ASTROPHYSICSIB PHYSICS | UNIT 13 | ASTROPHYSICS

The Scale of AstrophysicsIB PHYSICS | UNIT 13 | ASTROPHYSICS

13.1

IB Physics Data Booklet

Anything look familiar?

𝑃 = 𝑒𝜎𝐴𝑇4 λmax(metres) =2.90 × 10−3

T (kelvin)

𝐼 =𝑃

𝐴=

𝑃

4𝜋𝑟2

It’s a Great Big Universe…

The Solar System

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

Diameter[km]

4,880 12,104 12,756 6,787 142,800 120,000 51,800 49,500

Distance to Sun

[× 108 m]58 107.5 149.6 228 778 1,427 2,870 4,497

Poor Pluto… :’(

A celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.

Definition of a Planet (as of 2006)

Measuring Distances

Speed of Light: c = 3.00 × 108 m s-1

How far is one light year (ly)

3.00×108 𝑚

1 𝑠×

60 𝑠

1𝑚𝑖𝑛×

60𝑚𝑖𝑛

1 ℎ𝑟×

24 ℎ𝑟

1 𝑑𝑎𝑦×

365 𝑑𝑎𝑦𝑠

1 𝑦𝑟

= 9.46 × 1015 𝑚 𝑦𝑟−1

Measuring Distances

1.5 × 1011 m

Average Distance between the Earth and the Sun

Astronomical

Unit

[AU]

Measuring Distances

1 light year (ly) = 9.46 × 1015 m The distance that light travels in an earth year

1 astronomical unit (AU) = 1.50 × 1011 m The average distance between the earth and the sun

Measuring Angles

60 minutes in 1°

60 seconds in 1 minute

1° = 3600 arcseconds

1° = 3600 as

‘

“

Parsecs

1 parsec (pc) = 3.26 lydistance at which the mean radius of the earth's orbit subtends an angle of one second of arc.

parsec stands for parallax angle of one second

1 second

Parsecs

“It's the ship that made the Kessel run in less than 12 Parsecs.”

Why is this claim a little odd?

distance

IB Physics Data Booklet

Measuring the Universe

The Black Hole – by the numbers

55 million light years from Earth

6.5 billion times the mass of the Sun

Just bigger than our solar system

The Science Behind the Picture

Remember: 1° = 3600 arcseconds

1 arcsecond = 1,000,000 μas

*60 μas is about the same as resolving a picture of an orange sitting on the surface of the moon

The Science Behind the Picture

Stellar QuantitiesIB PHYSICS | UNIT 13 | ASTROPHYSICS

13.2

Measuring Distances

1 light year (ly) = 9.46 × 1015 m The distance that light travels in an earth year

1 astronomical unit (AU) = 1.50 × 1011 m The average distance between the earth and the sun

1 parsec (pc) = 3.26 lydistance at which the mean radius of the earth's orbit subtends an angle of one second of arc.

IB Physics Data Booklet

Stellar Parallax

Stellar Parallax

𝑑 𝑝𝑎𝑟𝑠𝑒𝑐 =1

𝑝 (𝑎𝑟𝑐 𝑠𝑒𝑐𝑜𝑛𝑑)

p (arc second)

Stellar Parallax

The angle must be measured to a very distant field of other stars

*The parallax method only works for stars that are relatively close to earth

Try This… | #1

The star Betelgeuse has a parallax angle of 7.7 × 10-3 arc seconds. Calculate its distance.

arc-seconds→ parsecs→ light years →meters

𝑑 =1

𝑝=

1

7.7 × 10−3 𝑎𝑟𝑐 𝑠𝑒𝑐𝑜𝑛𝑑𝑠= 129.9 𝑝𝑎𝑟𝑠𝑒𝑐𝑠

129.9 pc ×3.26 ly

1 pc×9.46 × 1015 m

1 ly= 4.0 × 1018 m

Luminosity vs Brightness

Luminosity L Brightness b

Intensity

W/m2 or W m-2

*Depends on the observer distance

Power Emitted

Watts [W]

Brightness

𝑏 =𝐿

4𝜋𝑑2Distance Brightness

x b

2x b/4

3x b/9

4x b/16

Distance

from star

Same Brightness, Different Stars

It is possible for stars to have the same brightness but have different distances and luminosities

Try This… | #2

The star Betelgeuse has an apparent brightness of 2.0 × 10-7 W m-2. Calculate its luminosity.

d = 4.0 × 1018 m𝑏 =

𝐿

4𝜋𝑑2

𝐿 = b 4𝜋𝑑2

= 2.0 × 10−7 4𝜋(4.0 × 1018)2

= 4.0 × 1031 W

Wien’s Displacement Law

*Note: This assumes perfect blackbody radiation

𝜆max𝑇 = 2.90 × 10−3 mK

λmax(metres) =2.90 × 10−3

T (kelvin)

This equation shows up in

subtopic 8.2 as

Try This… | #3

The star Betelgeuse has a max wavelength of 828.6 nm. What is its surface temperature?

𝜆max𝑇 = 2.90 × 10−3 mK

T =2.90 × 10−3 mK

λmax=2.90 × 10−3 mK

828.6 × 10−9= 3500 K

828.6 × 10−9 m

Solve for T

Luminosity

𝐿 = 𝜎𝐴𝑇4 Stefan-BoltzmannConstant σ 5.67 × 10-8 W m-2 K-4

Luminosity[W]

Temperature[K]

Surface Area (sphere)A=4πr2

Try This… | #4

Knowing everything else that we know about Betelgeuse, calculate the average radius of the star.

L = 4.0 × 1031 W

T = 3,500 K

𝐿 = 𝜎𝐴𝑇4

4.0 × 1031 = (5.67 × 10−8)(4𝜋𝑟2)(3500)4

𝑟 = 6.12 × 1011

Solve for r

IB Physics Data Booklet

All together now!

The brightest star in the sky is known as Sirius and has a parallax angle of 0.379 arc seconds, apparent brightness of 1.2 × 10-7

W m-2, and a max wavelength of 292 nm. Complete this table of stellar properties.

Brightness (W m-2) 1.2 × 10-7 W m-2

Max Wavelength (m) 292 × 10-9 m

Distance (m) 8.14 × 1016 m

Luminosity (W) 9.98 × 1027 W

Temperature (K) 9930 K

Radius (m) 1.2 × 109 m

𝑑 =1

𝑝=

1

0.379 𝑎𝑟𝑐 𝑠𝑒𝑐𝑜𝑛𝑑𝑠= 2.64 𝑝𝑎𝑟𝑠𝑒𝑐𝑠

2.64 pc ×3.26 ly

1 pc×9.46 × 1015 m

1 ly= 𝟖. 𝟏𝟒 × 𝟏𝟎𝟏𝟔 𝐦

𝐿 = b 4𝜋𝑑2 = 1.2 × 10−7 4𝜋(8.14 × 1016)2 = 𝟗. 𝟗𝟖 × 𝟏𝟎𝟐𝟕𝐖

T =2.90 × 10−3 mK

λmax=2.90 × 10−3 mK

292 × 10−9= 𝟗𝟗𝟑𝟎 𝐊

𝐿 = 𝜎𝐴𝑇4 9.98 × 1027 = (5.67 × 10−8)(4𝜋𝑟2)(9300)4

𝒓 = 𝟏. 𝟐 × 𝟏𝟎𝟗𝐦

𝟐𝟗𝟐× 𝟏𝟎−𝟗𝐦

Atomic SpectraIB PHYSICS | UNIT 13 | ASTROPHYSICS

13.3

Continuous Spectrum

When white light from the sun passes through a prism, the light is dispersed into its component colors in a continuous spectrum

Emission Spectrum

If an electric current is passed through an element in the form of a low-pressure gas, it will produce its own unique emission spectrum

Emission Spectrum

These spectra can be used to identify elements like a fingerprint

These are known as Line Spectra

Absorption Spectrum

If white light is passed through a sample of gaseous atoms or molecules, it is found that the light of certain wavelengths is missing

Absorption Spectrum

The emission and absorption spectra

are negative images of each other

What is Light?

Energy

Wave

Particle(photon)

Light is Quantized

Photons of light can only have certain ______________ values of energy

discrete

Energy of a Photon

𝐸 = ℎ𝑓

Planck’s Constant h 6.63 × 10-34 J s

Energy[J]

Frequency[Hz]

Energy of a Photon

𝐸 = ℎ𝑓 𝑐 = 𝑓𝜆

𝑓 =𝑐

𝜆

𝑐 = 3.00 × 108 𝑚 𝑠−1

𝐸 = ℎ𝑐

𝜆

𝜆 =ℎ𝑐

𝐸

Quick Recap of eV

𝑒𝑉 → 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 − 𝑣𝑜𝑙𝑡𝑠

Unit of Energy

𝐸𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 𝑒𝑉 =𝐸𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 𝐽

1.60 × 10−19

IB Physics Data Booklet

Try This…

Calculate the energy carried by one photon of microwaves of wavelength 10 cm (as might be used in a mobile phone) in J and eV

𝐸 =ℎ𝑐

𝜆=(6.63 × 10−34)(3 × 108)

(0.1)= 1.99 × 10−24 J

1.99 × 10−24

1.60 × 10−19= 1.24 × 10−5 eV

0.1 m

Shortcut time ☺

Since h and c are both constants, hc acts as a constant as well

𝐸 =ℎ𝑐

𝜆

1.99 × 10−24 J m

0.1 m= 1.99 × 10−24 J

1.24 × 10−6 eV m

0.1 m= 1.24 × 10−5 eV

Energy Levels

1

234567

Electrons in an atom exist at discrete energy levels

Absorbed

Emitted

Energy Levels

E4

E3

E2

E1

A photon is emitted whenever an electron transitions from one energy level down to a lower energy level

How many different transitions are possible between these four energy levels?

6

Energy Levels

n = 1 -13.6 eV

n = 2 -3.40 eV

n = 3 -1.51 eV

n = 4 -0.85 eVn = 5 -0.54 eV

n = 0.00 eV

Excited States

Ground State

Energy Transitions

n = 1 -13.6 eV

n = 2 -3.40 eV

n = 3 -1.51 eV

n = 4 -0.85 eVn = 5 -0.54 eV

n = 0.00 eV

Lyman Series (UV)

Balmer Series (Visible)

Paschen Series (IR)

Different Energy transitions result in different energies (wavelengths) of light that are absorbed or emitted

Calculating Wavelength Emitted

n = 1 -13.6 eV

n = 2 -3.40 eV

n = 3 -1.51 eV

n = 4 -0.85 eVn = 5 -0.54 eV

n = 0.00 eV What is the wavelength emitted?

𝜆 =ℎ𝑐

𝐸hc 1.99 × 10-25 J m 1.24 × 10-6 eV m

𝐸 = 13.6 − 1.51 = 12.09 eV

𝜆 =1.24 × 10−6 eV m

12.09 eV= 1.03 × 10−7 m

103 nm

Try This…

n = 1 -13.6 eV

n = 2 -3.40 eV

n = 3 -1.51 eV

n = 4 -0.85 eVn = 5 -0.54 eV

n = 0.00 eV What is the wavelength emitted?

𝜆 =ℎ𝑐

𝐸hc 1.99 × 10-25 J m 1.24 × 10-6 eV m

𝐸 = 3.40 − 0.85 = 2.55 eV

4.86 × 10−7 m

486 nm

𝜆 =1.24 × 10−6 eV m

2.55 eV=

Working Backwards…

What is the energy in eV for a 434 nm blue emission line?

𝜆 =ℎ𝑐

𝐸

hc 1.99 × 10-25 J m 1.24 × 10-6 eV m

2.86 eV𝐸 =ℎ𝑐

𝜆=1.24 × 10−6 eV m

434 × 10−9 m=

434 × 10-9 m

Working Backwards…

n = 1 -13.6 eV

n = 2 -3.40 eV

n = 3 -1.51 eV

n = 4 -0.85 eVn = 5 -0.54 eV

n = 0.00 eV Draw in the Energy Transition for a 434 nm blue emission line?

What transition has an energy difference of 2.86 eV?

𝐸 = 3.40 − 0.54 = 2.86 eV

Doppler Effect

As objects move away,

their wavelength is

stretched to be closer to

the red part of the

electromagnetic spectrum

Red Shift, Blue Shift

H-R Diagrams and Stellar SpectraIB PHYSICS | UNIT 13 | ASTROPHYSICS

13.4

Organizing the Stars

H-R Diagrams

Lsun

Lⵙ

H-R Diagrams

Our Sun

Large

Hot

Small

Hot

Large

Cool

Small

Cool

H-R Diagrams

Sizes of Stars

IB Physics Data Booklet

H-R Diagram for Calculating Distance

The maximum wavelength of a distant star is measured to be 600 nm, suggesting that it has a temperature of ~4800 K. If this star has a brightness of 1.0 × 10-12 W m-2, what is its distance from Earth?

Lsun = 3.84 × 1024 W

𝑏 =𝐿

4𝜋𝑑2𝑑 =

𝐿

4𝜋𝑏

𝑑 =1.92 × 1024

4𝜋(1.0 × 10−12)= 3.9 × 1017 m

4800 K

~0.5 Lsun

Lstar = 0.5 × (3.84 × 1024)

= 3.84 × 1024 W

Try This

The maximum wavelength of a distant star is measured to be 400 nm. If this star has a brightness of 0.5 × 10-12 W m-2, what is its distance from Earth?

Lsun = 3.84 × 1024 W

𝑏 =𝐿

4𝜋𝑑2𝑑 =

𝐿

4𝜋𝑏

𝑑 =3.84 × 1025

4𝜋(0.5 × 10−12)= 2.47 × 1018 m

𝑇 =2.90 × 10−3

400 × 10−9

𝑇 = 7250 K

7250 K

~10 Lsun

Lstar = 10 × (3.84 × 1024)

= 3.84 × 1025 W

Mass-Luminosity Relationship

𝐿 ∝ 𝑀3.5

For stars on the main sequence, there is a relationship between luminosity and mass

IB Physics Data Booklet

Stellar Spectra

Studying the Spectra of Stars can help determine what the stars are made of

Are these Emission Spectra or Absorption Spectra?

Stellar Spectra | Try it out

1 2 3 4 5 6 7 8 9 10

N N N N N N

N I N

I N N N N

N

N I N N I

N N I

Compare the spectra of the stars with the known absorption spectra of different elements to determine the composition of the stars

Redshifted

N → Neutral Spectra I → Ionic Spectra

Evolution of StarsIB PHYSICS | UNIT 13 | ASTROPHYSICS

13.5

Measuring the Age of the Stars

Life Span of the Stars

Which stars have the longest life span?

Red Dwarfs

**Hotter stars

burn fuel faster

and die quicker

Stellar Equilibrium

Increasing temperature increases pressure and causes expansion

What happens as Stars Age??

What happens as Stars Age??

H → He

• Heats up

• Hydrogen expands

• He → C

What happens as Stars Age??

Life Cycle of Sun-Sized Star

Life Cycle of our Sun

White Dwarf Stars

HOT but

SMALL

White Dwarf Stars

Chandrasekhar Limit = 1.4 Mʘ

There is a maximum mass of a core that can become a white dwarf

The core only makes up about 1/3 of the stars mass so a star with a total mass greater than about 4 Mʘ will not form white dwarfs

Sun’s

Mass

It will become

a White Dwarf

Star’s

Total

Mass< 4MʘIf

What about more massive stars?

Life Cycle of Massive Star

Life Cycle of Massive Star

Oppenheimer-Volkhoff Limit = 3 Mʘ

The mass of neutron stars are limited as well…

A Neutron star above the Oppenheimer-VolkhoffLimit will collapse and form a Black Hole.

Neutron Star< 3Mʘ→

Black Hole> 3Mʘ→

Life Cycle of Massive Star

Black Holes?

Life Cycle of a Star

> 3Mʘ

< 1.4Mʘ

The Expanding UniverseIB PHYSICS | UNIT 13 | ASTROPHYSICS

13.6

IB Physics Data Booklet

Henrietta Swan Leavitt

Cephid Variables

“Standard Candle”

Cephid Variables are with longer

brightness periods are more luminous

With this table, the luminosity of this

“standard candle” can be determined

as long as the period is known

Cephid Variables

What is the distance of the Cephid Variable with the period shown in the graph above? The brightness of this star is 8 × 10-10 W m-2.

1 Lsun = 3.84 × 1024 W

𝑏 =𝐿

4𝜋𝑑2𝑑 =

𝐿

4𝜋𝑏𝑑 =

5.76 × 1027

4𝜋(8 × 10−10)= 7.57 × 1017 m

Lstar = 1500 × (3.84 × 1024)

= 5.76 × 1027 W

5.4 days

~1500 Lsun

Type Ia Supernova

A type Ia Supernova forms when a white dwarf accretes matter from a companion star until it exceeds the Chandrasekhar limit and explodes

These supernovae have a constant luminosity so their brightness can be analyzed as a standard candle much like the Cephid Variables

Doppler Effect

Red-shifted

Red Shift, Blue Shift

Calculating Redshift

𝑧 =∆𝜆

𝜆𝑜≈𝑣

𝑐

Redshift

Change in

Wavelength

Wavelength

Emitted

Velocity of

the Source

Speed

of Light

(3.00×108 ms-1)

IB Physics Data Booklet

Calculating Redshift

A characteristic absorption line often seen in stars is due to ionized helium. It occurs at 468.6 nm. If the spectrum of a star has this line at a measured wavelength of 499.3 nm what is the recession speed of the star?

∆𝜆

𝜆𝑜≈𝑣

𝑐499.3 − 468.6

468.6≈

𝑣

3.00 × 108

𝑣 = 1.97 × 107 m s−1

Hubble’s Big Discovery

Edwin Hubble discovered that the amount of redshift changed by the distance

The Universe is Expanding

𝑣 = 𝐻0𝑑

Slope = H0 H0 ≈ 70 km s-1 Mpc-1

*current value is not necessarily constant

IB Physics Data Booklet

Using the Hubble “Constant”

Estimate the distance from the Earth to a galaxy with a recessional velocity of 150 km s-1

If a galaxy is 20 Mpc from Earth, how fast will it be receding?

𝑣 = 𝐻0𝑑 𝑑 =𝑣

𝐻0=

150 km s−1

70 km s−1 Mpc−1= 2.14 Mpc

𝑣 = 𝐻0𝑑 = 70 km s−1 Mpc−1 20 Mpc = 1400 km s−1

Calculating Redshift

Nothing can go faster than the speed of light so the Doppler effect can’t really hold up…

Calculating Redshift

Think of the wavelength change due to the stretching of space-time

𝑧 =𝑅

𝑅𝑜− 1

Cosmological

Redshift

Current

Universe Size

Cosmic Scale Factor

[𝑅 = 1]

Size of the universe at the time the light

was emitted (relative to the current size)

Calculating Redshift

If the redshift z = 3, what was the scale factor at the time that the light was emitted?

𝑧 =𝑅

𝑅𝑜− 1

3 =1

𝑅𝑜− 1

𝑅𝑜 =1

4= 0.25Note: This means that to result in this

cosmological redshift, the light had to have been emitted when the universe was a quarter of the size it is now

The Universe is Expanding

4 cm6 cm

2 cm

Think of a rubber band with marks when it is stretched out… Relative to the first dot, which dot moves the fastest?

8 cm

12 cm

4 cm

The farthest dot moves away the fastest

The Universe is Expanding

What comes before and after?

Pulling it all TogetherIB PHYSICS | UNIT 13 | ASTROPHYSICS

13.7

IB Physics Data Booklet

IB Physics Data Booklet

Measuring the Universe

Distance | Parallax

𝑑 𝑝𝑎𝑟𝑠𝑒𝑐 =1

𝑝 (𝑎𝑟𝑐 𝑠𝑒𝑐𝑜𝑛𝑑)1 parsec (pc) = 3.26 ly

Distance | Cepheid Variables

𝑏 =𝐿

4𝜋𝑑2

Distance | H-R Diagram

𝜆max𝑇 = 2.90 × 10−3 mK

𝑏 =𝐿

4𝜋𝑑2

Distance | Redshift & Hubble’s Law

𝑣 = 𝐻0𝑑

𝑧 =∆𝜆

𝜆𝑜≈𝑣

𝑐

How Old is the Universe?

𝑣 = 𝐻0𝑑

𝐻0 ≈ 70 km s−1 Mpc−1

𝑣 =𝑑

𝑡𝐻0𝑑 =

𝑑

𝑡𝑡 =

1

𝐻0

𝑡 =1

70 km s−1 Mpc−1= 0.0142 s Mpc km−1

0.0142 s Mpc

km×106 pc

1 Mpc×3.26 ly

1 pc×9.46 × 1015 m

1 ly×

1 km

1000 m= 𝟒. 𝟒 × 𝟏𝟎𝟏𝟕 𝐬

≈ 13.8 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑦𝑒𝑎𝑟𝑠

The “Big Bang”

Where’s the Evidence??

CMBCosmic Microwave Background

Cosmic Microwave Background

λmax = 1 mm

0.001 m

𝑇 =2.9 × 10−3

0.001= 𝟐. 𝟗 𝐊

Temperature of the Universe

What’s Next?

1. Static and Infinite2. Expanding and slowing to a stop3. Expanding, slowing, and

contracting in a “big crunch”4. Expanding and accelerating

There’s more out there

Age of the Universe Critical Density Neutron Degeneracy Satellite

Andromeda Galaxy Dark Energy Neutron Star Schwarzschild Radius

Binary Star Dark Matter Nucleosynthesis Stellar Cluster

Black Hole Galaxy OBAFGKM Supernova

Comet Meteor Planet White Dwarf Star

Constellation Milkyway Galaxy Pulsar

Cosmological Principle

Nebula Red Giant Star

Click Here for the Quizlet