Astrophysics | Go up

 

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

 

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Astrophysics problem set

Wien’s Law <<to top>>

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 AT4

Luminosity L = the star’s power output in watts

s = Stefan-Boltzmann constant = 5.67 x 10-8W/m2K4

A = The star’s surface area = 4p r2

T = The star’s surface temperature in Kelvins

 

Apparent Brightness <<to top>>

b = L/4p d2

b = apparent brightness in W/m2

L = luminosity (in Watts)

d = The distance to the star

 

Apparent Magnitude <<to top>>

m = 2.5log10 (2.52 x 10-8/b)

m = the stars apparent magnitude

b = the apparent brightness in W/m2

 

Absolute Magnitude <<to top>>

M = m – 5 log10(d/10)

M = The absolute magnitude

d = the distance to the star

m = The star's Apparent Magnitude

 

Astrophysics <<to top>>

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

  1. What is the wavelength of the blackbody peak for Doofus Major with temperature of 15,000? <<to problem list>>

    2. 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

     

     

  2. What is the temperature of the surface of Baldedurash if the black body peak is 450 nm? <<to problem list>>

    3. 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

     

     

  3. Sillius has a temperature of 8300 K, and a radius of 8.4 x 108 m. What is its luminosity? <<to problem list>>

    4. L = s AT4

    L = (5.67 x 10-8)(4p (8.4 x 108)2)(83004)

    L = 2.38596 x 1027

    L = 2.4 x 1027 W

     

     

  4. Alpo Yumae has a luminosity of 3.7 x 1029 W, and a Temperature of 5600 K. What is its radius? <<to problem list>>

    5. L = s AT4

    3.7 x 1029 = (5.67 x 10-8)( 4p (r2))(56004)

    (3.7 x 1029)/((5.67 x 10-8)( 4p )( 56004)) = r2

    Ö (5.28027 x 1020) = Ö (r2)

    r = 22978852983 m

    r = 2.3 x 1010 m

     

     

  5. Eta Peanut has a luminosity of 1.9 x 1027 W. What is its brightness if it is 130 Ly away? (convert Ly to m --- 9.46 x 1015m/1 Ly) <<to problem list>>

    6. b = L/4pd2

    b = (1.9 x 1027)/(4p (130 ly x (9.46 x 1015 m/ly)) 2)

    b = 9.9971 x 10-11

    b = 1.0 x 10-11 W/m2

     

     

  6. Weurmgeuse has a brightness of 1.3 x 10-12 W/m2, What is its luminosity if it is 65 Ly away? (convert Ly to m --- 9.46 x 1015m/1 Ly) <<to problem list>>

b = L/4pd2

1.3 x 10-12 = L/(4p (65 ly x (9.46 x 1015 m/ly)) 2)

1.3 x 10-12 = L/(4.751 x 1036)

(1.3 x 10-12 )( 4.751 x 1036 m) = L

L = 6.17678 x 1024

L = 6.2 x 1024 W

 

 

7) Canis Fetchus has a brightness of 4.5 x 10-12, and a luminosity of 2.3 x 1027 W. How far away is it? <<to problem list>>

b = L/4pd2

4.5 x 10-12 = 2.3 x 1027/4p d2

(4.5 x 10-12)( 4p d2) = 2.3 x 1027

d2 = 2.3 x 1027/((4.5 x 10-12)( 4p ))

Ö d2 = Ö 4.06729 x 1037

d = 6.3775 x 1018

d = 6.4 x 1018 m

 

 

  1. What is the apparent magnitude of Canopeas if it has a brightness of 3.5 x 10-17 W/m2? <<to problem list>>

    2. m = 2.5log10 (2.52 x 10-8/b)

    m = 2.5log10 (2.52 x 10-8/3.5 x 10-17)

    m = 22.1433

    m = 22

     

     

  2. Alpha Beta has an apparent magnitude of 23. What is its brightness? (lognx =y, ny = x) <<to problem list>>

    3. m = 2.5log10 (2.52 x 10-8/b)

    23 = 2.5log10 (2.52 x 10-8/b)

    23/2.5 = log10 (2.52 x 10-8/b)

    9.2 = log10 (2.52 x 10-8/b)

    109.2 = 2.52 x 10-8/b

    1.5848 x 109 = 2.52 x 10-8/b

    b = 2.52 x 10-8/1.5848 x 109

    b = 1.5901 x 10-17

    b = 1.6 x 10-17

     

     

  3. Twnentieth Centauri is 450 parsecs away, and has an apparent magnitude of 9. What is its absolute magnitude? <<to problem list>>

    4. M = m – 5 log10(d/10)

    M = 8 - 5 log10(450/10)

    M = -.266

    M = -.3

     

     

  4. Cepheid Variable has an absolute magnitude of -3. How far away is it if it has an apparent magnitude 17? (lognx =y, ny = x) <<to problem list>>

    5. M = m – 5 log10(d/10)

    -3 = 17 – 5 log10(d/10)

    (-3 – 17)/-5 = log10(d/10)

    4 = log10(d/10)

    104 = d/10

    104x10 = d

    d = 100000

    d = 1 x 105 pc

     

     

  5. 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) <<to problem list>>

    6. M = m – 5 log10(d/10)

    -2.5 = m – 5 log10((230/3.26)/10)

    -2.5 = m – 5 log10((70.552)/10)

    -2.5 = m – 5 log10(7.0552)

    -2.5 = m – 4.24355

    -2.5 + 4.24355

    m = 1.74255

    m = 1.7

     

     

  6. 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>>

    7. absolute magnitude = 6

     

    M = m – 5 log10(d/10)

    6 = 12 – 5 log10(d/10)

    (6 – 12)/-5 = log10(d/10)

    1.2 = log10(d/10)

    101.2 = d/10

    15.8489 = d/10

    d = 158.489

    d = 160 pc

     

     

  7. 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>>

absolute magnitude = 3

 

M = m – 5 log10(d/10)

3 = 21 – 5 log10(d/10)

(3 – 21)/-5 = log10(d/10)

3.6 = log10(d/10)

103.6 = d/10

3981.07 = d/10

d = 39810.7

d = 40000 pc

 

Created 6/8/00

By Brian Peterson

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