How is the Barycenter Determined in a Two-Body System?

[solarblast]

I'm looking at <http://en.wikipedia.org/wiki/Two-body_problem>. I'm
looking not too far down the page in the section:

Center of mass motion (1st one-body problem)

He computes easily R as the barycentric center. Why must
this be so? Can it be shown geometrically, or perhaps by forces that are zero there?

 

[tiny-tim]

hi solarblast!

(btw, if you just type a url, the pf software will auomatically make a link out of it, eg http://en.wikipedia.org/wiki/Two-body_problem#Center_of_mass_motion_.281st_one-body_problem.29 )

He computes easily R as the barycentric center. Why must
this be so? Can it be shown geometrically, or perhaps by forces that are zero there?

you mean R'' = (m₁x₁'' + m₂x₂'')/(m₁ + m₂) ?

that's just the double derivative of the standard vector formula for centre of mass …

R = (m₁x₁ + m₂x₂)/(m₁ + m₂)

(btw, i don't think anyone actually calls it "barycentre" )

 

[solarblast]

hi solarblast!

(btw, if you just type a url, the pf software will auomatically make a link out of it, eg http://en.wikipedia.org/wiki/Two-body_problem#Center_of_mass_motion_.281st_one-body_problem.29 )


you mean R'' = (m₁x₁'' + m₂x₂'')/(m₁ + m₂) ?

that's just the double derivative of the standard vector formula for centre of mass …

R = (m₁x₁ + m₂x₂)/(m₁ + m₂)

(btw, i don't think anyone actually calls it "barycentre" )

Thanks. I'll buy that with the caveats below about what I'm doing.

I think the writer of the page is likely British. They still cling to centre.

It's been a very long time since I've done anything substantive in physics. I'm grasping what I can in a book on celestial mechanics to get to the part I'm interested in, orbits. I skipped the preceding chapter which is really about center of mass and gravity, so maybe I might review what I need there.

While I'm at it, I'll ask another question on derivatives of vectors from an early chapter that I've skimmed through. He mentions r dot r = r**2, where the left side r's are vectors. (maybe I need some meta cmds to express these items as you have done. Where would I get them?) He differentiates that with respect to time and gets
r dot r-dot = rr-dot. r dot r = rr*cos(theta) = r**2 sort of gets me there, but I must be missing something. The derivative derivation he gives doesn't jump off the page to me.

 

[tiny-tim]

hi solarblast!

… He mentions r dot r = r**2, where the left side r's are vectors. (maybe I need some meta cmds to express these items as you have done. Where would I get them?) He differentiates that with respect to time and gets
r dot r-dot = rr-dot. r dot r = rr*cos(theta) = r**2 sort of gets me there, but I must be missing something. The derivative derivation he gives doesn't jump off the page to me.

the LHS is (r.r)' = 2(r.r') = 2|r||r'|cosθ

(for bold, use the B button just above the reply box, or use [noparse]…[/noparse] )

 

[pmacias]

The barycenter, or center of mass, is defined as the point at which the mass of a system can be considered to be concentrated and acts as if all the mass were located at that point. In the case of a two-body system, the barycenter is the point around which the two bodies orbit each other. This is because the barycenter is the point at which the gravitational forces between the two bodies cancel out, resulting in their mutual orbit around that point.

To understand why the barycenter is the center of mass in a two-body system, we can look at it mathematically. In the first one-body problem, the two bodies are considered to be point masses, meaning they have no physical size and all their mass is concentrated at a single point. In this case, the distance between the two bodies, R, is the sum of their individual distances from the barycenter, r1 and r2. This can be expressed as R = r1 + r2.

Since the two bodies are orbiting around the barycenter, we can also say that their velocities, v1 and v2, are equal in magnitude but opposite in direction. This means that the total momentum of the system is zero, as the two bodies have equal and opposite momenta. This can be expressed as m1v1 + m2v2 = 0, where m1 and m2 are the masses of the two bodies.

Using these equations, we can calculate the position of the barycenter using the formula R = (m1r1 + m2r2)/(m1 + m2). This shows that the barycenter is located at a point that is proportional to the masses of the two bodies and inversely proportional to their distances from the barycenter. In other words, the larger the mass of a body, the closer it is to the barycenter, and the smaller the mass, the further away it is from the barycenter.

In terms of forces, we can see that the barycenter is the point at which the forces between the two bodies are equal and opposite, resulting in a net force of zero. This is because the gravitational force between two bodies is directly proportional to their masses and inversely proportional to the square of the distance between them. At the barycenter, the distances and masses are such that the forces cancel out, resulting in a net force of zero.

Therefore, it can