<p>Isn't it just conservation of momentum? Without any external forces, the centre of mass of the binary system must stay in the same place.</p>

<p>For example, imagine that you had a binary system, where one star had a circular orbit and the other, with equal mass, was eccentric. Clearly, as the orbits proceeded, the centre of mass, which is half way between the stars, would have a position which oscillates with time. But with no external forces applied to the binary system, this is forbidden by the conservation of linear momentum.</p>

CC BY-SA 3.0
2015-12-08T07:49:03.770Z
2015-12-08T07:49:03.770Z
ProfRob
<p>In a binary star system, why does $$e_1 = e_2 = e$$</p>

<p>where $e_1$,$e_2$, and $e$ are the eccentricities of the three orbits of $m_1$,$m_2$, and the reduced mass, respectively.</p>

<celestial-mechanics>
In a binary star system, what relation determines the eccentricity of the three orbits (for $m_1$, $m_2$, and the reduced mass)?
Mohamed Ayman
# In a binary star system, what relation determines the eccentricity of the three orbits (for $m_1$, $m_2$, and the reduced mass)?

(Asked by: username_0 on 2015-12-07)

In a binary star system, why does $$e\_1 = e\_2 = e$$

where $e\_1$,$e\_2$, and $e$ are the eccentricities of the three orbits of $m\_1$,$m\_2$, and the reduced mass, respectively.

**Tags:** celestial-mechanics


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## 1 answers

### Answer by username_1 on 2015-12-08. Score: 0 | Accepted answer

Isn't it just conservation of momentum? Without any external forces, the centre of mass of the binary system must stay in the same place.

For example, imagine that you had a binary system, where one star had a circular orbit and the other, with equal mass, was eccentric. Clearly, as the orbits proceeded, the centre of mass, which is half way between the stars, would have a position which oscillates with time. But with no external forces applied to the binary system, this is forbidden by the conservation of linear momentum.


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