# 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?

• That is to say? – Andrew Mar 27 at 17:19
• I mean why you have used the sum of $M_{att}$and $M_1$ – Tojrah Mar 27 at 17:21
• link – Andrew Mar 27 at 17:35

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.

• Well, I don't see the need for a 3rd body... Just suppose the picture I made describes a "solar system" with a "sun" and only 1 planet: there is no third body. Sun and planet will be orbiting aroud G, apparently as if a fictitious attractive mass Matt was situated at G. This leads to the inconsistency I described. What's wrong ? – Andrew Mar 30 at 8:34
• @Andrew: in the case of a small planet orbiting the Sun, the mass of the planet will be much smaller than the mass of the Sun - so ignore it and see what you get. Note, for 2 objects, there's only 1 calculation for Kepler - not 2 calculations. And I recommend reviewing Kepler's laws. – Cinaed Simson Mar 31 at 5:40
• I appreciate that you get involved in answering, but this doesn't lead towards a solution... "much smaller" did you say: that's not mandatory, so there is no reason to ignore one of the 2 bodies orbiting, one around the other. Reviewing Kepler's laws is obviously a prerequisite; I did this and I still don't understand what kind of error in reasoning I make, when applying the 3rdlaw to each of the 2bodies rotating, as if a central attractive mass was as if it was the cause of their rotation around each other... – Andrew Mar 31 at 7:24
• One more time, you can't not apply the 3rd law to each body. Fix one body - call it the Sun. Then rotate the second body - call it the planet - around the Sun. The orbit of the planet around the Sun is an ellipse - with the Sun at it's foci. Can you draw a picture of the 1st law? – Cinaed Simson Apr 1 at 6:59
• Just a point. When you say "The orbit of the planet around the Sun is an ellipse - with the Sun at it's foci", i think it's not perfectly true; actually, for a single-planet system the focus of the ellipse followed by the planet is not at the center of the sun, but is at the joint center of mass, i.e. the barycenter. Do you agree? Both are actually orbiting elliptically around the barycenter. Placing the sun at the focus is an approximation. Hence my difficulty in understanding. Maybe the logic is simply : "if you use the barycenter with the 3rd law, you get an error, so don't do that"? OK? – Andrew Apr 2 at 7:57