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

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The discussion focuses on the determination of the barycenter in a two-body system, referencing the Wikipedia page on the two-body problem. Participants clarify the formula for the center of mass, R, as R = (m1x1 + m2x2)/(m1 + m2), and discuss its geometric implications and the conditions of zero net forces at this point. There is also a mention of the terminology differences, with one participant noting that "barycentre" is more commonly used in British English. Additionally, questions arise regarding vector derivatives and their applications in celestial mechanics, indicating a need for further review of foundational concepts. Overall, the conversation emphasizes the mathematical and conceptual understanding of barycenters in physics.
solarblast
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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?
 
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hi solarblast! :smile:

(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 :wink:)
solarblast said:
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) :smile:

(btw, i don't think anyone actually calls it "barycentre" :rolleyes:)
 
tiny-tim said:
hi solarblast! :smile:

(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 :wink:)


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) :smile:

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


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.
 
hi solarblast! :smile:
solarblast said:
… 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] :wink:)
 
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