Tides on a Super Close In Moon

Question

I have a moon covered in liquid water that orbits a super-Earth very closely. As you might imagine, the tides are incredibly strong on this moon. However, it is tidally locked and so does not experience tides like Earth does but, I'm unsure about one aspect of this.

Would the extreme tides cause the water to permanently drain around the side of the moon (forming a ring of land that wrapped around the world) and raise sea levels around where the Prime Meridian and equator meet, as well as on the opposite side of the moon?

The planet is 2.55 times the mass of Earth while the moon in 0.107 Earth masses and orbits at a distance of 82,300 km.

Answer 1

From some helpful posts on our sister site, Physics.SE...

There is no tidal bulge.¹ This was one of Newton's few mistakes. Newton did get the tidal forcing function correct, but the response to that forcing in the oceans: completely wrong. … What Newton got wrong, Laplace got right. Laplace's dynamic theory of the tides accounts for the problems mentioned above. It explains why it's always high tide somewhere in the North Sea (and Patagonia, and the coast of New Zealand, and a few other places on the Earth where tides are just completely whacko). … The vertical component is the driving force behind the response of the Earth as a whole to these tidal forces. This question isn't about the Earth tides. The question is about the oceanic tides, and there it's the horizontal component that is the driving force. … Oceanographers still teach Newton's equilibrium tide theory for a number of reasons. It does give a proper picture of the tidal forcing function. Moreover, many students do not understand how many places can have two tides a day. For that matter, most oceanography instructors and textbook authors don't understand! Many oceanographers and their texts still hold that the inner bulge is a consequence of gravity but the other bulge is a consequence of a so-called centrifugal force. This drives geophysicists and geodocists absolutely nuts. That's starting to change; in the last ten years or so, some oceanography texts have finally started teaching that the only force that is needed to explain the tides is gravitation. (Source)

In reality the "tidal bulge" model is very inaccurate for explaining how the oceans rise and fall. It is more accurate to consider the oceans as a very complex driven oscillator, where the water is slowly sloshing around. The periodic minor variations in gravity caused by the moon contribute energy to the sloshing, but the actual shape and timing of the sloshes is much more influenced by coastlines than than by the moon itself. (Source)

What this means is that the tidal bulge experienced by Earth isn't simply caused by the gravitational pull of the moon. In fact, the tidal pull of the moon at any given moment isn't even a fraction of what's causing the bulge. So, what does "oscillator" mean?

Grab a half-cup full of water in your hand and shake it in the same way you would if you were trying to mix your favorite flavor into it. The water "oscillates" around the interior of the cup. And once you get the "oscillation" going, you don't need to put as much energy into it because the momentum of the water is doing most of the work.

This is really important, because that bulge isn't due to the water being "squeezed" from the poles. That's not how gravity works. It's due to the mass of water "oscillating," which you can see in your cup. The water moves from the center to the outer edges. When you're dealing with a sphere, the only place the water can go is the equator (simplistically).

What adds to this oscillation? We already know the first two of the following bullets, but there's two more.

• The momentum of the water.
• The moon's gravity.
• The Earth's rotation.
• The sun's gravity.

Like the moon's gravity, at any given moment the rotation of the Earth isn't adding much. The Earth's rotation isn't pushing the water along (like a pump) any more than the moon's gravity or Sol's gravity are substantially lifting the water up (like a vacuum, or another pump). These three forces: Earth's rotation, Sol's gravity, and Luna's gravity, are the gentle energy that slowly got the water moving over a bazillion years. But the energy now is principally the momentum of the water itself.

And that momentum is centrifugal (as is Earth's rotation), which is why you see two bulges on Earth.

But what about your planet?

Earth rotates 365ish times in a single year. Your planet rotates but once. Worse, your moon is only 10.7% the mass (and, give-or-take, the size) of Earth. This means less rotational and centrifugal force contributing to the "oscillation."

Next, you're tidally locked, which (for the benefit of all readers) is a fancy way of saying the same face of the moon is always facing toward your planet. That means there is no gravitational component due to the planet contributing to the "oscillation."

You do have a sun, and you've told us nothing about it. We don't know the orbital period of your moon around its planet, the orbital period of the planet around the sun, or the distance of the sun from the planet. There will be a small contributor to the oscillation due to the sun.

Finally, let's talk about the effects of local gravity. You haven't given us enough information to calculate the gravity of either the moon or the planet (you need the radius of both). This means my next statement is inaccurate, let me show you why.

• The ratio of gravity between Luna (1.62 m/s²) and Earth (9.807 m/s²) is 0.165:1 or 16.5%.
• The ratio of mass between Luna (7.34767309 × 10²² kg) and Earth (5.972 × 10²⁴ kg) is 0.0123:1 or 1.2%.

That's an order-of-magnitude difference between the two ratios. So, let's calculate the ratio of mass in your situation and multiply by 10 and call it good.

• 1:238.3178 or 23,831.78%.

Your planet is exerting 1,444X the gravitational influence on your moon's oceans than Luna is on Earth's oceans.

And when you start plugging all those numbers into Laplace's equations, what you soon discover is there are no tides (excluding the small component from the sun, which is seriously overwhelmed by the planet) because everything is static.

Conclusion

• Insignificant centrifugal force.
• Insignificant rotational influence.
• Minor solar gravity influence.
• Massive planetary gravity influence.
• Tidally locked.

Fundamentally, no oscillation — no tides. This means you would not see a bulge of water on the back side of the moon. You would see a serious bulge on the planet-side of the moon. Considering how static everything is and how massive the gravitational influence is, I could be convinced that there isn't any surface water on the back side of your moon. It's all on the planet-side hemisphere.

Disclaimer: The missing sun and radius data could change this answer. But that planetary influence on a tidally locked planet is so great, that I have a hard time believing a sun large enough to cause an oscillation would permit the moon to be habitable.

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_{¹ What this respondent means is that the concept of a "tidal bulge" due to simple lunar gravitational effects is misunderstood. Read the entire answer to see what he means. I've included his title for dramatic effect — and to point out that the bulge isn't there simply due to gravity, but to a complex affect of gravity.}

Answer 2

tide /tʌɪd/ noun plural noun: tides

1. the alternate rising and falling of the sea, usually twice in each lunar day at a particular place, due to the attraction of the moon and sun.

On a tidally locked planet there would be no tides, according to the definition above. There would be a steady bulge peaked on the line conjugating the moon and the planet.

For more info, you can refer to this answer on Physics.SE