Is this definition of tidal locking really satisfying?

Question

We always are hearing this: "If a moon has equal rotational and orbital periods it's tidally locked to its host planet and always one side of it will face to the planet."

But what if, for example, the Moon had an orbital and rotational period of 24 hours? (with the same distance and the same rotational period of the Earth)

Would it be called tidally locked again?

Answer 1

Currently, the Moon is tidally locked to the Earth but the Earth is not tidally locked to the Moon.

In your scenario (which could be accomplished by raising the mass of the Earth), where the Moon has an orbital and rotational period of 24 hours, the two bodies are mutually tidally locked.

Indeed, the Earth-Moon system is gradually moving toward such a state due to tidal energy exchange -- albeit through slowing of the Earth's rotation rather than shortening of the Moon's orbit.

Answer 2

Rather than your question (which Sten has already answered well), let me reply to one of your comments under Sten's answer, since I think that's where the core of your misunderstanding lies:

In this scenario [where Moon's orbital period = Moon's rotational period = Earth's rotational period = 24 h], the Moon's orbital period is the same with the Earth's rotational period. So the Moon will be only visible to a certain region of Earth. Well, the Moon also rotates every 24 hours. So people of that certain region can see both sides of the Moon (12 hours one side, and another 12 hours the other side)

No, that's not what would happen in your scenario. It is, however, what would happen if the Moon did not rotate at all.

As it happens, Wikipedia's article on tidal locking already has an excellent animation illustrating this, so I'll just borrow it:

^{Animation by Wikipedia user Stigmatella aurantiaca, used under the CC By-SA 3.0 license.}

On the left, you can see the Moon orbiting the Earth (drawn as just a circle, and obviously not to scale) while also rotating at the same rate as it moves around the Earth, so that it completes one orbit around the Earth in the same time as it completes one rotation around its own axis — just as it (on average, neglecting libration) does in reality, and as it would also do in your scenario. You can clearly see that this results in the same side of the Moon always pointing towards the Earth.

On the right, meanwhile, is what would happen if the Moon did not rotate at all while still orbiting the Earth. In that case the side of the Moon facing the Earth does change: the Moon doesn't rotate, but it moves around the Earth, so that people watching it from the Earth see it from different directions.

(The Earth itself is drawn as just a featureless circle in this animation, so you'll have to imagine its rotation. But if you imagine it rotating at the same rate as the Moon in the left animation, so that the same side of the Earth always points towards the Moon, then you'll have exactly the mutual tidal locking scenario you envisioned.)

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For an alternative visualization, get a friend to help you and have them stand in one spot while you walk a full circle around them, with both of you looking each other in the eyes all the time. Obviously your friend will have to make a full 360° turn in order to maintain eye contact with you as you walk around them — but so will you! If you started the turn e.g. with your friend on your west side and went clockwise, then you'd end up looking first west, then north (after walking 90° around your friend), then east (after 180°), then south (after 270°) and finally west again (after walking a full 360° circle).

Or, if staring your friend in the eyes while walking around them feels too creepy and awkward, just hold hands with them instead. Or just hold your hand against a tree or a light pole while walking around it.

In any case, any time you walk a circle around something while keeping it on the same side of you, you'll inevitably end up making a 360° turn. And conversely, any time you walk around something and turn a full 360° at the same time, the same side of your body will always face the thing you walk around (at least on average).

Meanwhile, if your try to walk around your friend (or a signpost) without turning at all, and initially face towards them, you'll find that after walking 180° around them you now have you back towards them instead. (And if you practice your footwork until you can do the full circle like that smoothly without stumbling, you'll now know how to do a basic do-si-do and be one step closer to perfecting your folk dancing skills. Congratulations.)

Answer 3

I agree with the answer provided by Sten, but canot help from adding several details.

1. Historically, it was Immanuel Kant who proposed, on qualitative grounds, that eventually the tidal synchronism of the Earth and the Moon will become mutual -- i.e., that the Earth also will be showing the same face to the Moon.

Even among the scholars of Kant's heritage few people are aware of that work of his, because Kant published it not in a scientific journal or book, but in two consequtive issues of a local newspaper.

2. Whether the mutual synchronism will actually be achieved depends on how far the Moon will have to recede for that. If it has to tidally recede too far, the solar gravity will take over, and the Moon will fly away. In other words, the synchronism must be achieved while the Moon is still within the Hill sphere of the Earth. And this is not the end of story -- because in real life a satellite should not approach the Hill radius too closely, lest it motion becomes unstable. So the Moon should actually be inside the reduced Hill sphere, whose radius (for prograde satellites) is about 0.49 of the Hill radius.

It is possible to demonstrate that, yes, the mutual synchronism will be attained with the Moon still slightly below the reduced Hill radius.

3. In principle, mutual synchronism can be achieved not only by a tidally receding moon, but also by a tidally decending moon, provided it is sufficiently heavy, and the synchronism is reached before the moon crosses the planet's Roche radius.