Final answer:
Using Newton's reformulation of Kepler's third law, given a binary system with one star of mass 1/3 Msun and the other of mass 2/3 Msun with a period of 1 year, the distance between the two stars is found to be 2R, where R is the Earth-Sun distance.
Step-by-step explanation:
To calculate the distance between two stars in a binary star system, we can utilize Newton's version of Kepler's third law. The law relates the period (P) and the sum of the masses (M) with the semi-major axis (D) of the orbit.
Using the given that the lighter star is one-third the mass of the Sun (Msun), we designate its mass as M1 = 1/3 Msun. Accordingly, the companion star would be twice this mass, M2 = 2/3 Msun.
Adding both, we get the total mass of the system, M1 + M2 = Msun. Given a period (P) of 1 year (same as Earth's orbital period) and using Earth-Sun distance R as a scale, the law can be simplified as D3 = Msun P2, in which D represents the semi-major axis of the orbit of the binary system relative to the sun.
Since we are looking for the distance, not the semi-major axis, and both stars orbit their common center of mass, we seek the distance (2D) between the stars. Solving this equation, we find that D is equivalent to R, the earth-sun distance; thus, the total distance between the two stars, 2D, is 2R.