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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 Problem 15 Problem 16 Problem 17

  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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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 1.816x1014 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>>

  Ok... let’s think about this for a second.

  We have the h given to us which is 4.50 meters. Also, we know what “g” is ... 1.816x1014 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.

  After looking at our given data, we can just use the formula

  Df / f = gDh / c2

  We want to find the change in frequency we hear so we solve for the change in frequency... just plug in the given data...

  Df / 440 = 1.816x1014 (4.50) / (3x108) 2

  to solve this, you get ....

  Df = 3.9952 Hz which rounds to 4.00 Hz

  Now that we found the change, we need to add this to the frequency that we are hearing which is, obviously the one coming from the neutron star or the TOP.

  440.00 Hz + 4.00 Hz = 444.00 Hz

  We observe a frequency of 444.00 Hz with a frequency difference of 4.00 Hz

  Now that we found that is this sharp or flat... just remember that at the TOP it is fast/sharp and on the BOTTOM it is slow/flat... so there is your answer... it is sharper 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>>

  Well, this is basically the same concept as in number 15 just that we are finding a different variable... To be exact we want to know the value of one “g”...

  Here is the formula again :

  Df / f = gDh / c2

  Since we do not know the change in frequency yet, we need to figure that out which just takes a simply subtraction...

  Df = 91.7 – 90.2 = 1.5 MHz (so we need to multiply this answer by 106 )

  We want to know the “g” value at our elevation which means that we have to use our FM dial frequency as our frequency ... smart huh?!

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

  1.5x106 / 90.2x106 = g (35.6) / (3x108)2

  To solve for “g” is now easy

  Do the algebra and you get

  g = 4.2x1013 N/ kg

  Now are we higher or lower?! Let’s think about this just for a second... Just by looking at our given data, we can conclude that we must be 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...

  1 A.U. = 1.50x1011 m

  our change in frequency is easy to find... subtraction again

  Df = 423 – 417 = 6 nm

  use the formula Df / f = gDh / c2 and plug in the given info to find the value of “g”

  6 / 423 = (g) 1.50x1011 / (3x108)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

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