Solution homework Set V
                                                            
 
1)     Solution of the first problem. 
     
     The formula from which you obtain the radius is L = 4 pi R^2 sigma T^4.   
     For the star this means:
     L_star= 4 pi R_star^2  sigma T_star^4    (1)
     and the corresponding formula for the Sun is: 
     L_sun; = 4 pi R_sun^2 sigma T_sun^4;       (2).  
     The problem told you that R_star / R_sun;=  10, 
but the problem did not give you any value for the temperature. 
In  fact you need to calculate it as a function of the temperature of
the Sun using the Wien's law:     
  lambda_star= 0.029 / T_star     
  lambda_sun = 0.029 / T_sun 
     The problem also told you that the star emits at a wavelength
double the one of  the Sun, 
  this means that T_star / T_sun; = 1/2.  
And using equations (1) and (2), doing the ratio side by side, you
    obtain that  
L_star / L_sun = (100 * 1/ 16)^0.25 = 6.25       
 this means that the star has a luminosity 6.25 times the  one of the Sun. 
 
 
2)     Solution of the second problem.
 
    The formula to solve the problem is d = 1 / p, where d is the distance of the star in parsecs and p is the
    parallax of the star in arcsec. Inserting the numbers given by the problem in the formula, you obtain a distance d
    = 1.3 pc. Remembering that 1 pc = 3.26 ly, Proxima Centauri is at a distance d = 4.24 ly from the Earth, and
    considering that it has the largest parallax (except the Sun), it is also the closest known star. 
 
 
3)     Answer to the question # 3. 
 
    The theory of the planetary system formation has the system formed out of a large, spinning cloud of gas and dust.
    As the cloud begins to spin (because of factors not yet known), it flattens into a pancake shape, rotates faster
    and faster (because of the conservation of angular momentum), and forms a series of concentric rings that lie in
    the same plane and rotate in the same direction. Based on temperature, density and condensation factors, the
    central concentration originates the star, while each of these rings eventually forms a planet with unique
    characteristics, which depend on the distance from the central star.  
 
 
4)     Answer to the question # 4.   
     
    Since the two stars appear to have the same brightness, but B is intrinsically brighter (absolute magnitude 4),
    then B must also be more distant.  B is 5 magnitudes more luminous than A (9 - 4 = 5), which is exactly a factor
    of 100 in luminosity. Using the inverse square law, star B must be SQRT(100) = 10 times farther away than star A. 
 
 
5)     Answer to the question # 5.
     
    Using the inverse square law, if a Cepheid is 100 times brighter than an RR Lyrae variable, then it should be
    visible at 10 times the distance. In other words, placing the Cepheid 10 times farther away will make it 100 times
    fainter. Therefore it would equal the RR Lyrae star?s brightness.