Astronomy 12 - Spring 1999 (S.T. Myers)

Solutions to Problem Set #7

Solutions:
  1. We use the usual relation

    log( L / Lsun ) = ( 4.83 - Mv ) / 2.5

    so

    4.83 - Mv 2.5 [ 1.15 log( P / 1 day ) + 2.47 ]

    or

    Mv = -2.875 log( P / 1 day ) - 1.34

    for the absolute visual magnitude of a Cepheid. The distance modulus

    mv - Mv = 5 log( d / 1 pc ) - 5 + Av       ( no dust Av=0 )

    gives

    log( d / 1 pc ) = 0.2 [ mv + 2.875 log( P / 1 day ) + 6.34 ] = 0.2 mv + 0.575 log( P / 1 day ) + 1.27

    thus for Polaris ( P = 13 days, mv = 2.3 )

    log( d / 1 pc ) = 0.2 · 2.3 + 0.575 · log 13 + 1.27 = 2.37

    or d = 235 pc. Note that the absolute magnitude of Polaris is

    Mv = -2.875 log 13 - 1.34 = -4.54

    with a luminosity of 5611 Lsun!

  2. The mean density

    = ( 3 M ) / ( 4 R3 )

    and the period-density relation

    P = [ 3/ (2G) ]1/2

    give on substitution

    P = [ ( 2 2 R3 ) / ( G M ) ]1/2

    or numerically

    P = 6117 s · ( M / Msun )-1/2 ( R / Rsun )3/2 ( 3 / 4 )-1/2

    which fills in our table:

     M / Msun   R / Rsun    Mv    P (days)  
    5  36.3   -3.1   6.925 
    10  195.0   -5.9  60.96   

    The quantities of interest are:

    log P Mv
    0.840 -3.1
    1.785 -5.1

    and thus we solve the system of equations

    -3.1 = 0.840 · a   +   b
    -5.9 = 1.785 · a   +   b

    which gives by subtraction

    2.8 = - 0.945 · a   ==>   a = -2.963

    and then

    b = -3.1 - 0.840 · a   ==>   b = -0.611

    by substitution. Our period-magnitude relation is thus

    Mv = -2.963 log( P / 1 day ) - 0.61

    (which is similar to that used in the previous problem), and

    P = 20 days   ==>   log P = 1.301   ==>   Mv = -4.465

    and thus the distance modulus

    mv - Mv = 10.4 - (-4.465) = 14.865

    gives us

    log( d / 1 pc ) = 0.2 [ mv - Mv + 5 ] = 3.973   ==>   d = 9397 pc

    and the star is almost 10 kpc away.

  3. To locate Mira on the H-R diagram, we need its luminosity (in Lsun) and its effective temperature. For an absolute magnitude Mbol = -5 and the Sun's bolometric magnitude Mbol = +4.75 we get

    log( L / Lsun ) = ( 4.75 + 5 )/2.5 = 3.9   ==>   L = 7943 Lsun

    Looking at the H-R diagram on p.321 of ZG4 for example, the point

    log( L / Lsun ) = 3.9         log( T ) = 3.36

    is to the right of the asymptotic giant branch (AGB). Mira must be an evolved star (see also the diagram on p.353 of ZG4).

    For the average temperature T = 2300 K, the average radius is

    R / Rsun = ( L / Lsun)1/2 ( T / 5770 K )-2 = 439

    which is in line with it being a late M supergiant. The luminosity ratio is given by the magnitude difference

    Lmax / Lmin = 10 5.1/2.5 = 109.6

    and so

    Rmax / Rmin = ( Lmax / Lmin )1/2 ( Tmax / Tmin )-2 = 6.2

    and the radius changes by a factor of 6 during the pulsation cycle!

  4. For stars of the same luminosity L, they will appear to have a flux

    f = L / (4d2)

    and thus counting the number brighter than some limiting flux f is equivalent to counting the number closer than a limiting distance

    d = [ L / (4f) ]1/2.

    If the stars are distributed uniformly with density n per unit volume, then

    N( >f ) = n V( d3 ( f-1/2 )3 f-3/2.

    Note that this is true only for a uniform volume density of sources, but is true for any luminosity function of stars.

  5. The circular velocity is given by

    vcirc2 = G M / R   ==>   M v2 R

    and since the velocity v is constant

    M( < R ) R

    is the mass law. For a spherical mass "halo", the mass grows as

    dM/dr = 4R2   ==>   R-2 dM/dr R-2

    which is a common profile for the outer parts of massive galaxies and clusters of galaxies. Note you could have also obtained this scaling by

    M / R3 R / R3 R-2

    as before.


smyers@nrao.edu   Steven T. Myers