Classical two-body problem-understanding

1. May 22, 2007

neelakash

In a classical two body problem,when we superpose our origin of co-ordinates on any one of the two particles,the other particle(now we see it to have reduced mass [mM/(m+M)],moves w.r.t. me...

I saw in a book #Central Force Motion, that the particle is treated by 2nd law of motion...
i.e. [mM/(m+M)]*a=F where a and F are accelen. and force term.

My question is how do we know that this is indeed an inertial frame?After all the two particles are interacting among themselves and there may well be a radial accelen.(Like in the case of Binary stars).

Another point is that we invoke the concept of centrifugal energy...
(=L^2/(2*mu*r^2)) where mu denotes the reduced mass.

Atam P Arya says that here we are working from a rotating frame as a penalty of eliminating theta from the equations...i.e. we are being bound to put this centrifugal force term(and hence a centrifugal energy) as we are working from a rotating frame...

I am having the fragrance...But cannot see the reality...
Please highlight on this topic.

Last edited: May 22, 2007
2. May 22, 2007

StatMechGuy

If the entire universe is just these two bodies, and there are no external forces whatsoever, then there are no external forces. If you transform into a reference frame with the center of mass then with no external forces it is by definition an inertial reference frame.

However, this trick doesn't work if you're truly working with a many-body problem -- for example, if you wanted to calculate the effect of Jupiter's gravity on the Earth's orbit then transforming to center of mass only reduces your system by one degree of freedom but you're still left with too many degrees of freedom.

3. May 22, 2007

neelakash

So what?
I want to know why do not we cansider that we are working from a non-inaetial frame while describing the 2 body problem interms of centre of mass.