Both a parent star and its planet revolve around the center of mass of the system, the reason we see stellar wobble. But if we take this to be true, which it is, there can be a configuration in which two identical stars revolve around their center of mass, in a common orbit. What I find astonishing in this case is that they will be revolving around something having no mass at all, in a shared orbit, like two runners trying to catch each other but never quite being able to do so.

In that case, in a purely Newtonian system, the centripetal force must be provided by the gravitational attraction between the stars. Now if I assume both the stars to have a mass $m$, at a separation of $d$ from each other, revolving in the common orbit diametrically opposite to each other,

$$\frac{mv^2}{r} = \frac{Gm^2}{d^2}\; {\rm where}\; r = \frac{d}{2}$$

solving for $v$, we get a velocity where a stable orbit is formed, $v = \sqrt{\frac{Gm}{2d}}$

Note: I have failed to find such an expression, I am not certain about the math.

I did find a system described in such a way that two stars orbit a common point in separate ellipses though. Is my conventional wisdom correct, or is my derivation of this expression somehow intrinsically flawed?

Does this expression already exist? And I suppose the odds of observing such a system are 'astronomical', but has something like it ever been observed?

  • $\begingroup$ Observed, yes. en.wikipedia.org/wiki/Binary_star $\endgroup$ – Phil Frost Jul 24 '14 at 18:55
  • $\begingroup$ You are being ill-served by the remark that the earth revolves around the sun. That implies that there has to be something there to revolve around. In fact, all bodies of the solar system revolve around the common center of mass. Since the sun is so massive, it approximates the center of mass, which is where the remark comes from. Jupiter is about $1/1000$ the mass of the sun, so the sun-Jupiter CM is $1/2$ million miles from the center of the sun, which is slightly outside the sun. Does that meet your "observed? $\endgroup$ – Ross Millikan Jul 24 '14 at 21:27
  • $\begingroup$ @Ross Millikan, I do know about the stellar wobble. What I meant asking was if two identical stars revolving around each other had been observed. $\endgroup$ – Aayush Mahajan Jul 25 '14 at 16:08
  • $\begingroup$ How close do they have to be for you to consider them identical? I believe similar mass binaries have been observed, but don't have a catalog at hand. The Hulse-Taylor pulsar is two neutron stars-that is pretty similar $\endgroup$ – Ross Millikan Jul 25 '14 at 16:21

Yes, that can happen. It is somewhat realized in positronium, a bound electronic state where an electron and a positron revolve around each other. Both have the same mass, so they could (classically) have the same spherical orbit.

With Newton, you have an attractive force for two equal bodies of mass $m$ of $$ F = G \frac{m^2}{d^2}. $$

The centripetal force that would be needed in an orbit of radius $d/2$ is $$ F = m \frac{v^2}{d/2} = m \omega r^2, $$ where $v$ is the tangential velocity and $\omega$ the angular frequency.

Set them equal and you get: $$ G \frac{m^2}{d^2} = 2 m \frac{v^2}{d} \iff G \frac{m}{2d} = v^2 \iff v = \sqrt{\frac{Gm}{2d}}, $$ which is what you obtained.

You can get a deeper understanding if you look at the Jacobi method for the two body problem. There, you separate the center-of-mass motion of the relative motion. You define a relative distance $r$ between the two bodies. Then you need the reduced mass $$ \mu := \frac{m_1 m_2}{m_1 + m_2} $$ which is just $\mu = m/2$ in the case of two equal masses.

Then your problem is not two bodies on their mutual gravitational field, but one body of reduced mass in the field of the other particle. So we have a similar problem. The force is $$ G \frac{m \mu}{d^2}. $$

This force force is half as strong as before. However, the centripetal force has now to be calculated with $d$ being the radius, not $d/2$ as before. The force is half as strong as well. Therefore, the result is the exact same.

The derivation now goes: $$ G \frac{m \mu}{d^2} = m \frac{v^2}{d} \iff G \frac{m^2}{2 d^2} = m \frac{v^2}{d} \iff G \frac{m}{2 d} = v^2, $$ which I had before.


Two equal mass bodies orbiting around their centre of mass is exactly how Newtonian gravity works.

Your analogy of the runners is actually pretty good. Think about it this way - the two bodies are always attracted toward each other, so they accelerate a bit in that direction. Notice that "toward the other body" necessarily means "toward the centre of mass". But then the other body moves somewhere else, so the direction of the acceleration changes appropriately. But it's still toward the centre of mass. And we know the centre of mass must move with constant velocity (provided there are no other bodies around) since momentum is conserved, so it's easy and intuitive to transform to centre of mass coordinates. If the magnitude of the acceleration were always equal, you'd get circular orbit around the centre of mass, but in general the magnitude changes (since it depends on the distance between the bodies) so we get conic sections (ellipse, parabola, hyperbola or circle) instead.

As to an observed example... two exactly, or nearly, equal mass stars (or planets, moons, etc.) are hard to come by, see here for an explanation. But orbits around a centre of mass that is located outside either of the orbiting bodies are quite common, for instance many binary stars have this feature.


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