# Why is this the barycenter?

1. Feb 27, 2012

### solarblast

I'm looking at <http://en.wikipedia.org/wiki/Two-body_problem>. [Broken] 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?

Last edited by a moderator: May 5, 2017
2. Feb 27, 2012

### 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 )
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" )

3. Feb 29, 2012

### solarblast

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.

4. Feb 29, 2012

### tiny-tim

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

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