<blockquote>
  <p>I know they will be orbiting about a common center of mass, i.e. the
  barycenter. But, do the velocities have to be equal in magnitude and
  opposite in direction (normal to R when R is their distance from each
  other) for the orbit to be stable?</p>
</blockquote>

<p>The orbital velocities do not have to be and in general are not equal in magnitude. What is equal is the <em>angular velocity</em>, that is the angular rate (e.g., in $rad/sec$) that two bodies will be orbiting their common barycenter. The orbital radius, $r$, orbital velocity, $v$, and angular velocity, $\omega$, are related by the equation</p>

<p>$$\omega = v/r$$</p>

<p>Note that these are all scalar quantities and $v$ can be thought of as the component of the velocity vector which is perpendicular to $r$. Since conservation of angular momentum implies that $\omega$ remain constant for each body, individually, then we know that $v/r$ must also be constant which implies that bodies which orbit farther away from the barycenter must necessarily be orbiting faster and vice versa.</p>

<p>Now of course that simplifying explanation works for two bodies. Once you throw in more than two bodies, things become more complicated and the barycenter can move around, causing more complex motions.</p>

<blockquote>
  <p>I think, if the velocity of one mass were to vary with respect to the
  other it would create a moving barycenter in which the two masses
  would collide or throw on another out of the orbit</p>
</blockquote>

<p>This is a subtly different question. If one of the two masses had a varying orbital velocity, that would imply it was gaining or losing energy by some mechanism. This can occur through things like tidal interactions as it does for our Earth and Moon. As stated above, the angular velocity must remain constant which implies that the body whose velocity is changing must also migrate towards or away from the barycenter, potentially resulting in a collision or escape. Since the Moon's orbital velocity is being sped up through tidal interactions, it is moving farther away as a result. Another example might be two black holes emitting gravitational waves as they orbit which propagates away energy and thus they orbital velocities, causing them to get closer together until they collide.</p>

CC BY-SA 3.0
2016-04-04T20:26:00.383Z
2016-04-04T20:26:00.383Z
zephyr
<p>Let's use coordinates where the center of mass is at the origin.  Then in our two body system where the bodies are of equal mass, the center of mass is midway between the two bodies.  So
<span class="math-container">$$r_a = - r_b$$</span>
The motion of the center of mass the weighted sum of the motions of the constituents, x,<br />
<span class="math-container">$$v_\text{center} = \sum m_x v_x/ \sum m_x$$</span>
If we fix the center of mass for our two body system, so that it is 0 at all times, then we have <span class="math-container">$v_a = -v_b$</span> since the masses are the same.  The velocities are at all times opposite but equal in magnitude.</p>
<p>The previous poster may have been confused by the fact that the velocities of the 'stars' are not constant.  However in the center of mass frame, the two bodies execute their orbits synchronously, e.g., reaching periapsis at the same time.</p>

CC BY-SA 4.0
2021-10-06T12:23:12.980Z
2021-10-06T13:13:45.990Z
2021-10-06T13:13:45.990Z
Tom McGlynn
<p>Given two bodies of equal mass in an elliptical orbit:</p>

<p>I know they will be orbiting about a common center of mass, i.e. the barycenter. But, do the velocities have to be equal in magnitude and opposite in direction (normal to R when R is their distance from each other) for the orbit to be stable? I think, if the velocity of one mass were to vary with respect to the other it would create a moving barycenter in which the two masses would collide or throw on another out of the orbit, but I can't find any mathematical verification of this. </p>

<orbit><celestial-mechanics>
Two body orbit of equal masses
NotSoSN
# Two body orbit of equal masses

(Asked by: username_0 on 2016-04-04)

Given two bodies of equal mass in an elliptical orbit:

I know they will be orbiting about a common center of mass, i.e. the barycenter. But, do the velocities have to be equal in magnitude and opposite in direction (normal to R when R is their distance from each other) for the orbit to be stable? I think, if the velocity of one mass were to vary with respect to the other it would create a moving barycenter in which the two masses would collide or throw on another out of the orbit, but I can't find any mathematical verification of this.

**Tags:** orbit, celestial-mechanics


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## 2 answers

### Answer by username_1 on 2016-04-04. Score: 2 | Accepted answer

> I know they will be orbiting about a common center of mass, i.e. the
> barycenter. But, do the velocities have to be equal in magnitude and
> opposite in direction (normal to R when R is their distance from each
> other) for the orbit to be stable?

The orbital velocities do not have to be and in general are not equal in magnitude. What is equal is the *angular velocity*, that is the angular rate (e.g., in $rad/sec$) that two bodies will be orbiting their common barycenter. The orbital radius, $r$, orbital velocity, $v$, and angular velocity, $\omega$, are related by the equation

$$\omega = v/r$$

Note that these are all scalar quantities and $v$ can be thought of as the component of the velocity vector which is perpendicular to $r$. Since conservation of angular momentum implies that $\omega$ remain constant for each body, individually, then we know that $v/r$ must also be constant which implies that bodies which orbit farther away from the barycenter must necessarily be orbiting faster and vice versa.

Now of course that simplifying explanation works for two bodies. Once you throw in more than two bodies, things become more complicated and the barycenter can move around, causing more complex motions.

> I think, if the velocity of one mass were to vary with respect to the
> other it would create a moving barycenter in which the two masses
> would collide or throw on another out of the orbit

This is a subtly different question. If one of the two masses had a varying orbital velocity, that would imply it was gaining or losing energy by some mechanism. This can occur through things like tidal interactions as it does for our Earth and Moon. As stated above, the angular velocity must remain constant which implies that the body whose velocity is changing must also migrate towards or away from the barycenter, potentially resulting in a collision or escape. Since the Moon's orbital velocity is being sped up through tidal interactions, it is moving farther away as a result. Another example might be two black holes emitting gravitational waves as they orbit which propagates away energy and thus they orbital velocities, causing them to get closer together until they collide.


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### Answer by username_2 on 2021-10-06. Score: 1

Let's use coordinates where the center of mass is at the origin. Then in our two body system where the bodies are of equal mass, the center of mass is midway between the two bodies. So
$$r\_a = - r\_b$$
The motion of the center of mass the weighted sum of the motions of the constituents, x,  
$$v\_\text{center} = \sum m\_x v\_x/ \sum m\_x$$
If we fix the center of mass for our two body system, so that it is 0 at all times, then we have $v\_a = -v\_b$ since the masses are the same. The velocities are at all times opposite but equal in magnitude.

The previous poster may have been confused by the fact that the velocities of the 'stars' are not constant. However in the center of mass frame, the two bodies execute their orbits synchronously, e.g., reaching periapsis at the same time.


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