# Is an unchanging two star orbiting system in equilibrium (force wise)?

1. Oct 29, 2012

### student34

My professor actually told us that it isn't, but I don't see how it's not. If he meant that one of the stars is not in equilibrium, then that would make sense to me.

2. Oct 29, 2012

### sophiecentaur

If the orbits were circular then you 'could' argue that there is equilibrium but most orbits are elliptical, in which case there are unbalanced forces and consequent acceleration. So no.equilibrium, by definition.

3. Oct 29, 2012

### JustinRyan

Doesn't a circular orbit also have an acceleration? If the force between them is not constant but oscilates with a regular period, could this not be considered an equilibrium of sorts?

Student34, if you are not sure what your professor meant by this, I suggest you ask him/her.

4. Oct 29, 2012

### mikeph

Can you define what it means for a system to be in equilibrium? When this word is used in the context of forces, it is attributed to individual particles.

So neither planet is in equilibrium since each one has a resultant force acting on it.

5. Oct 29, 2012

### sophiecentaur

Strictly, there is no equilibrium in a circular orbit but, in the reference frame of one star, the forces are balanced and distances are not changing, so ....

6. Oct 29, 2012

### A.T.

In the inertial frame. But with circular orbits you could find a frame where both objects are in equilibrium. With elliptical orbits that is not possible.

7. Oct 29, 2012

### Staff: Mentor

If you consider general relativity, that system will emit gravitational waves and reduce the orbital radius slowly. But that is not an "unchanging" system any more...

8. Oct 29, 2012

### student34

What if the system is defined to be both stars; then don't the stars cancel out each other's force? Let's assume the orbits do not change in time, and assume gravitational radiation is negligible.

9. Oct 30, 2012

### JustinRyan

I think that has been presumed already.

Going by the definitions given and assuming the binary system is part of a galaxy or infuenced by gravity of other bodies in the universe, there is always going to be a net force on the system.

10. Oct 30, 2012

### mikeph

Why would you add the forces acting on two different particles?

This sum will be zero by the definition of Newton's third law.

Please define precisely what you mean when you ask whether a multiple particle system is "in equilibrium". This question is still vague to me.

11. Oct 30, 2012

### student34

My memory is fuzzy on the original question now, but I think you answered it in what I put in bold.

My argument was that the star system is in equilibrium because he was defining the system to be a two stars in an elliptical orbit. He probably meant to ask about whether or not one of the two stars is in equilibrium.

12. Oct 31, 2012

### sophiecentaur

If you were living on one of those stars (I guess it would need to be a rotating planet for this thought expt.) and it was covered in cloud, you would be aware of the other body because of the tiny tidal force as it rotated. Hence, I suppose you would be aware that you were not actually in equilibrium.
But the earlier question ("what is meant by equilibrium?") needs to be addressed if we really want closure on this.

13. Oct 31, 2012

### JustinRyan

My definition would be...

With the exclusion of inteference from outside the system, it would continue in its current state indefinitely. - I use the term indefinitely loosely, as I realise that will never be possible.

But the term the in the OP was "force wise" so I am not sure what that means.

14. Oct 31, 2012

### sophiecentaur

Using that definition, I'd say that a circular orbit would be in equilibrium but an elliptical orbit would not.

"Forcewise" you could come to the same conclusion as the forces (magnitude) would be unchanging in a circular orbit.

15. Oct 31, 2012

### mikeph

I'd say the word should not be used in this context, because it refers to a balance, neutralisation or cancellation. I simply don't see any two properties that can meaningfully be compared to test this condition, apart from summing all the forces (which is trivially zero).

16. Nov 1, 2012

### Staff: Mentor

You could view the elliptic orbits as "current state" and come to a different conclusion.
Or take the current position and velocity as "current state", and no orbit is in equilibrium.
So many different options unless "in equilibrium" gets a clear definition.

17. Nov 1, 2012

### student34

In terms of physics, does a mass in equilibrium mean that there are no net forces acting on it?

18. Nov 1, 2012

### sophiecentaur

I think that the classical approach would assume that, in equilibrium, you couldn't detect any acceleration. In the case of a rotating planet, because of the varying 'weight' forces for someone on the surface, they would detect acceleration change and I would say they would, in their own frame of reference, 'know' they were not in equilibrium. But, from another frame, it would be clear that they are constantly accelerating hence not in equilibrium.

I guess it's down to a more strict definition of the term than I'm capable of giving. It's been mentioned already that the term 'equilibrium' has different meanings in different situations.

19. Nov 2, 2012

### mikeph

By the definition we learnt, equilibrium = all forces cancel/no net force.

20. Nov 2, 2012

### sophiecentaur

No net force on what? - is the point. We all know that simple definition of equilibrium but I though we were exploring the various aspects of the system where equilibrium could be said to apply.
We could say that no energy is being exchanged with the system - so there is no work done on it and so the forces on the whole system must be in equilibrium. Looking at one of the pair, there is acceleration, whatever orbit, because there is motion in a curve. There must be unbalanced forces so No Equilibrium. But would the star detect any acceleration / force (if it could see no more than its partner star. Would the star think it was in equilibrium?
etc. etc. different considerations (reference frames) yield different conclusions.