The choice of origin in solving the two body problem

[AlonsoMcLaren]

In solving the two body problem we choose the center of mass of the system as the origin.

But is the frame of reference we are setting up inertial? I don't think so.

Suppose the central force is gravitation, A has mass 4kg and B has mass 1kg.

If we put an object at the center of mass, then obviously in the frame of the center of the mass this object is at rest. But an nonzero net force pointing towards A is acting on the mass. Therefore, Newton's First Law is violated and the frame in noninertial.

The correct frame to choose, in my opinion, is the frame centered at the "equilibrium position" (If we put an object at the equilibrium position no net force act on the object)

Any thoughts?

 

[genericusrnme]

The center of mass doesn't change in a completely closed off system.
If you put an object at the center of mass you have just changed the center of mass.

If your model is correct, ball A is going to do a lot more moving that ball B which should't be true since ball A has more mass than ball B, the acceleration on B will be greater than the acceleration on A.

If it helps, work out the equations of motion for the simple two particles attatched by a spring system and work out where the center of mass is for all t, you'll find that it remains stationary throught the whole motion, the same thing will happen with the 1/r^2 force as well.

 

[AlonsoMcLaren]

The center of mass doesn't change in a completely closed off system.
If you put an object at the center of mass you have just changed the center of mass.

If your model is correct, ball A is going to do a lot more moving that ball B which should't be true since ball A has more mass than ball B, the acceleration on B will be greater than the acceleration on A.

If it helps, work out the equations of motion for the simple two particles attatched by a spring system and work out where the center of mass is for all t, you'll find that it remains stationary throught the whole motion, the same thing will happen with the 1/r^2 force as well.

How does putting a mass at the center of mass of a system change the center of mass?

Even if it does change the center of mass, I can assume the mass I put be very small. (Just like the "test charge" in the definition of electric field is very small so it does not affect the electric field significantly)

 

[chrisbaird]

The center of mass frame is by definition the frame in which the center of mass is at rest. If the center of mass location changes as time progresses, all that means is that you have chosen the wrong frame. I think you are confusing the concept of a "frame" which includes all space, and a point in space. The center of mass point and the center of mass frame are not the same thing. The center of mass frame is all points in space in which the center of mass points appears to be at rest. Placing an object in the center of mass frame means placing it any where in space but with a speed that matches the center of mass speed so that in the center of mass frame it looks static. Placing an object at the center of mass "point" means placing it at a specific point. In order to get the new ball to not move, you have to place it in the center of mass frame (i.e. at rest with respect to the center of mass) and at the force-equilibrium point.

 

[genericusrnme]

okay, the center of mass is defined as the point

$r_{CoM} \equiv \frac{1}{M} \sum_n m_n r_n$

Adding a mass at the origin changes M, the total mass of the system, which will change the center of mass, unless all other masses are also at the origin.

If you make the mass so small that it dosen't effect the system then all you are doing is adding a particle which doesn't interact with your system, none of its motion affects the center of mass or the trajectories of any of the other masses.

If we take a center of mass frame, for it to be inertial we require that the center of mass has no acceleration, do we agree on this?

So let us look at $\frac{d^2 r_{CoM}}{dt^2}$ (for the 2 body problem)

$\frac{d^2 r_{CoM}}{dt^2} = \frac{d^2}{dt^2} \frac{1}{M} \sum_n m_n r_n = \frac{1}{M} ( m_1 \frac{d^2 r_1}{dt^2}+m_2 \frac{d^2 r_2}{dt^2}) = \frac{1}{M} (F_1 + F_2)$

Now, what do we know about forces?
Law no.3 - Every force has an equal and opposite reaction

The force on 1 due to 2 is equal and opposite to the force on 2 due to 1, so $F_1 = -F_2$

So we find that

$\frac{d^2 r_{CoM}}{dt^2} = \frac{1}{M} (F_1 + F_2) = \frac{1}{M} (F_1 - F_1) = 0$

I conclude that the center of mass frame is therefore inertial.

 

[AlonsoMcLaren]

So what about choosing mass A or mass B as frame of reference?

 

[genericusrnme]

Doing that will give you a non inertial frame of reference since A and B will both be changing directions during their trajectories, if we are dealing with a central force problem

 

[chrisbaird]

If object A was much greater in mass than object B, you could make the approximation that A is motionless, therefore its frame is inertial. This is typically the first approach done in teaching about Earth orbiting the sun or a classical picture of an electron orbiting a nucleus. Of course, assuming object A is very heavy so that object B's mass is negligible is the same thing as assuming that object A's center of mass is the same as the system's center of mass.