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What would be different for us if Earth and Moon revolved around each other like Pluto and Charon do?

Pluto and Charon

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    $\begingroup$ The reason this doesn't happen for the Earth-moon system is the different masses involved. Could you elaborate on your question? That is, in this hypothetical situation, is the Earth mass set equal to Pluto's? Or would you like to retain Earth's mass and just put it in a larger orbit around the system's c.o.m.? $\endgroup$ – user1991 Dec 7 '15 at 12:17
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    $\begingroup$ @John As gravitational potential of any given body decreases to zero only at infinity, a two-body system's center of mass is always at a location non-equal to the center of mass of the larger body. Hence, the above does happen for practically all systems (excluding some very unlikely, very specific cases) – like Earth-Moon, but the extent to which it happens varies with the mass ratio of the bodies involved. $\endgroup$ – V-J Dec 7 '15 at 13:58
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    $\begingroup$ @ V-J, yes, you are effectively repeating what I just said. The proposed question deals with a larger separation from the c.o.m. and thus either a different mass or a contrived, hypothetical situation. $\endgroup$ – user1991 Dec 7 '15 at 14:57
  • $\begingroup$ @cd1: What properties of the Pluto-Charon system is your focus on? Mutual tidal locking, a center of mass outside Pluto, tilted axis with respect to the ecliptic, larger moon, closer distance, moon of similar density, orbital period of the moon less than an Earth week? $\endgroup$ – Gerald Dec 8 '15 at 15:47
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They do, but due to the ratio of masses being vastly different, they seem like they would not to do so as moon seems to rotate just around (the centre of) Earth.

The ratio of Earth and Moon's masses is $\frac{M_{Earth}}{M_{Moon}} = 81.3$ whereas for Pluto and Charon the same ratio is $\frac{M_{Pluto}}{M_{Charon}} = 8.09$.

Because the ratio for Pluto and Charon is relatively small, the center of the system – barycenter, around which the two bodies orbit – is somewhere on a line drawn between the two celestial bodies's mass centers. But for Earth and Moon, because Earth is proportionally much heavier, the system's barycenter does not reach outside of Earth, but instead is about 4,500 kilometers from the center of Earth (see the photo below, too):

In cases where one of the two objects is considerably more massive than the other (and relatively close), the barycenter will typically be located within the more massive object. Rather than appearing to orbit a common center of mass with the smaller body, the larger will simply be seen to "wobble" slightly. This is the case for the Earth–Moon system, where the barycenter is located on average 4,671 km from the Earth's center, well within the planet's radius of 6,378 km. Source: Wikipedia - Barycenter

Earth-Moon System Barycenter: http://astronomy.stackexchange.com/questions/11246/how-long-until-the-earth-and-moon-become-a-binary-planet

The major effect of this co-rotational system is that Earth seems to "wobble" on its orbit, as mentioned in the quote from Wikipedia above.


@JeppeStigNielsen makes a good point about differences in tidal locking in the comments below. In the Earth-Moon system, only Moon is tidally locked (which causes us to see only one face of it, so only roughly a half of it, from Earth), whereas in Pluto-Charon both of the bodies are tidally locked. Earth is not tidally locked because of the higher mass ratio between it and Moon, but the lower mass ratio Pluto-Charon system is, as the lower mass Charon has slowly changed Pluto's rotation to match its orbital movement.

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    $\begingroup$ So Moon does not implicitly revolve around Earth, but instead both Moon and Earth revolve around a common point: the barycenter of Earth-Moon system. This applies to all celestial bodies, to a more or less negligible effect: for example, the planets in our Solar System do not implicitly revolve around Sun, but instead their respective systems' barycenters. This precision is not needed for most everyday cases however, so approximations like "Moon revolves around Earth" and "Planets revolve around Sun" are just fine. $\endgroup$ – V-J Dec 7 '15 at 13:31
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    $\begingroup$ There is another difference. In the Pluto-Charon system, the major body (Pluto) is tidally locked, while in our system, Earth is not tidally locked. Because of this we can enjoy the view of the moon from every longitude (from 180 degrees west to 180 degrees east) on Earth. If Earth had been tidally locked, Earth would have had a near side and a far side. As a consequence, Greenwich in England would not be an "arbitrary" origin of the longitude. Instead 0 degrees would be the defined as the meridian just under the Moon (on average). The most dramatic difference would be almost no tides. $\endgroup$ – Jeppe Stig Nielsen Dec 7 '15 at 14:27
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    $\begingroup$ @JeppeStigNielsen That's actually a really interesting point; never thought what Moon's path on the sky would look like, if seen at all, if Earth was tidally locked! $\endgroup$ – V-J Dec 7 '15 at 14:29
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    $\begingroup$ @John It seems to me that the question itself is actually based on a false assumption, namely that the way Pluto and Charon interact with respect to each other is fundamentally different from how the Earth and the Moon interact. Both systems revolve around their respective barycenters. The only important difference, as noted in Jeppe's comment, is that Pluto is tidally locked and the Earth is not. The question does not explicitly ask about tidal locking, but perhaps that is what the asker is really curious about. It's not clear. $\endgroup$ – Todd Wilcox Dec 7 '15 at 16:14
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    $\begingroup$ It might be worth mentioning that the Sun-Jupiter barycenter is not located within the Sun's radius. $\endgroup$ – Bobson Dec 7 '15 at 16:54
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On barycentres

The Pluto-Charon couple is not qualitatively different to the Earth-Moon couple with regards to orbits. As was pointed out in other answers, in both cases, the two bodies revolve around each other, i.e. they are best described as orbiting around their barycentre.

In more physical terms, the referential centred on the barycentre of the Earth-Moon system is "more Galilean" than the referential centred on the Earth geometrical centre: if you do high-precision measures of physical systems on Earth, you will see some "jitter" that shows that the Earth is not really Galilean. Most of which is due to the Earth rotation (the jitter being most famously demonstrated by Foucault's pendulum) but even if you account for the rotation, you still get some residual perturbations coming from the revolution of the Earth around the Earth-Moon barycentre. (And if you fix these ones, you still get some due to the Earth revolution around the Sun -- really, the revolution of the Earth around the Solar system barycentre -- and then some because of the rotation of the Galaxy, and so on, but they are increasingly difficult to detect.)


About tides

When two round bodies orbit each other, they have a tendency to go into "tidal locking": their individual rotation speed will synchronize with the revolution, so that, in fine, the two bodies always keep the same hemisphere toward each other. Pluto and Charon are at that step. The Moon is also tidally locked with the Earth: we always see the same hemisphere (in fact we see a tiny bit more than half of the Moon, because it wobbles a bit). The Earth is not tidally locked... yet. But it will ultimately be.

Indeed, the Earth and the Moon exert tidal forces on each other. This is most easily explained by considering orbital speed: when a very small satellite orbits around a big planet, it must go at a speed that depends on the satellite's altitude: the further the satellite is, the slowest it goes (e.g. low-orbit satellites zoom at around 8 km/s, while the Moon goes at a leisurely 1 km/s or so). But the Moon is quite bulky: its radius is a bit more than 1700 km. This means that if the Moon's centre goes at the right speed for its orbit, the rocks on the far side of the Moon are 1700 km further from the Earth, and thus are a wee bit too fast for that orbit, so they want to leave. Similarly, the rocks on the near side of the moon are 1700 km closer to the Earth, and are thus going too slowly: they tend to "fall" toward the Earth.

The phenomenon is symmetrical: the Earth also experiences tidal forces from the Moon. In fact, both Earth and Moon experience tidal forces from the gravitational Earth-Moon couple. This engenders tides, where water moves around in response to the forces; rocks do not because they're rocks, i.e. not very fluid under normal conditions -- they would like to move, but are too rigid to do so.

The tidal forces are somehow opposed to the Earth rotating faster than the 27 days for an Earth-Moon revolution, and the rotational energy is slowly dissipated: some of it happens to be injected in the Earth-Moon gravitational coupling, which drives them apart of each other (it has been measured thanks to reflectors from space probes and Apollo missions: the Moon is fleeing us at a rate of about 38 mm per year); the rest is lost in the friction from moving water, thus ultimately converted to heat radiated into space.

Bottom-line: the Earth rotation is slowing down. For instance, a day would have lasted about 22 hours at the time of the dinosaurs (the big ones, not the birds). The slowing-down is known around time-keeping circles as the ΔT.

However...

Even when the Earth becomes tidally locked with the Moon, there will still be tides (at least, if there is still liquid water at that time, which is not a given, since the Sun energy output is predicted to go down sharply 5 billion years from now). Indeed, the Sun-Earth couple also produces tidal forces. The Earth-Moon tidal forces are about twice stronger, so the Moon-induced tides are bigger, but in a tidally locked situation, we should still witness tides induced by the Sun -- but on a smaller scale.

(Without the Moon, the Earth would ultimately tidally lock with the Sun and a 365-days rotation -- using today's length for a day, of course. I am not entirely clear about what should become of the Sun-Earth-Moon system in the very long term; but it appears that this is still widely open research, notably because there are other planets in the mix, leading to a very complex situation.)

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TLDr answer:

Both answers are very good. There's a few more details to consider if we want to look at all the what-ifs in this amusing but wacky scenario.

Already mentioned, the ratio of size is 8 to 1, not 81 to 1, so for starters, the Charon like Moon would be much larger in the sky. The Moon, with roughly 10 times the mass, figuring slightly greater density due to some minor compacting, would still be 2.1 times as big across, assuming the same distance, that would make it 4 times as bright in the night sky. A full moon would be quite impressive. Perhaps (just barely) bright enough to read by if it was a large text book. (some people claim to be able to read by Moonlight now, most people can't, but 4 times brighter, a full moon might just be bright enough.

Solar eclipses would become more frequent and last about twice as long and you might think the Earth would be slightly colder due to the Moon blocking some sunlight, but the Moon, believe it or not, radiates more heat on the earth than it blocks because the lit up moon that faces us is nearly 400 degrees F in peak daytime and it's not hard to see that a surface that temperature radiates some heat. Not a lot, but some. A question on that here, so a little over 4 times the energy (ignoring solar eclipse losses), about 1/2,500th the heat from the sun, might work out to 1/10th of a degree at night during a full moon. Not a lot, certainly, but measurable to anyone with sensitive enough instruments. The brightness and size of the Moon would obviously be more noticeable than about 1/10th of 1 degree in temperature (C not F).

A moon of that mass would slow down the Earth's rotation significantly faster, already mentioned, but this one, we have to give some thought to. When the Moon formed it was much closer to the Earth, about 3-5 times the radius of the Earth away. Source. That's outside the Hill Sphere, and the formation of the Moon left the Earth rotating very rapidly so the effects (rapidly rotating earth, very powerful lunar tides) would still be there but the Lunar tides would be 10 times greater, so we're looking at earth-quake level tides when the moon, 10 times the mass, was 3-5 earth radius away. The Moon, because during formation it wouldn't have much angular momentum, would quickly settle into a tidally locked rotation around the Earth. The Tidal effect on the Earth, being 10 times greater would cause (roughly) 10 times the tidal bulge on the Earth would would push the Moon away from the Earth about 10 times faster, but, at the same time, the tidal drag slowing down the Earth, would be 10 times as great too (I assume that corresponds to about 10 times as fast).

So, basically, the Moon and Earth would follow the system they're in now, but it would proceed about 10 times as fast with a Moon with 10 times the mass. The estimate (here) is that it will take about 50 billion years for the Moon to slow down the Earth enough to enter into tidal locking, so divide that by 10, we would be very close to a tidal lock today. The Earth would rotate very slowly. The moon would also (likely) be a bit further from the Earth and probably have a more wobbly orbit due to solar perturbations, and perhaps, escaped completely. This is a complicated bit of mathematics that I'd prefer not to attempt (at the Moon's current mass, the Sun will go Red Giant long before either the Moon escapes or the Earth gets tidally locked but with a Moon 10 times as massive, that's probably no longer the case and either the Moon is gone or the Moon is more distant, has a more elongated orbit, and the Earth is at or close to tidally locked. If the Moon escapes, we'd have a near earth orbit object of enormous size that could later crash into us or swing past the earth and move our orbit - either effect and simply the effect of having no moon, would be enormous.

Discussion on the Moon/earth escape vs tidal locking here

If we assume full tidal locking, 29.5 days (synodic, not sidreal) and a moon a bit further out so we might be looking at 30 something to maybe 40 days for 1 earth rotation, that's 20 days of sunshine, 20 days of night. That would play absolute havoc on the weather systems and seasons. Day to night would have a bigger effect than Summer vs winter, and the summer days would be scorching, though some regions might due just fine because of rainfall. Evolution could probably adapt to that, but it doesn't sound fun to me. The further distance might make the Moon just 3 times as bright in the night sky instead of 4. still pretty bright though. You'd still get 6 months of sun and 6 months of night at the poles, but for most of the earth, this would be a radical change having days and nights that long.

Other possible effects, Obliquity (no moon, perhaps greater, a larger ice age driver), see here. Also, if the Earth still had the Moon but the Moon was in a more elongated orbit, we'd still have tides as the Moon moved in and out to appogee and perogee. see picture

enter image description here

Source

The bottom line, while we might not give it much thought, a different sized moon would actually change quite a lot. A smaller moon would move away from the Earth more slowly and earth might be able to have a 2nd moons perhaps by capture, if the moon was smaller, also we might have more aggressive ice ages and climate changes due to a greater obliquity variation and, assuming the giant impact still happens in a similar way but a smaller amount of debris (which doesn't make sense, but lets pretend), then a smaller moon wouldn't have slowed the Earth's rotation as much and the Earth might be rotating quite a bit faster, 10 or 15 hour days instead of 24. The effects would be pretty significant.

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