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alex41 [277]
3 years ago
15

How does the rotation of a galaxy result in spectral line broadening?

Physics
1 answer:
IRISSAK [1]3 years ago
4 0

Answer:

Explanation:

Normal galaxies are made up of stars and (in the case of spiral and irregular galaxies) gas and dust. Their spectra consist of the sum of the spectra of these components.

The optical spectra of normal stars are continuous spectra overlaid by absorption lines (Figure 1). There are two factors to consider when adding up the spectra of a number of stars to produce the spectrum of a galaxy:

Different types of star have different absorption lines in their spectra. When the spectra are added together, the absorption lines are 'diluted' because a line in the spectrum of one type of star may not appear in the spectra of other types.

Doppler shifts can affect all spectral lines. All lines from a galaxy share the red-shift of the galaxy, but Doppler shifts can also arise from motions of objects within the galaxy. As a result, the absorption lines become broader and shallower. We explain below how this Doppler broadening comes about.

HII regions in spiral and irregular galaxies (though not, of course, ellipticals) shine brightly and contribute significantly to the spectrum of the galaxy. The optical spectrum of an HII region consists mainly of emission lines, as in Figure 2. When the spectra of the HII regions and the stars of a galaxy are added together, the emission lines from the HII regions tend to remain as prominent features in the spectrum unless a line coincides with a stellar absorption line. There are Doppler shift effects, however, as described for stellar absorption lines, and hence emission lines too are broadened because of the motion of HII regions within a galaxy.

Box 1: Doppler Broadening

The Doppler effect causes wavelengths to be lengthened when the source is moving away from the observer (red-shifted) and shortened when the source is moving towards the observer (blue-shifted).

Light from an astrophysical source is the sum of many photons emitted by individual atoms. Each of these atoms is in motion and so their photons will be seen as blue- or red-shifted according to the relative speeds of the atom and the observer. For example, even though all hydrogen atoms emit H photons of precisely the same wavelength, an observer will see the photons arrive with a spread of wavelengths: the effect is to broaden the H spectral line - called Doppler broadening.

In general, if the emitting atoms are in motion with a range of speeds Δν along the line of sight to the observer (the velocity dispersion) then the Doppler broadening is given by

where c is the speed of light, and λ is the central wavelength of the spectral line.

Why would the atoms be in motion? An obvious reason is that they are 'hot'. Atoms in a hot gas, for example, will be moving randomly with a range of speeds related to the temperature of the gas. For a gas of atoms of mass m at a temperature T, the velocity dispersion is given by

where k is the Boltzmann constant (1.38 × 10−23 J K−1).

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Answer:

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Explanation:

Given:

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Velocity = Frequency x wavelength

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3 years ago
What is the relationship between the spring constant and the period in a mass hanging on a spring oscillation and why?
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Explanation:

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
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Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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Answer:

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