Calculate the Mass of the Milky Way Galaxy



Final Test Review

Kepler’s Laws

Terrestrial Planets

Olbers’ Paradox

Dark Matter

Universe

Review for Final Exam

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

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

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

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

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

Total Eclipse of Sun

Rainbows

Snowflakes

Liquid surface water

Life

Diversity

50 million species

Unique Earth

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

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

Earth Unique in the Solar System

Fossil fuels

Active plate tectonics

Stellar Spectral Classes

P. 390 caisson

Fig. 17.12

Notice absorption lines in the spectra

Astronomers get their info from the spectra

P. 391

Table 17.2

Familiar examples

Rigel, abs mag –6.8

Page A-5, Table 4

Ancient Astronomy

Gnomon

Klepsydra

Polos

Ephemerides

sun

Sundail

Water clock

Measure shadows

Table of future celestial events

king

Ancient Astronomy

Sirius (DOG STAR)

Eratosthenes

Gathered lots of star data

Hipparcus

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 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

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

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.

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





Edwin Hubble

Edwin Hubble

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.

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.

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

where

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

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

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

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

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