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 Problem 15 Problem 16 Problem 17

*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

15.
*A very strong concertmaster is playing 440.00 Hz at the top of an 4.50 m tall tower on a neutron star where the “g”
is 1.816x10^{14} N/kg. We are at the
bottom also playing 440.00 Hz. What is the beat frequency we hear? Do we
hear the player on the top of the tower as sharp or flat? What frequencies do we
observe?* <<to problem list>>

^{14} N/kg. Since we want to know what frequency we hear
when standing at the BOTTOM, we just use our given 440.00 Hz frequency.

^{2}

^{14} (4.50) / (3x10^{8})^{2}

__observe__ a frequency of **444.00
Hz** with a frequency difference of **4.00
Hz**

**sharp**er because it is at the __top__
coming toward the bottom so we hear a higher frequency...

*16. If
we are living on a neutron star, and we tune the local station “Neutrock
91.7 (MHz) in at 90.2 on our FM Dial. We know that we are at a different
elevation by 35.6 m. What is the “g” here? Are we higher or lower than the
broadcast antenna of “Neutrock”? * <<to problem list>>

^{2}

^{6}
)

The
height is given as well (35.6 m) so let’s plug it all in the formula:

^{6} / 90.2x10^{6} = g (35.6) / (3x10^{8})^{2}

^{13} N/ kg

**higher** than the antenna...

*17. A 417
nm spectral line is shifted to 423 nm through a distance of 1 A.U. What is
the change in frequency? What is the “g” in the vicinity of source? * <<to problem list>>

First,
we ought to know what 1 A.U. is when related to the distance...

^{11} m

^{2} and plug in the given info to find the value of “g”

^{11} / (3x10^{8})^{2}

**g
= 8510 N/kg** (yes,
this is right... the answer on the worksheet is wrong... trust ME)

**By
the crazy German ( Maike Scheller ) **

**Class
of 2004**