# Binary Star system with one star stationary?

Can a Binary Star system be possible where in one star is stationary and the other star revolves around it? (Just like a planet revolving a star. i.e planets in the system and the star revolving around their own center of masses to balance the system).

Well, motion is relative so you can choose a frame of reference where one is stationary. If you do though, it makes the equations of motion quite complicated.

Even in our solar system, the Sun isn't stationary. It orbits the center of mass of the whole solar system (barycenter), just as each planet orbits the center of mass.

The center of mass of our solar system moves relative to the sun due to the motion of the Sun and planets. Here is a graph:

(Source)

So you could potentially call a body in a system where the barycenter stays inside that body "stationary" but that's not technically correct, no matter how lopsided the masses of the bodies are. For example:

The larger mass will still always move relative to the system barycenter.

• OP has probably attached reference frame with heliocenter of the Sun. Commented Feb 6, 2014 at 8:08
• @SachinShekhar doing so makes the equations describing orbital motion much more complicated. Sure you can approximate it with the heliocenter but heck, you could choose the center of the Earth as a reference frame too. Point is, the barycenter of the system is the easiest (and I'd argue most correct) choice. The barycenter never coincides with the exact center of any body in any realistic n-body system. Commented Feb 6, 2014 at 8:11
• @BrandonEnright Klemperer rosette? Commented Feb 6, 2014 at 15:31
• @MrLister that's why I used the words "in any realistic n-body system" because those systems aren't stable. Commented Feb 6, 2014 at 15:33

As others explained: When two masses interact gravitationally, it's not like the smaller mass is orbiting the larger mass. Both bodies orbit the common barycenter. When one of the two masses is extremely large compared to the other, the barycenter of the system is almost in the center of the larger mass, so the effect on the larger mass is negligible (like a satellite orbiting the earth). But it still won't be completely stationary. The effect would just be too small to measure.

But I would like to get back to your original question, "Can a Binary Star system be possible where the mass difference is so large that the effect on the larger star is negligible": There isn't that much difference in the masses of stars. Stars need to be massive enough to generate enough pressure to maintain nuclear fusion, but not so massive that they collapse into black holes. Also, the larger the star, the faster does it undergo fusion and the shorter its lifespan. For that reason the heaviest known stars have just about 100 times the mass of our sun, while the lightest know stars have just one tenth of a solar mass. A mass difference of factor 1000 isn't small enough that the effect on the larger mass wouldn't be notable.

• Thanks Philipp. I'm confused now because of Sachin Shekar's answer. He gave an example of our sun sized star revolving around a large star like R136a1. Would the size difference be large enough for such configuration to be plausible? Commented Feb 6, 2014 at 16:21
• @user1324816 I think what Phillip and Brandon Enright are saying is that it is not possible for a body to remain stationary when another body orbits it. Sachin Shekhar is pointing out a situation in which one of the two stars appears to be stationary but it really isn't. Unless the mass of one star relative to the other is infinite, there will always be a small displacement from the center of the heavier body. It may be an inch or it may be hundreds of thousands of miles, but neither body is ever truly stationary. But it might be safe to assume it is if the displacement is negligible. Commented Feb 6, 2014 at 17:42

In any system, components revolve around center of mass of the system. In our solar system, Sun has 99.9% mass of the system. So, center of mass of system is inside the Sun (nearly coinciding with center of Sun).

Seeing the massive stars in the existence, yes, it's possible that a Sun type star can truly revolve around R136a1 type stars. R136a1 stars are 265 times solar mass. In a binary system with Sun, it'd contain 99.6% mass of the system. So, center of mass of the system would exist near its own center (so, size of star doesn't matter here).

This is an instance of the classical two body problem. We know from Classical Mechanics that the movement of two particles orbiting eachother can be reduced to the study of the movement of a single particle with a mass equal to the system's reduced mass, $$\mu=\frac{mM}{m+M}$$ moving in a conservative potencial (for more on this subject, you can refer to virtually any CM book; I specially recommend Golstein's "Classical Mechanics", although Thornton's book also gets the job done). As to your proposed problem, say you have two bodies orbiting each other, of masses $$m$$ and $$M$$, where $$m<. Then, the mechanical energy of the system is given by:

$$E = \frac{1}{2}\mu \dot r^2 + \frac{L^2}{2\mu r^2}+V$$

where $$r$$ is the distance between both bodies. Now, since the second star is much heavier than the first one, we could approximate the reduced mass as:

$$\mu = \frac{mM}{m+M}\cdot \frac{M}{M} = \frac{m}{1+m/M}\approx m$$

Therefore, Classical Mechanics correctly predicts that the heavier star remains stationary whilst the second one, which is much lighter, orbits it. We can introduce this approximation in the conservation of energy equation:

$$E \approx \frac{1}{2}m\dot r^2 + \frac{L^2}{2mr^2} + V$$

Where, in your problem, the potential will be keplerian:

$$V = -G\frac{M}{r}$$

Notice how the mechanical energy is conserved (since there are no external forces acting on the system), so you can actually calculate the orbit of the lighter star (more generally, you can reduce your problem to quadratures, which you can then integrate numerically):

$$dt = \sqrt{\frac{m}{2}}\cdot \frac{dr}{\sqrt{E-\frac{L^2}{2mr^2}-G\frac{M}{r}}}$$

Furhtermore, if you'd like to calculate the trajectory as a function of the angle (which I would recommend more than $$r(t)$$), you can use the conservation of angular momentum:

$$L = mr^2\dot \theta$$

Then, applying the chain rule to $$\dot r$$, you get:

$$\frac{dr}{d\theta}\cdot \frac{d\theta}{dt} = \frac{dr}{d\theta}\frac{L}{mr^2}$$

Therefore:

$$\frac{dr}{d\theta}\frac{L}{mr^2} = \sqrt{\frac{2}{m}(E-\frac{L^2}{2mr^2}-G\frac{M}{r})}$$

$$\boxed{d\theta = \sqrt{\frac{L^2}{2m}}\cdot \frac{dr}{r^2\sqrt{E-\frac{L^2}{2mr^2}-G\frac{M}{r}}}}$$

In short, your system would only be stable if one star were much lighter than the other, forming a binary system with an approximately circular orbit.