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For two bodies attracting to each other, we use the reduced mass and distance between them (r) to solve the equation of motion. The final result is an elliptical obit under certain conditions.

My understanding is that this obit is the obit of the reduced mass (so a virtual body) rotating about a point (focus) which is in a distance r from it. Is it correct?

In this case, if one of the bodies is much heavier than the other, we can say the heavier body is at the focus and the lighter one is orbiting around it. So the locus of the lighter mass is the same as the locus (orbit) of the virtual body with reduced mass. And this is the 1st Kepler's Law. Is my understanding correct?

If the masses of the two bodies are comparable, then we don't have straight forward information about the loci of the bodies, right? We only know how r (distance between the two bodies) changes but not their individual locus in the laboratory frame, right? And Kepler's 1st Law is not applicable any more (i.e. not rotating about the focus). Is it right? Are their orbits still elliptical?

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In this case, if one of the bodies is much heavier than the other, we can say the heavier body is at the focus and the lighter one is orbiting around it.

The concept of reduced mass doesn't dictate which of the two bodies is to be viewed as fixed. For example, one could look at things from the perspective of a very massive body orbiting a rather tiny body. For example, there's nothing wrong per se with looking at things from the perspective that the Sun orbits the Earth. (Modeling the motions of the other planets is of course a bit dicier from this geocentric perspective.)

In fact, one could pick any point along the line connecting the two bodies as being the fixed point and still wind up with each of the two bodies moving around that fixed point in the form of elliptical orbits. There are three special points of interest:

  1. The more massive body is deemed to be the fixed point.
    This is the point of view that leads to Kepler's laws in our solar system.
  2. The less massive body is deemed to be the fixed point.
    This is the geocentric point of view, which works fine from the perspective of the two body problem but requires the invocation of inertial forces in the n-body problem.
  3. The center of mass of the two bodies is deemed to be the fixed point.
    When extending to the n-body problem, this is the frame in which the equations of motion take on their simplest form, and this is why JPL, the Russian Academy, and the Paris Observatory all use this frame when they model the behavior of the solar system.

If the masses of the two bodies are comparable, then we don't have straight forward information about the loci of the bodies, right?

As mentioned above, which point one considers to be the focus of the two ellipses is a bit arbitrary. There are however three points that stand out.

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  • $\begingroup$ Thank you David! I want to make sure I understand correctly. So you mean any point on the line joining the two bodies can be made as a fixed point (and the focus) of the elliptical loci of the two bodies? And only the center of mass doesn't need the "invocation of inertial forces"? Because those points will be in accelerating frames? $\endgroup$ – HYW Oct 3 '16 at 3:23
  • $\begingroup$ Sorry, one more question. If we take the center of mass as the fixed point, but M is not much larger than m, we still cannot use the orbit derived for the reduced mass directly to describe the loci of m, right? Thanks! $\endgroup$ – HYW Oct 3 '16 at 3:33
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My understanding is that this obit is the obit of the reduced mass (so a virtual body) rotating about a point (focus) which is in a distance r from it. Is it correct?

Yes, that is correct.

In this case, if one of the bodies is much heavier than the other, we can say the heavier body is at the focus and the lighter one is orbiting around it.

That is a valid approximation because you have $$ \mu = \frac{mM}{m + M} \to \frac{mM}{M} = m \,.$$ in the limit of $M \to \infty$.

If the masses of the two bodies are comparable, then we don't have straight forward information about the loci of the bodies, right?

A (bad) example would be the positronium where a positron and an electron orbit each other. They have exactly the same mass. However this part of physics cannot be sensibly described using classical physics, one needs quantum mechanics to get the right energy levels. And even then one cannot describe the annihilation, quantum field theory is needed for that.

So as suggested in the comment below, take a binary star system where both stars have the same mass. But the mass should still be low enough that the emission of gravitational waves is negligible, otherwise we run into troubles with that example again.

Using any of the systems and looking at the classical motion only, we can look at your next question:

And Kepler's 1st Law is not applicable any more (i.e. not rotating about the focus). Is it right?

Take the equal masses example and set it up such that both particles have a circular orbit. They will both move on the same closed circle. Since both mode, neither is in the center of the circle, where both loci coincide. From this counter-example I conclude that Kepler's first law is made with the approximation.

However, in the relative frame where one uses the reduced mass, this is of course valid because the reduce mass orbits the center of the orbit.

Are their orbits still elliptical?

Yes, the orbits are always cone sections. This only fails when the bodies orbiting each other cannot be approximated by point charges.

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  • $\begingroup$ You can't really apply Newtonian mechanics to positronium. A far better example would be a binary star, or a system like Pluto and Charon. $\endgroup$ – Javier Oct 2 '16 at 17:11
  • $\begingroup$ Indeed, for that one needs QM to get the orbitals and QFT (QED) to get the annihilation. Binary stars might be problematic due to gravitational waves. I added the concerns to the question. $\endgroup$ – Martin Ueding Oct 2 '16 at 17:22
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    $\begingroup$ The effect of gravitational waves on binary stars is almost completely negligible. We've only measured orbital decay for one binary star system, and they're not going to to collide any time soon. $\endgroup$ – Javier Oct 2 '16 at 17:32
  • $\begingroup$ Thank you all! Can I say that all the orbits are cone sections, but only the orbits of m with M>>m (for 2 bodies) can be deduced by using reduced mass system? If so, can you let me know how people derive e.g. the orbits of the binary star system (not including gravitational waves and relativity)? Do you have any good books to recommend? $\endgroup$ – HYW Oct 3 '16 at 3:31
  • $\begingroup$ The orbits are always cone section in the relative coordinate system. You can solve the system there and then transform the coordinates back again with $x_1 = R + r/2$ and $x_2 = R - r/2$ where $x_1$ and $x_2$ are the coordinates of the two masses, $R$ is the position of the center of mass and $r$ is the relative displacement vector. $\endgroup$ – Martin Ueding Oct 3 '16 at 12:05

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