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Astronomy 2 Tutorials session 2002-03
Tutorial 7 Model Answer
1.1
TOT
From the notes on line emission, P21
 N 2A21h 21  N 2A21hc / 21 , and by the inverse square law with
TOT
R=1AU, P21
 4R 2F21 where F21 is the line flux (10-2 Wm-2) at the earth.
Hence
A21  4R 2F2121 /(N 2hc) .
Now the number of H atoms in corona NH is almost exactly equal to the mass of the corona divided by the mass
of a proton, 1016/1.7x10-27 .
Number of Fe atoms NFe = NH x abundance by number = NH x 7 x 10-5.
Number of Fe atoms with electron in upper level 2 is NFe x 10-4.
Thus A21  [4  (1.5  1011) 2  102  500  109 ] /[4.1 1034  6.6  1034  3  108 ]  1.75  105s1
and so  21  1 / A21  5.7  10 6 s , which is long, and so the line is forbidden.
The natural line width is
 NAT  221 /( 4c 21 )  (5  10 7 ) 2 /[ 4  3  108  5.7  10 6 ]  9.5  10 18 m
Stars and Their Spectra
Model Answer
Tutorial Question for 14/03/03
What are the primary advantages and disadvantages of using a colour-magnitude diagram as an
alternative to a standard H-R diagram based on spectral class?
Advantages:
- It relies on an objective classification – i.e. measurement-based classification
- It is possible to classify every object by measuring the flux through standard filters.
- Avoids ‘digitisation’ of the HR diagram, as classification not based on presence or absence of discrete features in
spectrum.
Disadvantages
- Requires accurate calibration of filters used, and good photometry
- Since classification is based on the flux in different parts of the spectrum, one must beware of the effects of
wavelength-dependent absorption (i.e. interstellar reddening) which would change the position of a star in the
colour magnitude diagram depending on its distance as well as spectral type
[5]
What is meant by the term ‘luminosity indicator’? Explain the difference between a positive and a
negative luminosity indicator.
A luminosity indicator is a property (e.g. width, depth) of a spectral feature (e.g. a line or a group of lines) which varies with
the star’s luminosity, within a single stellar spectral class (or at a given value of Teff). A positive L.I. is an indicator which is
enhanced with increasing luminosity whereas a negative L.I. is reduced with increasing luminosity.
[3]
Vega is an AOV star, of effective temperature T = 9520 K, and absolute bolometric magnitude Mbol = 0.3. Given that the
Sun has effective temperature 5800 K and Mbol = 4.72, estimate the radius of Vega in units of the solar radius, stating
clearly any assumptions that you make.
Assuming that the stars radiate as (spherical) blackbodies, we can write:
LBol,Vega
LBol, Sun
4Rvega TVega
2

4RSun TSun
2
4
4
Then use Pogson’s equation:
M Bol,Vega  M Bol, Sun
 RVega 2TVega 4 
 LBol,Vega 




 2.5 log 10 
  2.5 log 10  R 2T 4 
L
 Bol, Sun 
 Sun Sun 
Putting in numbers gives
 RVega 
  4.72  0.3  2.15  2.26 ,
5 log 10 
R
 Sun 
so that RVega=2.84 RSun
[5]
total
[13]