Download Eclipses, Distance, Parallax, Small Angle, and Magnitude (Professor

The Phases of the moon
New moon  First Quarter  Full moon
Evening Sky
The Phases of the moon
Full moon  Third Quarter  New moon
Morning Sky
Eclipses of the Sun and Moon
Solar Eclipses
•Solar eclipses only occur at New Moon.
The
Cause
of
Eclipses:
Shadows
•A solar eclipse occurs when the Moon comes
between the Earth and Sun. The Earth enters the
• The
Umbra & Penumbra
Moon’s
shadow.
Two types of shadows
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”
Some times the Moon rides above the Sun, sometimes below.
Solar Eclipse Occurs by Node
Solar Eclipse
Is the time of new moon within + or - 28 hours of node?
YES
NO
Is it within +or –
20 hrs of node ?
NO
Partial
No eclipse
YES
Central
Is it within + or – 8 days of apogee ?
YES
NO
Annular
Total
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:
Lunar Eclipse
Is the time of new moon within + or - 28 hours of node?
YES
NO
Is it within +or –
20 hrs of node ?
NO
Penumbral
No eclipse
YES
Umbral
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
Parallax ~
Baseline
Distance
1
1
Distance
Parallax
(Angle)
Distance
to Star
Parallax ~
“Parsec” is short for
parallax arcsecond
Baseline
(Earth’s orbit)
Calculating distance using Stellar Parallax
Take photos of a nearby star 6 months apart.
•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
Measure angle in arc seconds.
Take ½ of the angle, this is p.
The formula is this simple.
Sirius 28.036 pc or 8.6 ly
1
d 
p
1 pc = 3.26 ly
P is in parsecs (pc)
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.
• 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
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
ANDof
ANGULAR
SIZE OF
What is the
angular
size
the Sun
orOBJECTS
Moon?
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.

Diameter
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
Constant
from the earth.
To solve this problem use the θ = Dia ( 206,265)
distance
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)

d
(3, 476)(206, 265)

384, 400
  1,868arc sec s
1,868

 31arc min
60
31

 .51deg
60
The sharpest eye can distinguish objects
about .01 apart or 0.5 .
You could just tell if someone was
holding up one or two fingers at 100
meters
Magnitude
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 nonlinear detector, especially
at low light levels.
The Magnitude Scale
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)
•Apparent Magnitude
The magnitude of a star as you see it in the sky.
•The Magnitude Scales are backwards:
–A smaller number means brighter!
–A larger number means dimmer!
Brighter
Vega 0.0
61 Cygni 5.2
Procyon 0.38
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
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
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
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)
a
parsec
=
The distance formula:
m – M = 5 log(d)-5
(3.26 LY)
m is the apparent visual magnitude
M is the absolute magnitude
d is the distance to the star in parsec
•mv-Mv = -5 + 5 logd
Distance Modulus = mv-Mv
Problem #1
M=-(5log250)+5-0.09
Log250=2.39794
M= -5(2.3794)+5-0.09
M= -11.897+5-0.09= -6.987
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)- 5
1.46- 1.4=5log(d)-5
-2.86+5=5 log(d)
2.14/5 = log d or log d=0.428
d  10
d  2.679 pc or 8.7 ly
0.428
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