Basically there are about five things to know about stars and astrophysics.

Wien’s Law measures the **peak plack body wavelength **of a star**. **The formula is:

l _{max }= 2.9 x 10^{-3 m}/_{k}/T

l = Peak black body wavelength in meters

2.9 x 10^{-3 m}/_{k }= Wien’s constant

T = The star’s surface temperature in Kelvins

**Total power output (absolute luminosity) **<<to top>>

This is the **star’s power output in Watts**. The formula is:

L = s AT^{4}

Luminosity L = the star’s power output in watts

s = Stefan-Boltzmann constant = 5.67 x 10^{-8}W/m^{2}K^{4}

A = The star’s surface area = 4p r^{2 }

T = The star’s surface temperature in Kelvins

**Apparent Brightness **<<to top>>

b = L/4p d^{2}

b = apparent brightness in W/m^{2}

L = luminosity (in Watts)

d = The distance to the star

m = 2.5log_{10} (2.52 x 10^{-8}/b)

m = the stars apparent magnitude

b = the apparent brightness in W/m^{2}

**Absolute Magnitude **<<to top>>

M = m – 5 log_{10}(d/10)

M = The absolute magnitude

d = the distance to the star

m = The star's Apparent Magnitude

Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7

Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14

*What**is the wavelength of the blackbody peak for Doofus Major with temperature of 15,000?*<<to problem list>>*Wha**t is the temperature of the surface of Baldedurash if the black body peak is 450 nm?*<<to problem list>>*Sillius has**a temperature of 8300 K, and a radius of 8.4 x 10*<<to problem list>>^{8}m. What is its luminosity?*Alpo Yumae**has a luminosity of 3.7 x 10*<<to problem list>>^{29}W, and a Temperature of 5600 K. What is its radius?*Eta**Peanut has a luminosity of 1.9 x 10*<<to problem list>>^{27}W. What is its brightness if it is 130 Ly away?__(convert Ly to m --- 9.46 x 10__^{15}m/1 Ly__)__*Weurmgeuse**has a brightness of 1.3 x 10*<<to problem list>>^{-12}W/m^{2}, What is its luminosity if it is 65 Ly away?__(convert Ly to m --- 9.46 x 10__^{15}m/1 Ly)

l _{max} = 2.90 x 10^{-3 m}/_{K}/T

l = 2.90 x 10^{-3 }/ 15,000

l = 1.93 10^{-7} m

l = 193 nm

l _{max} = 2.90 x 10^{-3 m}/_{K}/T

450 x 10^{-9 }= 2.90 x 10^{-3 }/ T

(450 x 10^{-9})(T) = 2.90 x 10^{-3}

(2.90 x 10^{-3})(450 x 10^{-9}) = T

6400 K = T

L = s AT^{4}

L = (5.67 x 10^{-8})(4p (8.4 x 10^{8})^{2})(8300^{4})

L = 2.38596 x 10^{27}

L = 2.4 x 10^{27} W

L = s AT^{4}

3.7 x 10^{29} = (5.67 x 10^{-8})( 4p (r^{2}))(5600^{4})

(3.7 x 10^{29})/((5.67 x 10^{-8})( 4p )( 5600^{4})) = r^{2}

Ö (5.28027 x 10^{20}) = Ö (r^{2})

r = 22978852983 m

r = 2.3 x 10^{10} m

b = L/4pd^{2}

b = (1.9 x 10^{27})/(4p (130 ly x (9.46 x 10^{15 }m/ly))^{ 2})

b = 9.9971 x 10^{-11}

b = 1.0 x 10^{-11} W/m^{2}

b = L/4pd^{2}

1.3 x 10^{-12} = L/(4p (65 ly x (9.46 x 10^{15 }m/ly))^{ 2})

1.3 x 10^{-12} = L/(4.751 x 10^{36})

(1.3 x 10^{-12} )( 4.751 x 10^{36 }m) = L

L = 6.17678 x 10^{24}

L = 6.2 x 10^{24} W

*7) Canis Fetchus has a brightness of 4.5 ^{ }x 10^{-12}, and a luminosity of 2.3 x 10^{27} W. How far away is it? *<<to problem list>>

b = L/4pd^{2}

4.5^{ }x 10^{-12} = 2.3 x 10^{27}/4p d^{2}

(4.5^{ }x 10^{-12})( 4p d^{2}) = 2.3 x 10^{27}

d^{2} = 2.3 x 10^{27}/((4.5^{ }x 10^{-12})( 4p ))

Ö d^{2} = Ö 4.06729 x 10^{37}

d = 6.3775 x 10^{18}

d = 6.4 x 10^{18} m

*Wha**t is the apparent magnitude of Canopeas if it has a brightness of 3.5 x 10*<<to problem list>>^{-17}W/m^{2}?*Alpha**Beta has an apparent magnitude of 23. What is its brightness? (log*<<to problem list>>_{n}^{x}=y, n^{y}= x)*Twnentieth**Centauri is 450 parsecs away, and has an apparent magnitude of 9. What is its absolute magnitude?*<<to problem list>>*Cepheid**Variable has an absolute magnitude of -3. How far away is it if it has an apparent magnitude 17? (log*) <<to problem list>>_{n}^{x}=y, n^{y}= x*Cepheus**Firmea has an absolute magnitude of –2.5. What is its apparent magnitude if it is 230 Ly away from us.*__(convert Ly to pc --- 1Ly = 3.36 pc)__*Irregulus**with a surface temperature of 5000 K has what absolute magnitude according to the HR diagram on p. 885 in your book? Hoe far away is it if it has an apparent magnitude of 12?*<<to problem list>>*Cetus**Naue with a surface temperature of 7000 K has what absolute magnitude according to the HR diagram on p. 885 in your book? How far away is it if it has an apparent magnitude of 21?*<<to problem list>>

m = 2.5log_{10} (2.52 x 10^{-8}/b)

m = 2.5log_{10} (2.52 x 10^{-8}/3.5 x 10^{-17})

m = 22.1433

m = 22

m = 2.5log_{10} (2.52 x 10^{-8}/b)

23 = 2.5log_{10} (2.52 x 10^{-8}/b)

23/2.5 = log_{10} (2.52 x 10^{-8}/b)

9.2 = log_{10} (2.52 x 10^{-8}/b)

10^{9.2} = 2.52 x 10^{-8}/b

1.5848 x 10^{9} = 2.52 x 10^{-8}/b

b = 2.52 x 10^{-8}/1.5848 x 10^{9}

b = 1.5901 x 10^{-17}

b = 1.6 x 10^{-17}

M = m – 5 log_{10}(d/10)

M = 8 - 5 log_{10}(450/10)

M = -.266

M = -.3

M = m – 5 log_{10}(d/10)

-3 = 17 – 5 log_{10}(d/10)

(-3 – 17)/-5 = log_{10}(d/10)

4 = log_{10}(d/10)

10^{4} = d/10

10^{4}x10 = d

d = 100000

d = 1 x 10^{5} pc

M = m – 5 log_{10}(d/10)

-2.5 = m – 5 log_{10}((230/3.26)/10)

-2.5 = m – 5 log_{10}((70.552)/10)

-2.5 = m – 5 log_{10}(7.0552)

-2.5 = m – 4.24355

-2.5 + 4.24355

m = 1.74255

m = 1.7

absolute magnitude = 6

M = m – 5 log_{10}(d/10)

6 = 12 – 5 log_{10}(d/10)

(6 – 12)/-5 = log_{10}(d/10)

1.2 = log_{10}(d/10)

10^{1.2} = d/10

15.8489 = d/10

d = 158.489

d = 160 pc

absolute magnitude = 3

M = m – 5 log_{10}(d/10)

3 = 21 – 5 log_{10}(d/10)

(3 – 21)/-5 = log_{10}(d/10)

3.6 = log_{10}(d/10)

10^{3.6} = d/10

3981.07 = d/10

d = 39810.7

d = 40000 pc

Created 6/8/00

By Brian Peterson