar Parallax

•Observe a star when the Earth is at point A -Star is in front of Star A

•Observe it again 6 months later when the Earth is at point B -Star is in front of Star B

1d

pThe formula is this simple. P is in parsecs (pc)

Take photos of a nearby star 6 months apart.

Measure angle in arc seconds.

Take ½ of the angle, this is p.1 pc = 3.26 ly

Sirius 28.036 pc or 8.6 ly

• Best resolution from Earth:– Measure angles as small as P = 0.03”– Then d = 30 pc = 98 ly– Results: about 2,000 accurate distances

• Best resolution from satellite:– Measure angles as small as P =0.005”– Then d = 200 pc = 652 ly– Results: about 1 million accurate distances

Stellar Distances•Stellar Parallax is very small, a fraction of a second. 1 pc = 206,265 AU or about 3.26 ly

Why is it so important to know the distance to a star? By knowing the distance to a star, one can find out a star’s luminosity, diameter, and mass.

Worked Problems

Angles in Astronomy are usually measured in deg, min, sec.

There are 60 min in a degree and 60 sec in a minute.

25 deg, 35 min & 12 seconds can be written : 25 35 12

APPARENT LINEAR AND ANGULAR SIZE OF OBJECTS

Small angles measure the ratio of width/distance.

Small Angle Approximation

The angle , can be approximated as : Diameter/d. When angles are extremely small, then the sine and tangent of the angle are approximately equal to the angle itself.

Using the small angle formula, we can calculate the angular size.

What is the angular size of the Sun or Moon?

Diameter

To solve this problem use the formula .

The number 206,265 is a constant that defines the angle in arcseconds.

Dia = ( Diameter) linear size of an object θ = angular size of the object, in arcsec d = distance to the object

Dia ( 206,265)θ

distance

The Moon has a diameter of 3,476 km and is 384,400 km from Earth. Let’s use the small-angle formula to determine the angular size of the moon from the earth.

=

Constant

(206,265)

(3,476)(206,265)

384,400

1,868 sec

1,86831 min

6031

.51deg60

Dia

d

arc s

arc

The sharpest eye can distinguish objects about apart or ..01 0.5

You could just tell if someone was holding up one or two fingers at 100 meters

Magnitude

Magnitude Description

1st The 20 brightest stars

2nd stars less bright than the 20 brightest

3rd and so on...

4th getting dimmer each time

5th and more in each group, until

6th the dimmest stars (depending on your eyesight)

First introduced by Hipparchus (160 - 127 B.C.)

•Brightest stars: ~1st magnitude

•Faintest stars (unaided eye): 6th magnitude

The magnitude scale was originally defined by eye, but the eye is a non-linear detector, especially at low light levels.

The Magnitude Scale

•Apparent Magnitude

The magnitude of a star as you see it in the sky.

–A smaller number means brighter!

–A larger number means dimmer!

•The Magnitude Scales are backwards:

Brighter

Vega 0.0 Procyon 0.38

61 Cygni 5.2 faintest galaxies ~ 29

Inverse Squared Relationship

• The brightness of a light source is inversely proportional to the square of the distance.

Light Intensity with Distance

STELLAR PHOTOMETRY• Astronomers determine the brightness of stars

using an instrument called a photometer.

With modern equipment, we can measure more accurately.

1st mag. stars apear 100 times brighter than 6th mag. stars

If two stars differ by 1 mag. their apparent brightness differ by a of factor 2.512

So a 1st magnitude star is :

2.512 times brighter than a 2nd magnitude star

2.512^2 = 6.31 times brighter than a 3rd mag star

2.512^3 =15.9 times brighter than a 4th mag star

2.512^4 = 39.8 times brighter than a 5th mag star

2.512^5 = 100 times brighter than a 6th mag star

1st mag. stars appears 100 times brighter than 6th mag. stars

If two stars differ by 1 mag. their apparent brightness differ by a of factor 2.512

The apparent magnitude of a star depends upon two things.

(1) How far away the star is

(2) How large the star is

If all stars were the same distance away,we could use their Apparent Brightness to judge their Actual Brightness.

Finding Absolute Magnitude (M)

If we could line up all the stars at the same distance, we could observe the true brightness , Absolute Magnitude, of the Stars.

What distance ? It doesn’t matter but everyone needs to use the same distance.

So 10 parsecs was chosen , which is 32.5 light years.

10 parsecs is just right! Simple, small, numbers

Most Absolute Magnitudes are positive

Few are very large

Absolute Magnitude Absolute magnitude is the measure of the

true brightness of a star if it were 10 pc away. Nothing special about 10 pc. This magnitude is called the Absolute Magnitude (M)

The distance formula:

m – M = 5 log(d)-5m is the apparent visual magnitude

M is the absolute magnitude

d is the distance to the star in parsec

a parsec = (3.26 LY)

•mv-Mv = -5 + 5 logd Distance Modulus = mv-Mv

M=-(5log250)+5-0.09

Log250=2.39794

M= -5(2.3794)+5-0.09

M= -11.897+5-0.09= -6.987

Problem #1

Example 2

A star has an absolute magnitude of –5.0 and is located 420 ly from Earth. Find the apparent magnitude.

m – M = 5 log(d) - 5

m- (-5.0) = 5log(128.8) – 5

m = 5(2.1099) - 10

m = 0.55

Example 3

The bright star Sirius has an apparent magnitude of –1.46 and an absolute magnitude of +1.4. How far is the star from the Earth?

m – M = 5 log(d)- 51.46- 1.4=5log(d)-5

-2.86+5=5 log(d)

2.14/5 = log d or log d=0.4280.42810

2.679

d

d pc

or 8.7 ly

Thanks to the following for allowing

me to use information from their

web site :

Nick Stobel

Bill Keel

Richard Pogge

John Pratt

NASA, JPL, OSHO