"An orbit is the gravitationally curved path of one object around a point or another body, for example the gravitational orbit of a planet around a star" (from Wiki).

 

Under attraction, the naive point of vue is that bodies will bump.

Newton made a hard job on demonstrating that bodies falling to each other will not necesseraly bump, but orbit.

Under gravity, which is attractive.

 

Well, I was wondering, based on the general symmetry of physical laws, if under repulsive action, maybe the naive picture of bodies expelling each other away was right, or do they orbit ?

 

I couldn't find anything non-symmetric in the laws of motion. So I deduced that under repulsive force, objects must orbit too.

Right?

AFAIK there are no stable orbits under repulsion, with the possible exception of the trivial case of infinite separation. Attractive forces give rise to a negative mechanical energy, which is a bound state. These are unavailable with a repulsive force.

 

Attractive forces allow the possibility of exerting a force that does no work, which occurs when the force is perpendicular to the motion. That solution is not present for a repulsive force.

Are you suggesting that the laws of motion are not symmetric?

doesn't the symmetry usually have to do with time rather than direction?

Yes.

That's why I believe direction is not so important. The same laws must apply. Intuitively speaking.

you do realise the situations are totally different though,

 

there is a difference between reversing the direction of a force and keeping positive time than reversing time right?

Are you suggesting that the laws of motion are not symmetric?

 

They are certainly not symmetric in that sense, where you will always get a similar pattern of movement by reversing the forces.

Edited July 2, 201015 yr by J.C.MacSwell

Could you give an example?

Are you suggesting that the laws of motion are not symmetric?

What do you mean by that?

 

It is pretty obvious from the underlying mathematics that repulsive forces cannot result in closed paths.

It is obvious to you, not to me.

 

from Wiki laws of motion

"First Law: Every body will persist in its state of rest or of uniform motion (constant velocity) in a straight line unless it is compelled to change that state by forces impressed on it.[2][3][4] This means that in the absence of a non-zero net force, the center of mass of a body either is at rest or moves at a constant velocity.

 

Second Law: A body of mass m subject to a force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma. Alternatively, the total force applied on a body is equal to the time derivative of linear momentum of the body.

 

Third Law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction".

 

Where is it obvious?

It is obvious to you, not to me.

 

from Wiki laws of motion

"First Law: Every body will persist in its state of rest or of uniform motion (constant velocity) in a straight line unless it is compelled to change that state by forces impressed on it.[2][3][4] This means that in the absence of a non-zero net force, the center of mass of a body either is at rest or moves at a constant velocity.

 

Second Law: A body of mass m subject to a force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma. Alternatively, the total force applied on a body is equal to the time derivative of linear momentum of the body.

 

Third Law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction".

 

Where is it obvious?

 

Well, as stated in the third law, a repulsive force for one body will result in an equal but opposite, therefore repulsive, force to the other. They will be accelerated in opposite directions away from each other. There is no force available to facilitate an orbit.

It is obvious to you, not to me.

 

from Wiki laws of motion

"First Law: Every body will persist in its state of rest or of uniform motion (constant velocity) in a straight line unless it is compelled to change that state by forces impressed on it.[2][3][4] This means that in the absence of a non-zero net force, the center of mass of a body either is at rest or moves at a constant velocity.

 

Second Law: A body of mass m subject to a force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma. Alternatively, the total force applied on a body is equal to the time derivative of linear momentum of the body.

 

Third Law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction".

 

Where is it obvious?

 

Where is the analysis of underlying mathematics?

underlying mathematics have been mentionned by DH. Surely he knows better.

 

It is not in [math]

F = m a

[/math]

 

and not in [math]

F = \frac{d}{dt} (mv)

[/math]

 

Maybe in [math]

F = G \frac{m_1 m_2}{r^2}

[/math]

 

As I see it, there is no "obvious" orbital path coming from these equations.

DH said it's obvious when you look at the analysis of the underlying mathematics. Thus far you've gotten to the basic equations. Do the analysis. The wikipedia article you originally quoted has an analysis section. What happens when the force is repulsive?

That was my question.

Here?

Yes, there. You have a second-order differential equation. With an attractive force, you have cyclical solutions. Those are closed orbits. With a repulsive force, these solutions are not present.

In time reverse symmetry, the orbit would be in the opposite direction, but still incorporate an attractive force.

The paths of bodies with net attractive force will always be concave towards one another. The paths of bodies with net repulsive force will always be convex towards one another. (The trivial exception in both cases is the straight line directly towards or away that occurs with zero orthogonal velocity.) In order to make a circuit around a given point, your path has to be concave at some point.

Edited July 6, 201015 yr by Sisyphus

So I understand that you all agree upon the idea that physical laws of motion are asymmetric as concern the attractive-repulsive instance.

So I understand that you all agree upon the idea that physical laws of motion are asymmetric as concern the attractive-repulsive instance.

 

I don't know what you mean by asymmetric.

I mean when an object attracts another object, the result is totally different than when objects repulse each other. The path is different. It is not like the action-reaction concept for example, or like the equality of angle of incidence & reflection.

I mean when an object attracts another object, the result is totally different than when objects repulse each other. The path is different. It is not like the action-reaction concept for example, or like the equality of angle of incidence & reflection.

 

There are several symmetries. What symmetry is that supposed to represent? The closest example I can think of is charge conjugation, but it's not that, since C-symmetry requires changing the charge of both particles involved. So this is more of an un-symmetry, as it were. Trajectories are not symmetric under a change in the sign of the force.

Since it is about trajectory, I guess it is the original meaning of symmetry: geometric symmetry.

 

Trajectories are not symmetric under a change in the sign of the force.

 

Exactly, You have talent Swansont.

Just for the record, not only can't you have a stable orbit with repulsive forces, you can't get a stable orbit with anything other than an inverse square law.

So, trajectories are not symmetric under a change in the sign of the force. (thank you Swansont). In this case of course, ONLY the force have changed sign: attractive have becomed repulsive.

Now, the tricky question comes: what happens when both force and time change sign?