# Binary star systems

1. Sep 18, 2011

### SpY]

Application of the Doppler effect and Kepler's 3rd law

1. The problem statement, all variables and given/known data
(*This isn't the exact wording from a textbook, just as I had copied it down)
An eclipsing binary star system containing stars A and B in concentric orbits (about their common centre of mass) has it's maximum red- and blueshifts recorded.
The rest wavelength for the stars is 6000Å.
For star A, max-red occurs at 6001.3Å and max-blue at 6000.3Å.
For star B, max-red occurs at 6002.3Å and max-blue at 5999.3Å.
It takes 8.75 days to go from max-red to max-blue, and the stars orbit in the same period.

Find the following:
1. the radial velocity of the stars common centre of mass
2. the radial velocities of each star
3. the masses of each star

2. Relevant equations
Doppler Effect equation (for light):
$$v_{source} = c \frac{\lambda_{observed} - \lambda_{rest}}{\lambda_{rest}}$$

Newton's form of Kepler's 3rd law:
$$\frac{A^3}{T^2} = \frac{G}{4\pi^2} (M_A + M_B)$$
where A is the mean distance of separation (semimajor axis), T the orbital period, G the gravitational constant, and M the masses of each star. If A, T and M are expressed in astronomical units, years and solar masses respectively, $$\frac{G}{4\pi^2}$$ is unity.

3. The attempt at a solution
A very simple diagram of either orbit, would have the position of max-red at one end as the star recedes, and max-blue on the other end as the star approaches. The observer's line of sight would pass through the orbits' common centre.

1. To determine the shift for the centre of mass, I took the central (mean) wavelength for the stars:

$$\lambda_c = \frac{6000.3Å+6001.3Å}{2} = 6000.8Å$$

Then the radial velocity for the centre of mass (G)
$$v_G = 3 \cdot 10^5 \frac{6000.8Å-6000Å}{6000Å} = +40km/s$$
The positive implying it is receding from the observer.

2. For star A I compared wavelength at max-blue with the central wavelength,
$$v_A = 3 \cdot 10^5 \frac{6000.8Å-6000.3Å}{6000Å} = 25km/s$$

And for star B using max-red
$$v_B = 3 \cdot 10^5 \frac{6000.8Å-5999.3Å}{6000Å} = 75km/s$$

(these are the speeds of the stars relative to the centre of mass)

*I'm not entirely certain of my methods above, but those are the final answers that were given.

3. Not sure how to relate their masses or find their separation, all I can gather from the question is both stars have a period of 17.5 days.

thanks

Last edited: Sep 18, 2011
2. Sep 18, 2011

### Staff: Mentor

Doing nothing more than reword your own working may provide the answer: you have 3 concentric circles. You don't know their radii, but you do know that points on each circle, travelling at their respective radial speeds, all take 17.5 days to trace their circumference. So you can find the circumference of each circle, and thus their radii, and these provide the separation.

I would not have know how to solve this, but your working & provision of the equations seems to indicate this should work.

3. Sep 18, 2011

### SpY]

There's only 2 concentric circles, the centre of mass (G) is just a point about which the stars orbit.

But I suppose with your idea I can relate the period with the circumference from the orbital speed:
$$v = \frac{2\pi r}{T}$$
So for star A, and taking T in years (17.5days = 7/146 years) and 1km/s is approximately 0.21AU/year
$$r_A = \frac{T\cdot v_A}{2\pi} = \frac{\frac{7}{146}25\cdot 0.21}{2\pi} = 0.04 AU$$
And for star B
$$r_B = \frac{T\cdot v_A}{2\pi} = \frac{\frac{7}{146}75\cdot 0.21}{2\pi} = 0.12 AU$$

That's just the individual radii of the stars, how do I get the seperation/ semimajor axis from there?

4. Sep 19, 2011

### Staff: Mentor

Sketch 2 concentric circles of known radii, and mark the 2 dots representing the stars. Won't this show their distance apart?

5. Sep 20, 2011

### SpY]

In general terms: let R is the larger radii and r the smaller one. Their maximum distance apart is R+r (opposite side), and minimum R-r (same side) which average to R. However in this problem the stars orbit the same period, sweeping out the same angle per time, meaning the separation is constant - but I'm not sure if R+r or R-r.

Once I can be sure of that parameter, using K3 in astronomical units to get the sum of masses:
$$\frac{A^3}{{(17.5 \div 365)}^2} = M_A + M_B$$

Which is only the sum. I think the ratio of the masses will be the same as the ratio of the velocities (or inverse?), but not sure of the mathematical justification.

Last edited: Sep 20, 2011
6. Sep 21, 2011

### Staff: Mentor

The center of mass must be between the two stars. Otherwise, what force would provide the required centripetal force to enforce the circular orbit about the center of mass? Remember, the center of mass is only a location, it has no mass itself.

7. Sep 22, 2011

### SpY]

That is rather obvious, can't believe I missed it. Anyway putting in A=R+r I get the sum of the masses to be 1.78 (solar masses).

It makes sense to put the slower, heavier star on the inside and faster, smaller one on the outside. Not sure of the principle behind it, but if their masses are related by the inverse ratio of their radii:
$$\frac{M_A}{M_B} = \frac{r_B}{r_A} = 3$$
I get $$M_B = 0.45 M_{\odot} \; \;\;\;\;\;\; M_A = 1.35 M_{\odot}$$

thanks everyone for help