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

• 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

•

• 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• • •

Apparent Magnitude

What you see

Venus App Mag = ~ 4

Jupiter App Mag =~ 2

Saturn App Mag =~ 0

Polaris App Mag =~ 2

Limit of Naked Eye App Mag = 6

(In the country, away from light)

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 3rd 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 Intensity– (+1) 2.5

• (+1.5) 4

– (+2) 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

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•

•

•

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

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•

•

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

http://www.gi.alaska.edu/ScienceForum/aurora.html

The Northern LightsThe Northern Lights

By: Dawn Yanzuk

. 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 HomeArticle #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

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