Why is linear momentum not conserved for a particle in a central force? [duplicate]

I am studying a two-body central force system in which the two particles, one of mass $m$ and one of mass $M$, experience a force directed along the line connecting the two particles.

We can reduce this to a system of just one fictitious particle with reduced mass and a central force. Why is linear momentum not conserved in the CM frame when it was conserved before shifting to the CM frame?

marked as duplicate by sammy gerbil, stafusa, Kyle Kanos, rob♦Jun 22 '18 at 21:35

Suppose that I know that $$p$$ is conserved and $$q$$ is not. Now suppose I define the new variable $$p' = p + q.$$ If you weren't paying attention, you might conclude that $$p'$$ is also conserved, because it has the same letter as the conserved quantity $$p$$. But that's clearly wrong. The similarity is just superficial. Not every variable whose name looks like "$$p$$" has to be conserved.
Similarly, in the two-body problem, the conserved linear momentum is $$\mathbf{p} = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2.$$ To work in the reduced mass picture, we define, among other things $$m_{r} = \frac{m_1 m_2}{m_1 + m_2}, \quad \mathbf{v} = \mathbf{v}_1 - \mathbf{v}_2.$$ Your question is why the quantity $$\mathbf{p}' = m_{r} \mathbf{v}$$ is not conserved, while $$\mathbf{p}$$ is. But $$\mathbf{p}'$$ has nothing to do with the total linear momentum $$\mathbf{p}$$, the resemblance is totally superficial. There is no reason to expect that $$\mathbf{p}'$$ should be conserved.
Incidentally, the linear momentum is conserved in the reduced mass picture. The point of reduced mass is that the relative velocity between two bodies interacting by a central force, with masses $$m_1$$ and $$m_2$$ can also be computed by finding the relative velocity between two bodies with masses $$m_r$$ and infinity attracting by the same force; this second mass stays fixed at the origin. The total linear momentum in this one-body problem is the sum of $$\mathbf{p}'$$ and the momentum of the second mass, and is indeed conserved. However, it has nothing to do with the total linear momentum in the original two-body problem.