<p>You can't. There are an infinite number of stable orbits that are possible depending on the angular momentum and total energy of the system - which must be negative for a bound orbit.</p>
<p>If you specify the eccentricity (there is only one eccentricity value for the orbit) and orbital period then Kepler's third law will give you the semi-major axis, from which, the apoapsis can be calculated. Or if you specify the apoapsis then the eccentricity can be calculated. If you wish to consider the centre of mass frame then each body executes an ellipse with the same eccentricity (<a href="https://physics.stackexchange.com/questions/746748/proof-for-equal-eccentricity-in-a-binary-star-system">see here</a> for example) and with equal semi-major axes (if the masses are equal), where their sum is equal to the semi-major axis of the system.</p>
<p>The speed at apoapsis can be obtained from the <a href="https://en.wikipedia.org/wiki/Vis-viva_equation" rel="nofollow noreferrer">vis viva equation</a> once the orbital parameters are established.</p>

CC BY-SA 4.0
2023-03-22T15:49:17.017Z
This is work for "two ellipse" system?
Domahidi Péter
CC BY-SA 4.0
2023-03-22T15:51:47.847Z
The parameters of a binary orbit include *one* eccentricity and *one* semi-major axis. @DomahidiPéter
ProfRob
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2023-03-22T16:05:34.007Z
I mean orbit system like in picture
Domahidi Péter
CC BY-SA 4.0
2023-03-22T18:19:15.043Z
@DomahidiPéter all bound two body orbits are "two ellipse systems" since circles are just a special case of an ellipse. The only other options are two hyperbolas for unbound, and two parabolas for the borderline zero total energy case (zero velocity at infinity). Because the ratio of their distances to their center of mass must always be (by definition) inverse to their ratio of mass, they have no choice. They are always opposite each other relative to their center of mass. Whatever Fred Astaire does, Ginger Rogers has to backwards, and in high heels!
uhoh
CC BY-SA 4.0
2023-03-22T11:36:56.247Z
2023-03-22T16:21:51.667Z
2023-03-22T16:21:51.667Z
ProfRob
<p><a href="https://i.sstatic.net/Oi0Qr.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Oi0Qr.gif" alt="enter image description here" /></a></p>
<p>If I know the masses of the two bodies (they have equal mass), how do I calculate the two ellipse eccentricities, length of the apoapsis and speed at the apoapsis, that make stable system, as in this picture.</p>

<orbit><binary-star><orbital-elements>
Calculating the parameters of binary systems?
Domahidi Péter
# Calculating the parameters of binary systems?

(Asked by: username_0 on 2023-03-22)

[![enter image description here](https://i.sstatic.net/Oi0Qr.gif)](https://i.sstatic.net/Oi0Qr.gif)

If I know the masses of the two bodies (they have equal mass), how do I calculate the two ellipse eccentricities, length of the apoapsis and speed at the apoapsis, that make stable system, as in this picture.

**Tags:** orbit, binary-star, orbital-elements


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## 1 answers

### Answer by username_1 on 2023-03-22. Score: 5

You can't. There are an infinite number of stable orbits that are possible depending on the angular momentum and total energy of the system - which must be negative for a bound orbit.

If you specify the eccentricity (there is only one eccentricity value for the orbit) and orbital period then Kepler's third law will give you the semi-major axis, from which, the apoapsis can be calculated. Or if you specify the apoapsis then the eccentricity can be calculated. If you wish to consider the centre of mass frame then each body executes an ellipse with the same eccentricity ([see here](https://physics.stackexchange.com/questions/746748/proof-for-equal-eccentricity-in-a-binary-star-system) for example) and with equal semi-major axes (if the masses are equal), where their sum is equal to the semi-major axis of the system.

The speed at apoapsis can be obtained from the [vis viva equation](https://en.wikipedia.org/wiki/Vis-viva_equation) once the orbital parameters are established.

> **Comments:**
> **username_0**: This is work for "two ellipse" system?  
> **username_1**: The parameters of a binary orbit include *one* eccentricity and *one* semi-major axis. @username_0  
> **username_0**: I mean orbit system like in picture  
> **username_2**: @username_0 all bound two body orbits are "two ellipse systems" since circles are just a special case of an ellipse. The only other options are two hyperbolas for unbound, and two parabolas for the borderline zero total energy case (zero velocity at infinity). Because the ratio of their distances to their center of mass must always be (by definition) inverse to their ratio of mass, they have no choice. They are always opposite each other relative to their center of mass. Whatever Fred Astaire does, Ginger Rogers has to backwards, and in high heels!  

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