I just stumbled on this rather confused QA. The point seems to be if the Earth-Moon rotation period was identical to the Earth-year, then ...

enter image description here

... all three would stay in a line at all times.

Could this actually happen? (ie: they stay in a line at all times.) It seems "wrong" somehow.

Or do we simply not know ... the solution would only be numeric since it's a three-body problem?

Assuming the solar system was otherwise completely empty, so no other perturbances; assuming the actual masses of the three.

{If the answer is "yes", where the hell is the CG, what figure does it make?}

  • $\begingroup$ Three-body problem is generally only numerically solvable, BUT these cases have - relatively simple - analytical solutions, too. Check for "lagrange points" on the google. $\endgroup$ – peterh says reinstate Monica Feb 26 '16 at 13:55

That configuration is not stable, as the moon would have to be at Lagrangian point $L_1$ or $L_2$, which are not stable. Any perturbation would cause an exaggerated deviation from those orbits, ending either in an earth moon system rotating around each other, or the Earth and moon orbiting the sun separately.

  • $\begingroup$ Rick thanks - just for clarification, notice I mentioned "Assuming the solar system was otherwise completely empty, so no other perturbances" For clarity, in an empty universe are L1 and L2 stable? $\endgroup$ – Fattie Feb 26 '16 at 13:48
  • $\begingroup$ @JoeBlow The variability in solar winds would be a source of perturbations that could trigger the instability. I don't know the time scale over which such a perturbation would be amplified into separate orbits though. It might take centuries, but I don't have a good guess. $\endgroup$ – Rick Feb 26 '16 at 13:59
  • $\begingroup$ Rick - fair enough. I understand from you now it's inconceivable in the real universe. Thanks again. I guess I was further asking about the 3body problem, in the abstract (point masses) where they are in a line ... but that's a different question eh! $\endgroup$ – Fattie Feb 26 '16 at 14:18
  • $\begingroup$ @JoeBlow They are "semi-stable" so if the point masses were placed within $\epsilon$ of the ideal positions, then as $\epsilon$ decreases, the length of time that the objects would remain in orbit would increase. As $\epsilon$ approaches zero, the length of time approaches infinity. $\endgroup$ – Rick Feb 26 '16 at 14:58
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    $\begingroup$ @CarlWitthoft sure you could put something on that orbital trajectory, but it wouldn't follow that orbit for long. The $L_1$ and $L_2$ equilibrium points are saddle points, nothing will stay in an orbit there for long. $\endgroup$ – Rick Feb 29 '16 at 14:16

This is the phenomena seen in spiral galaxies, where circular speed at different distances from the center of galaxy is same (pretty much). To explain this phenomena, physicists had to come up with dark matter hypothesis. So, may be with bunch of dark matter with right kind of spread, it is theoretically possible.

Because physicists had to come up with dark matter hypothesis to explain this phenomena, you can safely bet, it is not explainable otherwise.

  • $\begingroup$ heh a good point. $\endgroup$ – Fattie Feb 26 '16 at 14:17

It could work - while it is a three body problem, it is a solvable one.

You can make a hand-waving approximation by just looking at the earth-moon system and figuring out the distance the moon needs to be from the earth to have an orbital period of 1 year. This turns out to be about 0.014 AU - which is large enough to suggest that the sun's gravitational field is enough to perturb things by maybe a few percent between opposite ends of the orbit.

But that's a very messy way of dealing with it. There is a cleaner solution. All three masses (which I'll treat as point masses) are in a line, and all are orbiting the center of mass of the earth-moon-sun system. To simplify things, lets pretend the sun is right at the center of mass; this isn't going to make much difference.

Lets pick the scenario where the earth is closer to the sun than the moon is. In effect then, we have the earth at distance $r_{e}$ and the moon at distance $r_{m}$ orbiting with the same period, and all the forces involved are conveniently completely radial.enter image description here

The earth is orbiting at $r_e$ with a centripetal force given by the gravitational force from the sun minus that from the moon. The moon is orbiting at $r_m$ with a centripetal force given by the gravitational force from the sun plus the earth.

$m_e\omega^2 r_e=\frac{Gm_s m_e}{{r_e}^{2}}-\frac{Gm_m m_e}{(r_m-r_e)^{2}}$

$m_m\omega^2 r_m=\frac{Gm_s m_m}{{r_m}^{2}}+\frac{Gm_m m_e}{(r_m-r_e)^{2}}$

Since the various masses are known, and $\omega$ is the angular velocity which is the same for both objects (by definition of the problem) and $\omega=\frac{2\pi}{T}$ where $T$ is the orbital period in seconds (1 year, for the sake of argument). Then you need to solve the pair of equations to find $r_e$ and $r_m$.

You can throw in the orbit of the sun around the center of mass of the solar system to give you an exact solution for point particles. It is an nice equilibrium solution, but may or may not be a stable equilibrium.

The real world is more complex of course, with real bodies not being point masses, other bodies existing in the solar system, magnetic fields, tidal effects and other fun things that might slowly make a difference.

  • $\begingroup$ That's the $L_2$ Lagrangian Point, which is indeed "semi-stable". $\endgroup$ – Rick Feb 26 '16 at 15:13
  • $\begingroup$ @Rick - The L2 Lagrange point is an unstable equilibrium point, very much like a pencil standing perfectly upright, on its tip. $\endgroup$ – David Hammen Feb 26 '16 at 16:08

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