Stars Part One:Brightness and Distance

Concept -1 – Temperature

λmax (metres) = 2.90 x 10-3 m kT (Kelvin)

λmax = Peak black body wavelengthT = The star’s surface temperature in Kelvins

So HOT = BLUE, COOL = RED

λmax = 2.90 x 10-3 m KT

From Jay Pasachoff’s Contemporary Astronomy

λmax

Put this in your notes: A star has a surface temperature of 5787 K, what is its λmax?

λmax = (2.90 x 10-3 m K)/T = (2.90 x 10-3 m K)/5787 = 501 nm

501 nm

A star has a λmax of 940 nm, whatis its surface temperature?

λmax = (2.90 x 10-3 m K)/T, T = (2.90 x 10-3 m K)/ λmax = (2.90 x 10-3 m K)/ (940E-9) = 3100 K

3100 K

Concept 0 – Total power output

Luminosity L = σAT4

Luminosity L = The star’s power output in Wattsσ = Stefan Boltzmann constant = 5.67 x 10-8W/m2K4

A = The star’s surface area = 4πr2

T = The star’s surface temperature in Kelvins

Put this in your notes: Our Sun has a surface temp of about 5787 K, and a radius of 6.96 x 108 m. What is its Luminosity?

L = σAT4 = 5.67x10-8(4π(7x108 ) 2 )(5787)4 = 3.87x1026 Watts

3.87x1026 Watts

A star has a radius of 5 x 108 m, and Luminosity of 4.2 x 1026 Watts, What is its surface temperature?

Luminosity L = σAT4 , T =(L/(σ4(5x108)2)).25 =6968 K = 7.0x103 K

7.0x103 K

Concept 1 – Apparent Brightness

Apparent Brightness b = L 4πd2

b = The apparent brightness in W/m2

L = The star’s Luminosity (in Watts)d = The distance to the star

L is spread out over a sphere..

Apparent Brightness b = L/4πd2

The inverse square relationshipFrom Jay Pasachoff’s Contemporary Astronomy

Put this in your notes: Our Sun puts out about 3.87 x 1026 Watts of power, and we are 1.50 x 1011 m from it. What is the Apparent brightness of the Sun from the Earth? (The Solar “Constant”)

b = L/4πd2 = 3.87E26/ 4π1.50E112

= 1370 W/m2

1370 W/m2

Another star has a luminosity of 3.2 x 1026 Watts. We measure an apparent brightness of 1.4 x 10-9 W/m2. How far are we from it?

b = L/4πd2 , d = (L/4πb).5 = 1.3x1017 m

1.3x1017 m

Concept 2 – Apparent Magnitude

Apparent Magnitude: m = 2.5log10 (2.52 x 10-8W/m2/b)

b = The apparent brightness in W/m2

m = The star’s Apparent MagnitudeNote that the smaller b is, the bigger m is.Logarithmic scale: x100 in b = -5 in m

Put this in your notes: What is the apparent magnitude of a star with an apparent brightness of 7.2x10-10 Wm-2? What is that of a star with an apparent brightness of 7.2x10-12 Wm-

2?

Apparent Magnitude: m = 2.5log10 (2.52 x 10-8W/m2/b)

b = The apparent brightness in W/m2

m = The star’s Apparent MagnitudeNote that the smaller b is, the bigger m is.Logarithmic scale: x100 in b = -5 in m

Example: What is the apparent magnitude of a star with an apparent brightness of 7.2x10 -10 Wm-2? What is that of a star with an apparent brightness of 7.2x10-12 Wm-2?

For the first: m = 2.5log(2.52E-8/7.2E-10) = 3.9 (no units)For the second: m = 2.5log(2.52E-8/7.2E-12) = 8.9

Notice that when it got dimmer by 100x, the magnitude went up 5. yeah – it’s weird.

From Jay Pasachoff’s Contemporary Astronomy

So apparent magnitude is a counter-intuitive scale

-27 is burn your eyes out,

1 is a normal bright star, and

6 is barely visible to a naked, or unclothed eye. (You can’t see anything if you clothe your eyes!!)

A magnitude 1 star delivers 100 times more W/m2 than a magnitude 6 star.

What is the Apparent Magnitudeof a star that has an apparent brightness of 1.4 x 10-9 W/m2 ?

m = 2.5log10 (2.52 x 10-8W/m2/b) = 2.5log10 (2.52 x 10-8W/m2/ 1.4 x 10-9 W/m2) = 3.1

3.1

The Hubble can see an object with an apparent magnitude of 28. Whatis the apparent brightness of sucha star or galaxy?m = 2.5log10 (2.52 x 10-8W/m2/b) , b = 2.52 x 10-8W/m2 /10(m/2.5)

= 1.6 x10-19 W/m2

1.6 x10-19 W/m2

Concept 3 – Absolute Magnitude

Absolute Magnitude: m - M = 5 log10(d/10)

M = The Absolute Magnituded = The distance to the star in parsecsm = The star’s Apparent Magnitude

The absolute magnitude of a star is defined as what its apparent magnitude would be if you were 10 parsecs from it.

Absolute Magnitude: m - M = 5 log10(d/10)M = The Absolute Magnituded = The distance to the star in parsecsm = The star’s Apparent Magnitude

Example: 100 pc from an m = 6 star, M = ?(10x closer = 100x the light = -5 for m)

M = 6 - 5 log10(100/10) = 1

Put this in your notes: The Sun has an apparent magnitude of -26.8, we are 1.5x108

km or 4.9 x 10-6 pc from the sun. What is the sun’s absolute magnitude?

M = m - 5 log10(d/10) = -26.8 - 5 log10(4.86 x 10-6 /10) = 4.7

4.7

You are 320 pc from a star with an absolute magnitude of 6.3. What is its apparent magnitude?

M = m - 5 log10(d/10), m = M + 5 log10(d/10) = 6.3 + 5 log10(320/10) = 14

14

Concept 4 – H-R diagramsIn 1910, Enjar Hertzsprung of Denmark, and Henry Norris Russell at Princeton plotted M vs T in independentresearch.

From Douglas Giancoli’s Physics

Main Sequence

Hot Cooler

Dim

Bright

From Douglas Giancoli’s Physics

Hot Cooler

Dim

Bright

From Jay Pasachoff’s Contemporary Astronomy

O Oh (Hot)B beA aF fineG girl,K kissM me! (Cooler)

The Star

The atmosphere

Light from star is filtered by atmosphere

From Jay Pasachoff’s Contemporary Astronomy

Spectral types:

Suffixes 0 (hottest) - 9 (coolest) so O0, O1…O9, then B0, B1…

O -

B -

A -

F -

G -

K -

M -

30,000 - 60,000 K, ionized H, weak H lines, spectral lines arespread out. O types are rare and gigantic.10,000 - 30,000K, H lines are stronger, lines are less spread out(Rigel, Spica are type B stars)

7,500 - 10,000K, strong H lines, Mg, Ca lines appear (H and K)(Sirius, Deneb and Vega are A type stars)6,000 - 7,500K, weaker H lines than in type A, strong Ca lines (Canopus (S.H.) and Polaris are type F)5,000 - 6,000K, yellow stars like the sun. Strongest H and Klines of Ca appear in this star.

3,500 - 5,000K, spectrum has many lines from neutral metals.Reddish stars (Arcturus and Aldebaran are type K stars)3,500 or less, molecular spectra appear. Titanium oxide linesappear. Red stars (Betelgeuse is a prominent type M)

SO…

Hot stars are:Big, Bright, Brief and Blue

Cool stars are: Diminuitive, Dim, and Durable and um… reD

More about brief and durable next time…

Dim

Bright

Spectroscopic “parallax”

Spectrum = M(from diagram)Measure mFind d

From Douglas Giancoli’s Physics

Put in your notes: How far is an B0 that has an m of 8?

M = m - 5 log10(d/10)5 log10(d/10) = m - Mlog10(d/10) = (m - M)/5d/10 = 10((m - M)/5)

d = (10 pc)10((8 - -4.3)/5)

d = 2884 pc

2884 pc

How far is an K0 that has an m of 14?

d = (10 pc)10((14 - 5.8)/5)

d = 437 pc

437 pc

Ok, now…

Drool.