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?
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'' = (m_{1}x_{1}'' + m_{2}x_{2}'')/(m_{1} + m_{2}) ? 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" )
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! 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] )