Astronomy 11 - Fall 1998 (S.T. Myers)

Solutions to Problem Set #6

Solutions:

  1. (a)  One 1 AU subtends angle p at distance r - this is our usual skinny triangle! If p is in radians, then r = 1/p AU. If p is in arcseconds, then r = 206265 AU / p. The distance 206265 AU is defined to be the parsec: the distance at which the parallax is one arcsecond! Thus, if p is in arcseconds, r = 1/p parsecs (pc). For EPS451, p=0.05" so r = 1/p = 20 pc (or 4.125 x 10^6 AU).

    (b)  The speed of light (3x10^5 km/s) is equivalent to 0.3068 pc/year, and thus one light-year (ly) equals 3.26 pc. Thus, the 20 pc to EPS451 equals 65.2 ly.

    (c)  Wien's Law gives the peak wavelength to be 2.898 mm/ T(K), so for a blackbody peak at 5882 Angstroms = 5.882 x 10^-7 m = 5.882 x 10^-4 mm, we find a temperature of 4927 K.

    (c)  If we take the ratio of the flux from EPS451 at distance r to the flux from the Sun at 1 AU, we find

    L / Lsun = ( F / Fsun ) x ( r / 1 AU )2 = (2.35x10-14)(4.125x106)2 = 0.4

    (c)  We have the Stefan-Boltzmann Law for the surface flux from the star, which when scaled to the luminosity, temperature, and radius of the Sun gives

    L / Lsun = ( R / Rsun )2   ( T / Tsun )4

    so

    R / Rsun = ( L / Lsun )1/2   ( T / Tsun )-2 = (0.4)1/2   (4927/5770)-2 = 0.87

    is the radius of EPS451.

  2. The equilibrium blackbody temperature of a sphere (for A=G) is given by

    Teq = Tss/21/2 = ( L / 16r2 )1/4 = 279 K ( r / 1 AU )-1/2 ( L / Lsun )1/4

    when scaled to our solar system. Thus, for L = 0.4 Lsun

    Teq = 221 K ( r / 1 AU )-1/2

    in the EPS451 system. We find, since EPS451 is fainter than our Sun, the habitable zone is smaller and closer in: 0.35 - 1.65 AU versus 0.56 - 2.60 AU for our Sun.

  3. (a)  From the calculation we did for the parallax in #1 we know that at the distance of EPS451, 1 AU subtends 0.05". Thus, any angular distance we just divide by 0.05" to get AU! Therefore, 0.35" = 7 AU. Kepler's 3rd Law, when scaled to our solar system with the mass of the Sun gives

    M / Msun = ( a / 1 AU )3 ( P / 1 yr )-2 = (7)3/(20.706)2 = 0.8

    and thus the star is 0.8 Msun.

    (b)  We did the equilibrium temperature above, and found

    Teq = 221 K ( r / 1 AU )-1/2

    and so

    Tss = 21/2 Teq = 313 K ( r / 1 AU )-1/2

    easily.

    (c)  We fill in the table:

    Orbit Semimajor Axis Period      Teq         probable class     
    no. (arcsec) (AU)   (K)  
    1 0.005 0.1 12.9 d 699 terrestrial, like Mercury?
    2 0.080 1.6 945.4 d 175 terrestrial, in habitable zone
    3 0.125 2.5 4 y 153 d 140 terrestrial
    4 0.20  4 8.9443 y 110 terrestrial or Jovian?
    5 0.35  7 20.706 y  84 Jovian
    6   13 52.4 y  61 Jovian
    7   25 140 y  44 Jovian?
    8   ~50 ~400 y  31 iceworld? like Pluto?

smyers@nrao.edu   Steven T. Myers