1

Lecture 24

Characteristics of Stars

January 10d, 2014

2

Stars come in

many

varieties!

Distance

Size

Temperature

Luminosity

Composition

Color

Age

3

Distances to the

Stars

• Closest stars measured

using parallax

• 1 parsec = 3.26 light

years

4

Parallax Calculations

baseline (km)distance (km) 57.3

parallax angle ( )

1distance (pc)

parallax angle (arcseconds)

or

5

What is the distance to a star that has

a parallax angle of 0.40 arc seconds?

A. 0.40 ly

B. 2.5 ly

C. 0.13 ly

D. 8.2 ly

6

What is the distance to a star that has

a parallax angle of 0.40 arc seconds?

A. 0.40 ly

B. 2.5 ly

C. 0.13 ly

D. 8.2 ly

pc

pc

1 1

0.40 arcsec

2.5 pc 3.26 ly/pc

8.2 ly

dp

d

7

4 2

4 2

Luminosity Temperature Radius

4L T R

Luminosity

• Total energy emitted by a star each second

• Luminosity of Sun represented by L

• Luminosities range from 106 L to 10–5 L

• Depends on size and temperature

8

Apparent Brightness • The brightness an object appears to have.

• The further away the object, the dimmer it looks

• Often described in terms of apparent magnitude m

2

LuminosityApparent Brightness

4 d d = distance

9

Two stars each have the same luminosity but star

A is 25 pc away and star B is 75 pc away. What is

the ratio of their apparent brightness, bB/bA?

A. 1/75

B. 1/25

C. 1/9

D. 1/5

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Two stars each have the same luminosity but star

A is 25 pc away and star B is 75 pc away. What is

the ratio of their apparent brightness, bB/bA?

A. 1/75

B. 1/25

C. 1/9 D. 1/5

2 2

B B A

2 2

A A B

2

4

4

25 pc 1

75 pc 9

b L d d

b L d d

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Apparent Magnitude Scale

Several stars in and around the

constellation Orion labeled with

their names and apparent

magnitudes

Stars visible to the naked eye have

apparent magnitudes between

m = –1.44 and about m = +6.

Star A has a +3.0 apparent magnitude. If star B

appears 100 times brighter than star A, what is its

apparent magnitude?

0 of 78

A. −2.0

B. +5.5

C. +7.5 D. +8.0

12

120

Star A has a +3.0 apparent magnitude. If star B is 100

times brighter than star A, what is its apparent magnitude?

A. −2.0 B. +5.5

C. +7.5 D. +8.0

5.0 3.0 5.0 2.0

or

2

3.0 2.5log 100

2

.5

0

o

.

l g

B A

B A B A

B

m m

m m B B

m

13

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Determining Distance

• Measure apparent brightness

– Often written in terms of apparent magnitude m

• Measure intrinsic brightness (luminosity)

– Need temperature

– Need size of star

– Write in terms of absolute magnitude M

• Compare intrinsic brightness to apparent brightness to determine distance d.

–

• Problem: size of star is rarely known and is hard to measure.

4 24L T R

5log 5m M d

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A star has an apparent magnitude of +4.5 and an

absolute magnitude of –1.8. How far away is it?

A. 182 pc

B. 315 pc

C. 450 pc

D. 900 pc

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A star has an apparent magnitude of +4.5 and an

absolute magnitude of –1.8. How far away is it?

A. 182 pc B. 315 pc

C. 450 pc

D. 900 pc

1 15 5

5 4.5 1.8 5

2.26

5log 5

10 10

10 pc 182 pc

m M

m M d

d

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Temperature

• Rough estimate: Color

– Red = cool

– Blue = hot

– Wien’s Law relates surface temperature to color

6

max

2.9 10K

nmT

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Temperature and Color

The intensity of light emitted by three hypothetical stars is plotted against

wavelength. The range of visible wavelengths is indicated. Where the peak

of a star’s intensity curve lies relative to the visible light band determines

the apparent color of its visible light. The insets show stars of about these

surface temperatures.

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Temperature • Problem:

– stars have absorption lines

– may not be able to observe peak of spectrum

– Interstellar dust may distort the spectrum

• Solution: Use spectral classification

Appendix

• The following are pages from a different

textbook and a handout I’ve created to

explore the math behind luminosity and

distance calculations a little better.

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Star Brightness and Distance

In principle, if we know the actual luminosity of a star and how bright it appears to us, we can figure out how far away it is. If two stars have the same luminosity, the brighter one will be closer to us and the dimmer one will be further away. But how can we find the distance in practice? We must employ some mathematical tools. First some definitions:

b = apparent brightness (introduced on page 412). This is the intensity of the starlight in watts per square meter falling upon the Earth. It is related to the luminosity of a star and its distance away from us:

24Lbdπ

= . For instance, we know the apparent brightness of the closest star, the Sun, is

measured to be 1370 W/m2 at the top of Earth’s atmosphere. The distance to the sun is 1.496×1011 m. Therefore we can find the Sun’s luminosity:

( ) ( )22 11 2 264 4 1.496 10 m 1370 W/m 3.85 10 WL d bπ π= = × = ×

m = apparent magnitude (introduced page 416-417). This is a convenient way to handle the numerically very small apparent brightness values of most stars, as well as the very large range of b values among the stars we can detect. It is based on the fact that the brightest stars that could be seen with the naked eye are about 100 times brighter than the faintest star. This range of brightnesses was divided into five “magnitudes” by Ptolemy, with a magnitude of 1.00 corresponding to the brightest stars and a magnitude of 6.00 corresponding to the dimmest stars. There is a math formula that connects apparent magnitude to apparent brightness:

5 log2

bm mb

⎛ ⎞= − ⎜ ⎟

⎝ ⎠. The system was built around the star Vega (*see note below) from which

we have determined the Sun’s magnitude is -26.74. The starlight from Vega is measured as a brightness of b = 2.76×10-8 W/m2 and therefore Vega’s magnitude is:

8 2

2

5 5 2.76 10 W/mlog 26.74 log 0.002 2 1370 W/m

bm mb

−⎛ ⎞ ⎛ ⎞×= − = − − =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠. The faintest stars visible to the

naked eye have brightnesses of around 1.10×10-10 W/m2, or magnitudes of 10 2

2

5 5 1.10 10 W/mlog 26.74 log 6.002 2 1370 W/m

bm mb

−⎛ ⎞ ⎛ ⎞×= − = − − = +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

Note that dimmer stars have larger magnitudes in this scheme!

*Note from http://curious.astro.cornell.edu/question.php?number=569 :

In the modern day, we've tweaked Hipparchus's definitions a bit to make them more precise and more convenient. Thus, we now define a difference of five "magnitudes" as exactly equal to a brightness ratio of 100. Furthermore, we tie the whole scale to the star Vega, which is defined to have a magnitude of 0 (or very close to it, at any rate -- modern, precise readjustments of the magnitude scale actually now put Vega at closer to 0.03, but that's a technical point). Finally, even though Hipparchus's scale only went from 1 to 6, we make no restrictions on how far the scale can go in either direction -- anything brighter than Vega simply has a negative number as its apparent magnitude! (An astronomy professor of mine in college used to speculate that maybe Hipparchus was only looking in one direction of the sky when he made up his magnitude scale; then he turned around and saw a brighter star like Vega and therefore said, "Whoops, I guess we'll have to make that magnitude 0...")

Interlude: Why does the text keep talking about the number 2.512 at the top of p. 417? Well, solve the above equation for brightness instead of magnitude:

( )

( )2 22 15 5

22 1

1

22 1

1

2 22 1

1 1

5 log2

2 log5

10 so if 1, 10 2.512m m

bm mb

bm mb

b bm mb b

− − +

⎛ ⎞= − ⎜ ⎟

⎝ ⎠⎛ ⎞

− = − ⎜ ⎟⎝ ⎠

= − = − = =

which means, for instance, that a star with a magnitude of +3 will be 2.512 times brighter than a star of magnitude +4.

M = absolute magnitude (introduced p.417). This the apparent magnitude a star would have if it were located exactly 10.0 parsecs away from Earth. It is directly related to the luminosity of the star, so it is a way of comparing luminosities of various stars. The absolute magnitude of the Sun is +4.832. We can thus find the absolute magnitude of a star from its luminosity:

5 5log 4.832 log2 2

L LM ML L

⎛ ⎞ ⎛ ⎞= − = + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Only for the curious: Where does the equation in box 19-3 come from? Now use these ideas to determine the distance to a star from its apparent magnitude and its absolute magnitude. We’ll use the relationship between brightness, luminosity, and distance stated above:

2

2 2

4 so that 4 4

L L d bbd L d b

ππ π

= = .

If we also remember that 5 log2

bm mb

⎛ ⎞− = − ⎜ ⎟

⎝ ⎠ we can figure out how they got the formula in box 19-3

page 429:

( )

( )

2

2

2 2

2

5 5 4log log but log log log so that2 2 4

5 5 5log log log log2 2 2

5 log now more about logs2

L d bM M M ab a bL d b

d b d bM M Md b d b

dM M m md

ππ

⎛ ⎞⎛ ⎞= − = − = +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥= − + = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞= − + −⎜ ⎟

⎝ ⎠

( ) ( )

2

16 11

: log 2 log and log log log :

5 log 5log 5log Finally, rearrange:

5 log 5log 26.74 5log 3.0857 10 m/pc 5log 1.496 10 m 4.832

5log 5

ax x a bb

dM m m M m m M d dd

m M m d d M d

m M d

⎛ ⎞= = −⎜ ⎟⎝ ⎠

⎛ ⎞= − + − = − + − +⎜ ⎟

⎝ ⎠

− = + − − = − + × × − × −

− = −

where d is the distance to the star in parsecs.