|Feb27-12, 10:14 AM||#1|
Why is this the barycenter?
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?
|Feb27-12, 10:53 AM||#2|
(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-bod...ody_problem.29 )
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" )
|Feb29-12, 11:00 AM||#3|
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.
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