Solutions To Binary Star Problems

The problems again:

(1) Find, approximately, the periods of revolution of the following binary star systems in which each star has the same mass as the Sun, and in which the semi-major axis of the relative orbits has the value:

(a) 1 AU

(b) 6 AU

(c) 100 AU

(2) For each of the systems in (1), at what distance would the two stars appear to have an angular separation of 1"?

(3) The true relative orbit of Epsilon Ursae Majoris has a semi-major axis of 2½" and the parallax of the system is 0."127. If its period is 60 years, find the sum of the components in solar mass units.

(4) A hypothetical spectroscopic-eclipsing binary system is observed and its period is 3 years. The maximum radial velocities with respect to the center of mass of the system are:

Star A: 4π/3 AU/yr   and  Star B: 2π/3 AU/yr

(a) Find the ratio of the masses of the components.

(b) Find the mass of each star in solar units. 

(Assume the eclipses are central)

Solutions:

1)   For each system we have: m1 = m2 so total mass = 2M_s (solar masses)

Then:  m1+ m2 = (a)³/ P² and

P = [(a³)/ (2)]^½

(a) IF  a = 1 AU

P = [1/2]^½ = 0.707 yr.

(b)IF  a = 6AU

P = [(6)³ / 2]^½ = 10.3 yr.

(c) IF a = 100 AU

P = [(100)³/2]^½ = 707.1 yrs.

(2)We need to determine the distances for the two stars in each system of (1) to appear to have an angular separation of 1". 

We require:

m1 + m2 = (d a")³/ P²

and need to find d for different a" = 1"

Simplifying:

(a) P = 0.707 yr.

(d a") = [(m1 + m2) P²]^1/3

Then: d = [(2) (0.707)²)]^1/3 = 1 pc

(b) P = 10.39 yr.

Then: d = [(2) (10.3)²)]^1/3 = 6 pc

c) P = 707.1 yr

Then: d = [(2) (707.1)²)]^1/3 = 100 pc

(3)We note the true relative orbit of Epsilon Ursae Majoris has a semi-major axis of 2½" and the parallax of the system is 0."127. The period of 60 years then allows us to find the sum of the components in solar mass units. 

First, find the distance d from the parallax method:

d = 1/p"  =  1/ 0."127 = 7.87 pc

Apply Kepler’s 3^rd law for binaries separated by a”.

m(A) + m(B) = (d a")³/ P² 

where a" = 2½" and d = 7.87 pc with P = 60 yrs.

Then:

m(A) + m(B) = ((7.87) (2½" )]³/ 60² = 2.1 solar masses 

(4)The graphs of the radial velocity curves for the stars, matching the physical situation of each, are given in the diagram below. The important thing is to have the maxima and minima in the correct directions at the key points in their respective orbits. 

Positions (1) and (2) in the graphs allow us to obtain the maximum relative velocity for the system, for which:

V = (4π/3 + 2π/3)AU/yr = 6π/3 AU/yr = 2π AU/yr

P = 3 yrs.


(a) Find the ratio of the masses of the components. 

The ratio of the masses will be in the ratio of the radial velocities or: 

4π/3: 2π/3 =  4π/ 2π  =  2: 1

(b) To get the mass of each star in solar units.(Assume the eclipses are central)

Recall that to get distance:

a = (V x P)/ 2π = (2π AU/yr x 3 yr)/ 2π = 3 AU

and: m1 + m2 = (3AU)³/(3 yr) ² = 3 solar masses

The masses are inversely proportional to the radial velocities in their orbits. Since:

Star A has v_A = 4π/3 AU/yr = 2 v _B where v _B = 2π/3 AU/yr

then:  m(A)/ m(B) = v _B / v _A = 1/2

or m(A) = m(B)/2 or m(B) = 2m(A)

But: m(A) + m(B) = 3  so: m(A) + 2m(A) = 3

3 m(A) = 3

and m(A) = 1 solar mass, m(B) = 2m(A) or, 2 solar masses