Download Calculate the Mass of the Milky Way Galaxy

Final Test Review
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Kepler’s Laws
Terrestrial Planets
Olbers’ Paradox
Dark Matter
Universe
Review for Final Exam
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Hubble Law
Cepheid Variables
Mass of Galaxy
Videos
Terraforming Mars
Hazardous Objects in Space
Review for Final Exam
• Moon’s existence
• Plate Tectonics
• Old Friends from Test 1
Review for Final Exam
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
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 = 18.9E8 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
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Mass = (distance)^3 divided by (time)^2
M=(18.9E8)^3 / (2.5E8)^2
M=6.75E27 / 6.25 E 16
M= 1.0 E 11 Solar Masses
Translation, there are 100,000,000,000 stars in our galaxy!
~100 billion Suns in the Milky Way!
REDSHIFT AND RECESSIONAL VELOCITY
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RECESSIONAL VELOCITY,v,
CAN BE FOUND FROM HUBBLES LAW:
p. 590 Caisson, A-T
EXAMPLE:
REDSHIFT =
6,500 km/s
DIVIDED BY:
300,000 km/s
= 6.5/300 =
= 0.022 REDSHIFT
CEPHEID VARIABLES ANOTHER STANDARD CANDLE
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A CEPHEID HAS A PERIOD OF 3 DAYS (p. 524 Caisson)
LUMINOSITY = 1,000 Solar Units (Fig. 23.7)
“F” STAR - Fig. 23.6
Absolute Mag = 0
OBSERVED APPARENT BRIGHTNESS = LUMINOSITY (Abs. Mag.)
FROM CEPHEID PEROD
• DIVIDED BY
• DISTANCE
HUBBLE’S LAW
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CURRENT HUBBLE CONSTANT = 65 km/s/Mpc
CHANGES AS WE SPEAK
P. 568
GALAXY IS 100 Mpc DISTANCE
RECESSIONAL VELOCITY (BY HUBBLE’S LAW) IS:
65 km/s/Mpc x 100 Mpc = 6500 km/s
Earth Unique in the Solar System
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Total Eclipse of Sun
Rainbows
Snowflakes
Liquid surface water
Life
– Diversity
– 50 million species
Unique Earth
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1 AU
Ozone layer
Technology society
Elements > 92
Hailstones
Solar constant 750w/sqyd
icebergs
Complex fossils
Fossil fuels
Salt mines
Gypsum
2 inferior planets
6 exterior planets
Extinct species
Earth Unique in the Solar System
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Total Eclipse of Sun
Rainbows
Snowflakes
Liquid surface water
Life
– Diversity
– 50 million species
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Air (79%N2, 21%O2)
1 bar
50 deg F mean Temp
3rd Rock from Sun
365 days to revolve
Blue Sky
Earth Unique in the Solar System
• Fossil fuels
• Active plate tectonics
Stellar Spectral Classes
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P. 390 caisson
Fig. 17.12
Notice absorption lines in the spectra
Astronomers get their info from the spectra
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P. 391
Table 17.2
Familiar examples
Rigel, abs mag –6.8
Page A-5, Table 4
Ancient Astronomy
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Gnomon
Klepsydra
Polos
Ephemerides
sun
Sundail
Water clock
Measure shadows
Table of future celestial events
king
Ancient Astronomy
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Sirius (DOG STAR)
Eratosthenes
Gathered lots of star data
Hipparcus
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Rises before the flooding of the Nile
Measured
circumference of Earth
Mesopotamians/Egyptians
Precession of the equinoxes (Astrology downfall ) Earth wobbles once/25000years.
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
Terraforming Mars
Terraforming Mars
• Oxygen
– Plants
Atmospheric pressure = 1 bar
CO2, O2,
Work on ozone layer
Work on bad storms (underground cities)
Work on soil compostion
Cows,concrete, corn, tomato
Cost= > savings and loan bailout
The further an object is from the Earth,
the faster it is traveling away
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Galaxies are not merely stationary objects in space, they move. American
astronomer Edwin Hubble (1889-1953) discovered an important
characteristic of their motion -- that the Universe itself is expanding and
galaxies are also traveling away. Hubble discovered in 1929 that the
further a galaxy was from the Earth, the faster it was moving away and
that the closer a galaxy was to Earth the slower it moved.
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Hubble discovered that the more distant a
galaxy was,
the greater its redshift, and therefore the
higher its velocity
• While at the Wilson Observatory, Hubble studied galaxies, measuring the
length of light waves in distant galaxies and discovering that the waves
were longer and redder (redshift). He then calculated the distance to the
galaxy and its redshift and found that there was a proportional relationship
between the two. This relationship was that the more distant a galaxy, the
greater its velocity. This discovery also confirmed the Universe is
expanding.
Hubble’s law
• http://spaceboy.nasda.go.jp/note/Kagaku/E/kag101_hubble_e.ht
ml
http://spaceboy.nasda.go.jp/note/Kagaku/E/Kag1
3_e.html
Edwin Hubble
• Edwin Hubble
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• Lawyer-turned-astronomer
who photographed the galaxies
• Edwin Hubble (1889-1953) was an American astronomer. He was originally a lawyer
but turned his attention to astronomy after taking postgraduate studies at the
University of Chicago. After World War I, Hubble started work at the Wilson
Observatory, taking photographs of galaxies through the lens of a massive telescope.
At about the same time, another American astronomer discovered that the Andromeda
Nebula was moving away from the Earth.
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• Ripples caused by Einstein's space theory
• In 1916, Albert Einstein announced his general theory of relativity and the following
year produced his model of space based on that theory. Einstein argued that the
universe was immobile, but Dutch astronomer Willem de Sitter calculated Einstein's
equation and proved that the universe was actually expanding. In 1922, Russian
physicist Alexander Friedmann used Einstein's equations to prove that the universe
could either shrink or expand.
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• Hubble's rule of an expanding universe
• During the uncertainties of the era, Hubble was able to observe galaxies at distances
up to 7 million light years away. By doing so he was able to come up with Hubble's
Law, which said that the further galaxies were away from earth the faster they moved
away from our planet. Hubble's rule proved the universe was expanding like a big
balloon. In 1930, Einstein visited Wilson Observatory and viewed photos of galaxies
taken by Hubble. After seeing the photographs, Einstein gave up his theory of an
immobile universe for all time. The orbiting space telescope observing the universe is
named after Hubble.
Expansion Factor
• In 1929, Hubble estimated the value of the expansion factor,
now called the Hubble constant, to be about 50 km/sec/Mpc.
Today the value is still rather uncertain, but is generally believed
to be in the range of 45-90 km/sec/Mpc.
Hubble's Law
• The dominant motion in the universe is the smooth expansion known as
Hubble's Law.
• Recessional Velocity = Hubble's constant times distance
• V = Ho D
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V is the observed velocity of the galaxy away from us, usually in km/sec
H is Hubble's "constant", in km/sec/Mpc
D is the distance to the galaxy in Mpc
10 Mpc, V=50*10=500 km/sec
V=50*20 =1000 km/sec
Example
• V = Ho D
• V = 75 km/sec/Mpc times .306Mpc equals 23 km/sec or 82,800
km/h
• Where 1000000 ly = .3 Mpc
• 1Mly =1000000 ly
• 3.26 ly = 1 pc
• 1/3.26 =.3
Hubble’s Law Graph
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Few of the points fall exactly on the line. This is because all galaxies have some additional residual
motion in addition to the pure expansion. This is referred to as the "cosmic velocity dispersion" or
"cosmic scatter" and is probably due to the fact that the gas clouds that formed the galaxies all had some
small additional motion of their own. The recessional velocity of a galaxy at a particular distance
inferred from Hubble's law is called the "Hubble velocity".
• The Hubble Constant is generally expressed in units of
kilometers per second per megaparsec: between 100 and 50, or
approximately 70,000-35,000 mph/mly. In other words, a galaxy
that's 1 million light years away will appear to recede from us at
35,000-70,000 miles per hour.
• Another manner of expressing the Hubble Constant is:
• H = (0.5 to 1)/(10^10)years
Olbers’s Paradox
• Pixel – picture element
• Night sky should be bright as a star since there are 100 billion
billon stars
• Observational evidence says it is not.
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
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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
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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
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Test 1 Review
Black Hole
Test 1 Review
More
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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
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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
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$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
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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
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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
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Black Hole
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Extremely dense point = Singularity –
Billion tons/golf ball
Enormous gravity field
Light cannot escape from a black hole
Looks like a funnel
Top of Funnel = Event Horizon
Bottom of the funnel
Black Holes come from Collapsed Massive Stars
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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
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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
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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:
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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
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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:
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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
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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
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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
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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
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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
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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!
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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!
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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
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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
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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
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•
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•
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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)
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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
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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
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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
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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