I was willing to believe Chris on this one, but I figured I'd try it in my head.... And, sure enough, Chris is right!
One thing that may be confusing people here is that in all ordinary circumstances, the direction of F and a are the same. Not so for a negative mass. You have to be very careful of the signs, and you have to keep track of the unit vectors...
I'm going to set it up slightly differently from Chris's post #4, but the result is the same (also note that this is much easier to do on paper by drawing pictures and noting all of the vectors, so if you are still confused, try that!):
(origin at center)
Recall that F=m a r^ (I don't have an r-hat symbol, so that'll have to do for the unit vector), and Fg = - G m1 m2 r^a->b / r2.
1 --------- 2
(m) ------ (-m)
So, for the positive mass:
m a r^ = - G m (-m) r^1->2 / r2 where r^ is pointing from the negative mass to the positive mass.
and for the negative mass:
(-m) a r^ = - G m (-m) r^2->1 / r2 where r^ is pointing from the positive mass to the negative mass.
So, what is the magnitude and direction of the acceleration of the negative mass? And do the same for the positive mass...
(I hope I got all the vectors right... so much easier with a pencil...)