Download Summary of Astronomy

Summary of Astronomy
•
•
•
•
Test 1 Review
Kepler’s Laws
Test 1 Review
Spectra
– Continuous
– Absorption
– Continuous
• Test 1 Review
• Compute distance to a star using parallax
– arcsec
Apparent Magnitude
What you see
Venus App Mag = ~ 4
Jupiter App Mag =~ 2
Saturn App Mag =~ 0
Polaris App Mag =~ 2
Apparent Magnitude is Unfair
• If a dwarf star is nearby, it seems brighter
• If a supergiant star is far away, it appears dimmer
• These apparent sights are governed by the Inverse
Square Law of Light Intensity
– As you get closer to a star, it gets really bright, really
fast
– As you move away from a star, it gets really dim, really
fast
– Recall doubling time example
Doubling Time (An Example of a
square function)
• Would you take a job
which paid a penny on
the first day, and the
pay would double
every day for 30 days?
• 1 penny on day 1
• 2 pennies on day 2
• 4 pennies on day 3
•
•
•
•
•
•
•
8 pennies on day 4
16 pennies on day 5
32 pennies on day 6
64 pennies on day 7
$1.28 on day 8
$2.56 on day 9
$5.12 on day 10
• $10.24 on day 11
• $20- on day 12
(rounded)
• $40- on day 13
• $80- on day 14
• $160- on day 15
•
•
•
•
•
Binary Star Systems
• Most stars in the universe have companion stars –
these are called binary stars
• The bigger star is called the primary while the
smaller star is the companion
• We are able to measure the time (years) it takes for
the companion to orbit the primary simply by
observation
• We can also measure the separation distance
between the primary and the companion
Barycenter
• Other names are fulcrum or center of mass
• The barycenter is located between the two stars
• The barycenter is closer to the primary (larger
mass) star
• One can find the barycenter from photographic
plates
• One can measure the distance from the barycenter
to the primary star (shorter than the distance from
the barycenter to the companion star
Binary Star Separation Distance
• One can measure the separation distance
between the primary and companion stars
• One can compute the distance between the
barycenter and the primary star
Compute the distance from the
barycenter to the primary star
Compute the distance (x) from
the barycenter to the primary star
• The separation distance between the 2 stars
is 12 AU
• We know the companion is 5 times farther
from the barycenter than the primary
• 5 times the distance from the barycenter to
the primary star is (5 times x) or 5x.
• The distance from the barycenter to the
companion must be equal to (12-x)
The computation
•
•
•
•
•
5x = 12 – x
5x +1 x = 12 – x + x
6x = 12
X = 2 AU (answer)
The primary star is 2 AU from the
barycenter
• What is the distance from the barycenter to
the companion?
10 AU
• Checks: 5x = 5 * 2 = 10 AU
•
• .
• .
•
Black Hole
• Extremely dense point = Singularity –
Bottom of the funnel
• Billion tons/golf ball
• Enormous gravity field
• Light cannot escape from a black hole
• Looks like a funnel
• Top of Funnel = Event Horizon
Black Holes come from
Collapsed Massive Stars
• Massive > 8 solar masses
• F=(M1xM2)/dist^2 (Newton’s universal law of
gravitation
• Massive star has to go supernova
• Supernova is a star which explodes
• Proton-proton chain reaction = Nuclear Fusion
• Burn H into He – Sun wants to explode all the
time
Proton-Proton (Nuclear Force) v
Gravity
• Nuclear furnace goes out when runs out of fuel
• H->He->O->…->Fe
• Star core fusion produces energy until it starts to
burn Fe . . . Now it requires energy!
• The core no longer can win against gravity
• Star collapses into the core
• Rebounds
• Spews elements 1-92 into space
Neutron Star
• Something left over in the core is either a
neutron star or , if the progenitor star was
very massive, a black hole.
Element Synthesis
• Elements 1-26 are forged in ordinary stars
• Heavy elements up to 92 are formed by
supernovas
• Strong Nuclear Force (double-sided sticky tape)
holds the protons (+) together
• Protons that get smashed together stay together
• Supernovas occur every second in the universe
Space-Time Continuum (the
fourth dimension)
• Arrow of time points forward
• Law of Entropy – disorder increases with time
(things left to themselves)
• Masses on a space-time continuum for an indent
called a gravity well and it loks like a funnel
•
•
•
•
•
Calculate the Mass of the Milky
Way Galaxy
• Use Newton’s form of Kepler’s third law
• Mass of Galaxy = (Distance from Sun to
Center of the Milky Way)^3 divided by the
(Time for the Sun to orbit the center of the
Milky Way)^2
Distance from the Sun to the
Center of the Milky Way Galaxy
• Distance must be in Astronomical Units
(AU)
• 1 AU = 93,000,000 miles
• Distance from the Sun to the Center of the
Milky Way Galaxy = 9,000 PC
• 9,000 PC * 2.1E5 = 18.5E8 AU
Orbital Period of the Sun about
the Center of the Milky Way
– 2.5E8 years for the Sun to orbit once about the
center of the Milky Way Galaxy
– Time squared = (2.5E8)^2 = 6.25E16
Newton’s form of Kepler’s Third
Law
•
•
•
•
•
Mass = (distance)^3 divided by (time)^2
M=(18.7E8)^3 / (2,5E8)^2
M=6.35E27 / 6.25 E 16
M= 1.1 E 11 Solar Masses
Translation, there are 110,000,000,000 stars in our
galaxy!
• ~100 billion Suns in the Milky Way!
•
•
•
Determination of Stellar Masses
• Given:
–
–
–
–
Orbital Period = 30 years
Maximum separation = 3” (arcsec)
Trigonometric Parallax = 0.1”
Companion is 5 times farther from barycenter
Determination of Stellar Masses
• Find:
– Combined mass of companion and primary
– Individual mass of primary, M1, and
companion, M2
Determination of Stellar Masses
• M1+M2=(3/0.1)^3/(30)^2
• M1+M2 = 30 Solar Masses
• By lever logic:
– M1 = 25 Solar Masses
– M2= 5 Solar Masses
–
–
–
–
–
Kepler’s Laws
• 1. Planets have an elliptical orbit
• 2. Equal time sweeps out equal area
• 3. Time squared = distance cubed
Kepler’s Third Law
• Time squared = distance cubed
• Time = years to revolve around the Sun
• Distance is from the Sun to the planet of
interest (Astronomical Units – AU)
• Example:
– Next Slide Please
Kepler’s
rd
3
Law
• Example:
– Jupiter takes ~12 years to revolve one time around the
Sun
– 12 squared = 144
(Time squared)
– Dist^3 = 144
– Cube root of both sides
– Distance from the Sun = cube root of 144 or
– 5.2 Astronomical Units (AU)
–
–
–
Absolute Magnitude
• “True” magnitude since distance to the
Earth is eliminated
• Standard distance from Earth is 10 parsecs
(pc)
• We pretend the star is at 10 pc from Earth
when we assign an absolute magnitude
Absolute mag used for finding
Distance
• By comparing absolute and apparent mags,
we can find the distance of the star from
Earth
• Find difference in magnitudes
• Find difference in luninosity
• Take square root
• Multiply by 10 pc to get distance
Absolute Magnitude
• “Moving” a star from 2 pc to 10 pc would
make the star seem dimmer to earthlings
• “Moving” a star from 20 pc to 10 pc would
make the star seem brighter to earthlings
Absolute Magnitude
• “Moving” a star from 2 pc to 10 pc would make
the star seem dimmer to earthlings
• It would seem inverse of 5 squared (1/25th) as
bright (luminosity) at 10 pc
• This star moves 5 times its original distance (5 *
2pc = 10pc)
• That (25) is equivalent to between 3 (16) and 4
(40) magnitudes dimmer (say 3.5 magnitudes)
• So if its apparent mag was 2, the absolute mag
would be 5.5 (2+3.5=5.5)
Find Distance using Delta Mags
• The apparent mag of a star is 2 and the absoute
mag is 5.5. Find the distance to the star.
• Since the absolute mag is dimmer than the
apparent mag, we know the star has to be closer to
us than 10 pc
• Delta mags is 5.5-2 = 3.5
• Pogson scale says 3.5 mags is ~ 25
• Square root of 25 is 5
• Star is 1/5th of 10 pc from earth or 2 pc distance
•
•
Consideration of Mass in a
Binary Star System
• The larger mass star (Primary) is closer to
the barycenter than the companion
• One can use ratios similar to lever ratios
• If the primary is 2 units from the barycenter
and the companion is 10 units from the
barycenter, what is the lever ratio?
• 10/2 = 5
• 1:5 is the ratio
Applying the lever ratio to
masses
• If we have a lever ratio of 1:5, then the
distance from the barycenter to the companion
would be 5 times more than the distance
from the barycenter to the PRIMARY.
• In order to be in equilibrium, the mass of
the companion x 5 must be equal to the
mass of the PRIMARY x 1.
Example
• The total mass of a binary star system is 24
solar masses (given).
• The lever ratio is 1:5 (PRIMARY:companion)
• Mass of PRIMARYx1 = mass of companionx5
• Mass of PRIMARY must be 5 times more
than the mass of the companion
• Notice that 1:5 is the same ratio as 4:20
Equivalent Ratios
•
•
•
•
•
•
•
1:5
2:10
3:15
4:20
5:25
6:____?
7:____?
Apply Ratio to Masses
• 24 solar masses (given)
• 1:5 lever ratio (which you calculated)
• What 2 numbers which add up to 24 will
also match the 1:5 lever ratio?
• Try 2&10. No, adds up to 12
• Try 3&15. No, adds up to 18
• Try 4&20. Yes! Adds up to 24!
Answer
• Given: 24 solar masses is the total mass of
the binary star system
• Lever ratio is 1:5 (You computed this)
• Mass ratio is 4:20 (Based on 1:5 & 24)
• Companion mass must be 4 solar masses &
Primary mass must be 20 solar masses.
Check the Answer
• Distance from the PRIMARY to the
barycenter was 2 AU.
• Mass of PRIMARY was 20 solar masses
• Product of PRIMARY mass x distance =
• 20 x 2 = 40
• Remember the 40!
• Distance from the companion to the barycenter
was 10 AU.
• Mass of companion was 4 solar masses
• Product of companion mass x distance =
• 4 x 10 = 40
• Remember the 40?
• It checks!
•
•
•
Parallax
• Used for determining distances
• Your eyes are a few inches apart which
allows you to judge distances
• Earth is at different positions in its orbit eg
January and June
• Distance (pc) = inverse of parallax angle
(arcsec)
Find the distance knowing the
parallax angle
•
•
•
•
•
•
•
•
•
A star has a parallax angle of 0.2 arcsec
Find the distance to the star:
Distance = 1/parallax angle
Distance = 1/0.2
Distance = 5 pc
Earth Unique in the Solar System
•
•
•
•
•
Total Eclipse of Sun
Rainbows
Snowflakes
Liquid surface water
Life
– Diversity
– 50 million species
Earth Unique in the Solar System
•
•
•
•
•
Total Eclipse of Sun
Rainbows
Snowflakes
Liquid surface water
Life
– Diversity
– 50 million species
•
•
•
•
•
•
Air (79%N2, 21%O2)
1 bar
50 deg F mean Temp
3rd Rock from Sun
365 days to revolve
Blue Sky
Practice problem
• . You measure a separation of 0.5
arcsec between two images of the same
star. You took the photo images 6
months apart. How far from us is the
star?
Answer
• Distance = 1/separation (arcsec)
• Distance = 1/0.5
• Distance = 2 pc
Practice #2
• . A star has an apparent magnitude of
5. It is 20 pc from us. What is its
absolute magnitude?
Answer - #2
• The absolute magnitude will be a smaller number
than 5 because the star will “move” to 10 pc (the
standard distance from Earth) from 20 pc.
• Since the star will be “closer”, it will be brighter.
• A brighter star has a smaller magnitude
• Thus, we expect an absolute magnitude less than
5.
Answer #2 – cont.
• Since the star “moves” to a distance 1/2 of
its original distance, the luminous intensity
will be the square of ½ or ¼.
• Take the 4 to the Pogson scale
– Mag
– (+1)
Intensity
2.5
• (+1.5)
– (+2)
4
6
Answer # 2 – cont.
• Therefore, we subtract 1.5 magnitudes from the
original apparent magnitude of 5.
• 5 – 1.5 = 3.5 ( the absolute magnitude)
•
•
•
•
•
Newton’s Form of Kepler’s Third
Law
• Combine Mass with Distance and Time
• Mass = (distance)^3 divided by (time)^2
•
•
•
•
Proving that the Earth Revolves
• Nearby stars exhibit stellar parallax
• Nearby stars are less than 100 PC away
• Only way to get parallax is if Earth has a
baseline (It does and it is equal to 2 AU)
• Ergo, Earth must revolve about the Sun
Alternate Proof of Revolution
• Roemers experiment about the speed of
light
•
•
•
Proving that the Earth Rotates
•
•
•
•
•
•
•
•
Foucault Pendulum
Pins are on the Earth
Bob always moves in a North-South plane
Pins get knocked over, ergo, the EARTH
ROTATES
Scientific Method
•
•
•
•
•
•
•
•
•
•
1. Observations, data
2. Hypothesis
3. More test
4. Occam’s Razor
5. Form a theory
6. Publish, Test of Time
7. Law of Science
Terraforming Mars
• Raise Mean Temp
• Polar caps
•
•
•
•
•
•
•
•
Test 1 Review
• Kepler’s Laws
Test 1 Review
• Spectra
– Continuous
– Absorption
– Continuous
Test 1 Review
• Compute distance to a star using parallax
– arcsec
Test 1 Review
• Telescopes
– Refracting
– Reflecting
Test 1 Review
• Fundamental forces
Test 1 Review
• Proton-Proton Chain reaction in Sun
Test 1 Review
• Apparent Magnitude
Test 1 Review
• Absolute Magnitude
Test 1 Review
• Compute distance knowing absolute and
apparent magnitudes
• Compute apparent magnitude knowing
absolute magnitude and distance.
• Etc.
Test 1 Review
• Compute distance of PRIMARY to
barycenter
• Compute distance of companion to
barycenter
Test 1 Review
• Compute mass of PRIMARY
• Compute mass of companion
Test 1 Review
• Quantum theory
• Electron energy levels
• Spectral lines
Test 1 Review
• Local Star Time
Test 1 Review
• Black Hole
Test 1 Review
• More
The Northern Lights
By:
Dawn Yanzuk
http://www.gi.alaska.edu/ScienceForum/aurora.html
The image provides an estimate of
location, extent, and intensity of the
aurora borealis.
.
The image provides an estimate
of location, extent, and intensity
of the aurora australis.
Alaska Science Forum
May 19, 1999
If the Light's on, Somebody May be Home
Article #1441
by Ned Rozell
The recent discovery of three planets orbiting a sun-like star once again raises one of
mankind's favorite questions: can life exist on other planets?
Someday, the aurora might help find an answer, according to Syun-Ichi Akasofu,
director of both the Geophysical Institute and the International Arctic Research Center
at the University of Alaska Fairbanks. Akasofu has studied the aurora since 1957, and
he thinks other planets' auroras could be used to detect life elsewhere in the universe.
The idea was inspired by the discovery of three planets circling a sun that's much like
ours, though a bit bigger. The star, Upsilon Andromedea, is 44 light years away, which
means a spacecraft moving at the speed of light would take 44 years to get there. Still,
it's not too far away by astronomical standards (the diameter of the Milky Way is about
100,000 light years).
The most significant part of the discovery is that it shows our solar system is not unique:
maybe there's a planet out there just like Earth, providing a home for people or dinosaurs
or beetles or bacteria or something we can't even imagine.
The possible connection between extraterrestrial life and aurora first came to Akasofu
when he spoke to an Elderhostel group. Akasofu fielded a question from a man who
wanted to know if the aurora would change color if humans polluted the atmosphere
enough to change the concentration of its gases. The question is valid because gases
create the colors of the aurora. Green is the most common color in the aurora because
Earth has a lot of oxygen in its atmosphere. This oxygen, exhaled by plants and animals,
finds its way to the upper reaches of the atmosphere. When oxygen molecules and atoms
floating about 60 miles above Earth are struck by particles from the sun, they glow
green, and people lucky enough to live near the poles see aurora.
Earth has aurora because it meets the two required conditions; it has a magnetic field
(shaped like the one surrounding a bar magnet) and that magnetic field is struck by the
solar wind, a stream of particles and gas from the sun. The interaction between the
magnetic field and the solar wind creates the magnetosphere, an invisible, comet-like
structure around Earth.
In the early days of space physics research, there were hints that other planets had
auroras, too. In 1955, scientists discovered that Jupiter was giving off radio emissions,
which suggested the planet might have a magnetic field. In 1974, the Pioneer-10
spacecraft had instruments aboard that detected a magnetosphere on Jupiter. Five years
later, the spacecraft Voyager captured the first image of Jupiter's aurora.
Since that time, scientists found aurora on Saturn. The Hubble Space
Telescope allowed scientists to see that both Jupiter and Saturn have
auroras colored pink from hydrogen in those planets' atmospheres. Neither
planet has green auroras because neither planet has oxygen, and oxygen, as
far as we know, is only given off by living organisms.
If life in the forms we're familiar with exists on some other planet, that
planet will have oxygen in the upper reaches of its atmosphere. If that
planet has the magnetism to support an aurora, it should at least
occasionally be green. And a green aurora 100 light years away will be
much easier to detect than a wandering stegosaurus at the same distance.
.
THE END
.
.