The moon has an elliptical orbit (of course), but as it inches away from the Earth each year, does its orbit become more circular and how come? Io's orbit is becoming more circular due to tidal heating effects, but why?

  • $\begingroup$ Suggestion: go to HORIZONS, compute the Moon's osculating elements wrt Earth for the largest possible time range and see if the eccentricity changes a lot. $\endgroup$ – barrycarter Jan 1 '19 at 16:45

Short answer: YES. I will assume you mean the eccentricity in the orbit of the Moon around the Earth.

In general, tidal forces in binary systems (like the Earth-Moon system, or a binary star, etc...) effect the binary in three main ways: in order of longest timescale to shortest timescale

1) circularization of the orbit (eccentricity goes to zero, binary separation goes to minimum).

2) alignment of the binary components' spin angular momenta with the orbital angular momentum (the directions of $S$ and $L$ are the same).

3) synchronization of the binary components' rotational frequency with the orbital frequency.

but why?

There's different ways of answering "why," and here is a great conceptual answer given by the God Father of astrophysical tides himself, Z. P. Zahn:

A fundamental property of closed mechanical systems is that they conserve their total momentum. This is true in particular for binary stars, star-planet(s) systems, whether they possess or not a circumstellar disc, if one can ignore the angular momentum that is carried away by winds and by gravitational waves. Through tidal interaction, kinetic energy and angular momentum are exchanged between the rotation of the components their orbital motion and the disc. In the absence of such a disc, which is the case that we shall consider here, they evolve due to viscous and radiative dissipation to the state of minimum kinetic energy, in which the orbit is circular, the rotation of both stars is synchronized with the orbital motion, and their spin axis are perpendicular to the orbital plane. How rapidly the system tends to that state is determined by the strength of the tidal interaction, and thus by the separation of the two components...

So, basically, the tidal torque drives dissipation, and this dissipation brings the binary to a minimum kinetic energy state, i.e. circular orbit, synchronized spins with the orbit, aligned spins with the orbit.

Here's a contour plot of the timescales provided by equations #$9 - 13$ from Hut's 1981 seminal paper, assuming the separation does not change much relative to the other quantities, for a binary composed of a black hole and a Wolf-Rayet star, which is a similar system to a planet-satellite system, where the timescales are parameterized in terms of the mass of the WR star and the separation of the binary:

enter image description here

The dotted black line is the merger timescale for the binary due to gravitational waves, meaning below that line you are binary that mergers within the lifetime of the universe. The synchronization timescale is independent of the initial spin of the WR star, which is why there is only one synchronization line in the plot, but the alignemnt timescale does depend on the initial spin of the component feeling the tides. Points below the contours achieve that process (below the red dashed line are synchronized). The quantity $f_{B}$ is the break-up fraction parameter, is between $0$ and $1$, and the choice of which determines the initial WR star spin as a fraction of its break-up spin. (the tidal timescale contours were made by equating the WR lifetime with the tidal timescale). I did not include the circularization timescale because it is essentially always less than the sync. timescale (seen simply from the fact that generally, $S << L$).

Lastly, it must be said that a WR star is quite different from a moon/satellite, however the effects of tides on each is similar enough to make a conceptual comparison (since a WR star is much less massive than the BH, as is the moon than the Earth, the hierarchy of tidal timescales is the same, but you must keep in mind that the WR's apsidal response to the tidal gradient is very different than a rocky moon's).

As you pointed out, the Moon's orbit is slightly eccentric, but the moon is also tidally synchronized with the Earth's rotation (which are both sync'ed to the orbital rotation). This should make sense since it takes longer for tides to circularize than to synchronize.

I'm sure somebody can post an example of using tides in an actual planet-moon system, rather than my black hole-star system. :)

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    $\begingroup$ I would like to add to this, that if not for the Sun, an Earth-Moon system, then the Moon would likely have a very near circular orbit. The Sun's pull on the Moon is what causes the Moon's eccentricity as well as what pulls the moon 6 degrees off an equatorial orbit. The tidal forces do circularize the orbit, but the Sun is the bigger factor, so as the Moon moved away from the Earth it's orbit has likely become less circular over time for much of it's history. $\endgroup$ – userLTK Jan 1 '19 at 23:32
  • $\begingroup$ Re "the magnitudes of $S$ and $L$ are the same": shouldn't it be the frequencies that are the same? For example, the Moon is already tidally locked to the Earth, but its orbital angular momentum $L$ is much larger than its spin angular momentum $S$ because the relevant moments of inertia are quite different. $\endgroup$ – rob Jan 2 '19 at 0:38
  • $\begingroup$ @userLTK that is a very good point, thank you for making it. According to observation, the moon's orbit has increasing eccentricity due to the Sun which is well understood, and there is an anomalous increase that is currently unexplainable ( arxiv.org/abs/1102.0212 ). And so clearly, the Sun's effect is dominating over the tides, since the eccentricity is increasing. I was merely trying to explain how tides act on the moon's orbit to cause circularization, but in reality our Moon's eccentricity is increasing. $\endgroup$ – N. Steinle Jan 2 '19 at 2:30
  • $\begingroup$ @rob thank you for pointing this out rob, I agree I should be talking about the rotational frequency of the binary component and the orbital frequency in terms of synchronization. $\endgroup$ – N. Steinle Jan 2 '19 at 2:33
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    $\begingroup$ @userLTK The Moon's orbital plane is slightly more than 5° off an ecliptic orbit. Its orbital plane precesses quite quickly, and its inclination to the celestial equator varies between 18° & 28°. If you look at the Moon's solar orbit on a computer screen, it looks quite circular, since the eccentricity of the Earth-Moon system is fairly small, and the perturbations on the Moon's solar orbit due to the Earth are hard to see, amounting to no more than a pixel or 2 on a typical monitor. $\endgroup$ – PM 2Ring Jan 2 '19 at 15:58

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