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Why is this the barycenter?by solarblast
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#1
Feb2712, 10:14 AM

P: 136

I'm looking at <http://en.wikipedia.org/wiki/Twobody_problem>. I'm
looking not too far down the page in the section: Center of mass motion (1st onebody 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? 


#2
Feb2712, 10:53 AM

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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/Twobod...ody_problem.29 ) that's just the double derivative of the standard vector formula for centre of mass … R = (m_{1}x_{1} + m_{2}x_{2})/(m_{1} + m_{2}) (btw, i don't think anyone actually calls it "barycentre" ) 


#3
Feb2912, 11:00 AM

P: 136

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 rdot = rrdot. 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. 


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