Messier Marathon

• Assignment 15 complete (10% penalty/day if you're late)

• Second mid-term• Wednesday April 23rd• Covers chapters 5, 10 and 11• Revise!• NOTE: today I will mainly go through math - but

most of the points on the exam come from the multiple choice sections - re-read these Chapters!!!

• extra office hours 2pm - 4pm tomorrow (Tuesday)

• Lab as usual this week• Honours section meet after this lecture

Admin:

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Review: powers of 10

• 101=10• 102=1010=100• 103=101010=1000• 104=10101010=10000

powers of 10

• 101=10• 100=10/10=1• 10-1=10/10/10=0.1• 10-2=10/10/10/10=0.01

An example: powers of 10

• To multiply, add the exponents• 103104=107

• To divide, subtract the exponents• 103 / 104=10-1

• To raise to a power, multiply exponents• (103)4 = 1012

• sqrt(x) = x1/2

Scientific notation

• 3000=31000=3103

• 0.0002 = 20.0001=210-4

• 210-4 3103

= 2 3 10-4 103 = 6 10-1

• (3 108)2

= 9 1016

Finding a ratio(See Appendix C.5)

• What if you want to compare two quantities?

• e.g. The average density of Earth is 5.52 g/cm3, that of Jupiter is 1.33 g/cm3. How much more dense is Earth than Jupiter?

• 5.52/1.33=4.15 times as dense.

Finding a ratio

• What if the two quantities involve an equation?

• e.g. the mass-energy of a 1kg mass m1, is 91016 Joules. What is the mass-energy of a 3kg mass?

1

3

1

2

21

22

1

2

m

m

m

m

E

E

cm

cm

E

E

does it make sense?

425.0

1

1

11

1

4

4

2

22

2

1

1

2

2

2

12

1222

1

22

21

22

1

2

d

d

b

b

d

dd

dd

d

dL

dL

b

b

d

d

d

d

• Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2 . Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at 1/2 the distance?

24 d

Lb

does it make sense???

d1=1d2=0.5

What if it were located 7 times the distance?

often don't need to memorize equations!!!

Ch. 5 Summary of Mathematics I f = c

= wavelength, f = frequencyc = 3.00 108 m/s = speed of light

E = h f = photon energyh = 6.626 10−34 joule s

Given any one of E, f, or you should be able to calculate the others

example: what is the energy of a photon with a wavelength of 300nm ?

E=h c/(1eV=1.610-19J)

Properties of Thermal Radiation1. Hotter objects emit more light at all frequencies per

unit area.

2. Hotter objects emit photons with a higher average energy (shorter wavelength).

Ch.5 Summary of Mathematics II1. Hotter objects emit more light at all frequencies per

unit area.

2. Hotter objects emit photons with a higher average energy.

)Kelvinwatt/(m107.5

area power/unit emitted428

4

T

nm)Kelvin(

000,900,2max T

Given the temperature of an object, you should be able to calculate the emitted power/unit area and the peak wavelength (see "cosmic calculations 5.1"). Given the peak wavelength, you should be able to calculate the temperature. E.g. Sun~6000K

Ch. 5 Summary of Mathematics III

• Doppler shift

rest

restshift

c

v

Given the wavelength of spectral lines, you should be able to tell if an object is moving away or towards you, and at what speed (see "cosmic calculations 5.2").

• In hydrogen, the transition from level 1 to level 2 has a rest wavelength of 121.6 nm. Suppose you see this line at a wavelength of 120.5nm in Star A and at 121.3nm in Star B.

• Calculate speed of Star A. • 2710 km/s

• Is it moving toward or away from us• toward us

• Calculate speed of star B• 740 km/s

• toward us

Math question

Features of a Spectrum

• By carefully studying the features in a spectrum, we can learn a great deal about the object that created it.

What is this object?

Reflected Sunlight: Continuous spectrum of visible light is like the Sun’s except that some of the blue light has been absorbed—object must look red

What is this object?

Thermal Radiation: Infrared spectrum peaks at a wavelength corresponding to a temperature of 225 K

What is this object?

Carbon Dioxide: Absorption lines are the fingerprint of CO2 in the atmosphere

What is this object?

Ultraviolet Emission Lines: Indicate a hot upper atmosphere

What is this object?

Mars!

The relationship between apparent brightness and luminosity depends on distance:

Luminosity Brightness = 4π (distance)2

We can determine a star’s luminosity if we can measure its distance and apparent brightness:

Luminosity = 4π (distance)2 (Brightness)

Ch. 11 Summary of Mathematics I

Thought Question

How would the apparent brightness of Alpha Centauri change if it were three times farther away?

A. It would be only 1/3 as bright.B. It would be only 1/6 as bright.C. It would be only 1/9 as bright.D. It would be three times as bright.

Thought Question

How would the apparent brightness of Alpha Centauri change if it were three times farther away?

A. It would be only 1/3 as bright.B. It would be only 1/6 as bright.C. It would be only 1/9 as bright.D. It would be three times as bright.

Math Question

The sun's measured apparent brightness is 1.36103 W/m2 at Earth's distance from the sun (1 AU = 1.51011m). What is the Sun's Luminosity?

24 d

Lb

Math Question

The sun's measured apparent brightness is 1.36103 W/m2 at Earth's distance from the sun (1 AU = 1.51011m). What is the Sun's Luminosity?

24 d

Lb

= 3.81026 W

Parallax and Distance

p = parallax angle

d (in parsecs) = 1

p (in arcseconds)

d (in light-years) = 3.26 1

p (in arcseconds)

(by definition)

Ch. 11 Summary of Mathematics II

Parallax and Distance

What's the distance of a star with a parallax angle of 1 arcsecond?

d= 1parsec =3.26 light years

What's the distance of a star with a parallax angle of 2 arcseconds?

d= 0.5parsecs =1.63 light years

NOTE: bigger angle means star is closer

Alpha Centauri: parallax angle 0.7420.

Ch. 11 Summary of Mathematics II

Sirius A has a luminosity of 26LSun and a surface temperature of about 9400K. What is its radius?

LSun=3.81026W

Ch. 11 Summary of Mathematics II

9

157

27

48

26

4

42

1030.1

108.7102.7

109.9

9400107.514.34

108.326

4

4

r

r

T

Lr

TrL

The Magnitude Scale

m apparent magnitude M absolute magnitude

apparent brightness of Star 1

apparent brightness of Star 2(1001/5)m1 m2

luminosity of Star 1

luminosity of Star 2(1001/5)M1 M2

related to apparent brightness related to luminosity

5 magnitudes difference = a factor of 100 in brightnessNote that a lower number means a brighter star

Ch. 11 Summary of Mathematics III

The Magnitude Scale

Note that a lower number means a brighter star

How much brighter is a 3rd magnitude star than an 8th magnitude star?

m1=8, m2=321

5

1

2

1 100

mm

b

b