Question 3.) You observe a eclipses with a period of 4.0 days. ingle-lined spect

Question

Question 3.) You observe a eclipses with a period of 4.0 days. ingle-lined spectroscopic binary system, which also undergoes !bi a) if we assume both stars are main sequence stars, what does the presence of only one set of lines tell us about the relative masses of the two stars? b) Imaging the bright member (Star A) of this system in B and V flters yields a (B-) color of -0.33 for the star. What is its surface temperature? c) Using this temperature and Table A.5 in your book, what is the spectral type and mass MA (in Mo) of this star? d) Spectroscopy of Star A yields a maximum Doppler shift of =001 87 nm for a line with ,-700.0 nm. What is the star's orbital speed v e) Using the assumption from part a), and the values in parts c) and d), what is the mass MB of the fainter star? f) Which eclipse will be deeper: when Star A blocks Star B, or when Star B blocks Star A? Why?

Explanation / Answer

a) we have the only information of eclipsing period = 4 days in this case and we can see only one set of lines that means one of the star is too faint while the other produces the single set of lines. That means the line producing star has the maximum mass while fainted star has the negligible mass.

b) using B-V index we can calculate the surface temperature using Ballestros' formula

T= 4600 [ (1 /(.92 * B-V +1.7)) + (1/ (.92 B-V +.62))]

using value for te index -.33 we get

T = 4600 [1/ 1.3964 + 1/.3164]

or T =4600 [.7161 + 3.16]

or T =4600 * 3.87 = 17832.61 K

c) Using the temperature we can say its spectral type is class B (Temerature in the range of 10000-30000K) and mass in the range of 2.1- 16 M.

d) The orbital speed Va= (doppler shift/ rest wavelength ) * speed of light =( .0187/700 ) * 3 * 10 ^8 m/s =8014.287 m/s

e) Using the formula P= (4pie^2 *a^3/G(Ma +Mb)) ^1/2

where P= 4 =(4 * 3.14^2 * a^3 / 6.67 * 10^-11 (Ma+Mb))^1/2

using Ma value from the range 2.1 -16 M we can get the mass of other star.

f) The deeper eclipse will occur when star B will block A as A is the hotter star between the stars.

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