Two bodies orbiting around common barycenter Assume $2$ bodies of mass $M_1$ and $M_2$ orbiting in a plane around their common barycenter $G$. It means the axis $(M_1M_2)$ is rotating in the plane around a vertical axis through $G$. [Please refer to picture]. Assume uniform circular motion ($\omega$ is the angular velicity). According to the situation, the period of rotation of $M_1$ and $M_2$ around $G$ is the same, say $T$.
Suppose that we assign an attractive mass to the point $G$, say $M_{att}$, so as to say that $M_1$ is orbiting around some body of mass $M_{att}$, at distance $d_1$, with period $T$, and $M_2$ is orbiting around some body of mass $M_{att}$, at distance $d_2$, with period $T$...
In that case, the $3^{rd}$ Kepler's Law applied to $M_1$ and $M_2$ says, if I am not mistaking:
$$\frac{d_1^3}{T^2}=\mathcal{G}\frac{M_{att}+M_1}{4\pi^2} \quad \frac{d_2^3}{T^2}=\mathcal{G}\frac{M_{att}+M_2}{4\pi^2}$$
So that:
$$\frac{d_1^3}{d_2^3}=\frac{M_{att}+M_1}{M_{att}+M_2}$$
By definition of $G$ as barycenter of $M_1$ and $M_2$, we have:
$$M_1d_1=M_2d_2 \Longrightarrow \frac{d_1}{d_2}=\frac{M_2}{M_1} \Longrightarrow \frac{M_2^3}{M_1^3}=\frac{M_{att}+M_1}{M_{att}+M_2}$$ 
And we get:
$$M_{att}=-\frac{M_2^4-M_1^4}{M_2^3-M_1^3}$$
which is clearly inconsistent, due to the $-$ sign...
Does it mean that we cannot assign an attractive mass to the barycenter, or is there a way out of this inconsistency? Is it impossible to consider a "central point" with a "central mass", around which $M_1$ and $M_2$ would be orbiting? Is it a non-sense question?

 A: for mass $M_1$: apparent mass at COM (G) is  $M_{att1}=\frac{M_2 d_1^2}{(d_1+d_2)^2}$
for mass $M_2$: apparent mass at COM is $M_{att2}=\frac{M_1 d_2^2}{(d_1+d_2)^2}$
This can be obtained by writing Newton's 2nd Law for COM, taking its acceleration as zero (COM is at rest):
$$M_1 a_1+M_2a_2=0$$
Then from there we can find acceleration of each mass:
$a_1=-\frac{GM_2}{(d_1+d_2)^2}~~~$ and   $~~~a_2=-\frac{GM_1}{(d_1+d_2)^2}$
and then equate each of these accelerations to centripetal acceleration (orbits are circular):
$a_1=-\frac{GM_2}{(d_1+d_2)^2}=-\frac{v_1^2}{d_1}~~~$ and $a_2=-\frac{GM_1}{(d_1+d_2)^2}=-\frac{v_2^2}{d_2}$
From there we can find orbital speeds of $M_1$ and $M_2$:  $v_1$ and $v_2$. 
Then equating each of these speeds to Keplerian speeds for each circular orbit, e.g. $v_1=\sqrt{\frac{G M_{att1}}{d_1}}$, $v_2=\sqrt{\frac{G M_{att2}}{d_2}}$ and finding $M_{att1}$ and $M_{att2}$ from there, see answers at the top.
$M_{att}$ is not the same for both $M_1$ and $M_2$, otherwise they will not orbit with the same orbital period.
Pluto+Charon system is example of such motion.
A: If I understand you correctly, what you're trying to do won't work.
In the center of gravity frame, gravity will act on the center of mass as if all mass was concentrated at that point.
In short, you need a 3rd body. Then you can use Kepler's laws to find the center of mass motion around the 3rd body.
For instance, the center of mass of the Earth and the Moon follows an elliptical orbit around the Sun which can be calculated using Kepler's laws.
