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In an Earth-like world with a moon as large as this:

Large Moon

Would the larger tides discourage oceanic shipping?

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    $\begingroup$ The Moon does not cause waves. $\endgroup$ – HDE 226868 Apr 18 '15 at 16:46
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    $\begingroup$ I would like to add that even now, ships regularly have to wait for tide conditions to allow passage. Also, as a submariner, most of our mass was underwater when we were surfaced - we would intentionally ride tidal currents upriver when we were going home, because it made docking easier. $\endgroup$ – Sean Boddy Apr 19 '15 at 16:47
  • $\begingroup$ Tangential: you may find Amphidromic point combined with this animated gif interesting. Tides are not simple things. $\endgroup$ – user487 Apr 20 '15 at 3:35
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    $\begingroup$ Is that moon in the picture actually bigger than ours? Or is it just an illusion? $\endgroup$ – Tom Anderson Apr 20 '15 at 3:49
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    $\begingroup$ We can’t tell how big the moon is in the picture, as depends on the focal length of the lens used to take the picture. $\endgroup$ – JDługosz Nov 20 '16 at 21:28
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As HDE 226868 notes, the moon does not cause waves. It causes tides. A larger moon, that is, a more massive moon, would have a larger mass. This means larger tides, as the gravitational force of the moon on the earth's water would be larger. (If the overall mass of the moon was the same, I would easily expect no change in tides. Same mass means the same gravitational force.)

Would very strong tides deter ocean shipping? Not that I can see. A larger moon would discourage a lot of coastal cities. Dikes/Levee/Floodbanks/Stopbanks would need to be larger. Some areas which would be used as ports would not be usable as ports, but then other areas would open up.

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    $\begingroup$ Even if the moon mass was kept constant? Increase the volume, but decrease the density. It wouldn't affect anything with those changes, right? $\endgroup$ – mikhailcazi Apr 19 '15 at 7:14
  • $\begingroup$ @mikhailcazi edited to reflect your comment. $\endgroup$ – PipperChip Apr 19 '15 at 22:16
  • $\begingroup$ A larger moon might have a slightly higher gravitational pull because it's mass would be more distributed, so the stuff further away would have less of a pull and the stuff closer would have more, but the overall effect I think would be positive given that the relationship is 1/r^2 and not 1/r. Probably negligible though. $\endgroup$ – k_g Apr 19 '15 at 23:51
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    $\begingroup$ @k_g This is very negligble. The radius of our current moon is about 1 738 km, with a perigee of 362 600 km and apogee of 405 400 km. The gravitational effects from one side of the moon when compared to the other from those distances is very small. I can use symmetry and the huge distances involved to approximate this detail away: since the stuff closer contributes more than the stuff further away, I can just put all the mass in the middle and treat it as one disk of mass. $\endgroup$ – PipperChip Apr 20 '15 at 16:43
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Higher tides would make shipping more challenging, but (probably) not insurmountably. Certainly shallow-water boating would be affected.

Why do I say "not insurmountably"? Consider the Bay of Fundy, with normal tides in excess of 50 feet. Fishing is common, although there are a few challenges

enter image description here enter image description here

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How high do you think these tides would be? In many parts of the world, tides are not particularly significant. However, Liverpool has a tidal range of up to about 9m (30ft) and was, at one time, the shipping hub of the British Empire and one of the largest ports in the world. So it's perfectly possible to have a functional port with a tidal range that's much higher than today's world average. Unless you're imagining much larger tides than this, the answer to the question is No, larger tidal ranges would not discourage ocean shipping.

(The high tidal range at Liverpool is a resonance effect, similar to the Bay of Fundy, which has been mentioned in some of the other answers.)

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  • $\begingroup$ Indeed I (along with many others) used to commute across the river at its narrowest point - the huge tidal range was obvious, but did not stop a regular ferry service operating. What is more ferries have existed since Mediaeval times so you don't need all that much by way of tech. to make it work. $\endgroup$ – Francis Davey Apr 19 '15 at 13:27
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No, but it would change how harbors are built. Currently the tides only cause a relatively small height change of 60 cm in average. The high tides in the Fundy Bay and Rance are due to resonance effects, there are also places on Earth where no tides are observable.

If you decrease the distance, the tides will be higher. Harbors will be then more likely build near rivers (using the river as transportation to the sea) or when they are build, the harbors will have swimming pontons accessed by ramps, compensating the height differences. So not a problem at all.

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If you want a detailed analysis, simply solve Laplace's tidal equations. These are not easy to do. If you're curious, I can get you part of the way to what might be an answer. Also, I'm curious as to what the math will turn up.

Achille Hui has a very helpful spoiler (regular ones don't work for $\LaTeX$, and take up too much space), which I've used to cover up the math. Click on it to show some of my work, or skip it to read more.

$$\require{action} \toggle{ \begin{array}{cl} & \bbox[2pt,color:black;border-radius:3px;box-shadow:4px 4px 8px black]{ \verb/Click to show math/} \end{array} }{ \begin{array}{cl} A=\frac{1}{a\cos\varphi}\frac{\partial}{\partial\lambda}(g \zeta+U) \\ B=\frac{1}{a}\frac{\partial}{\partial\varphi}(g \zeta+U) \\ C=2\Omega\sin\varphi \\ \\ \text{We now have} \\ \\ \frac{du}{dt}-vC+A=0 \\ \frac{dv}{dt}+uC+B=0 \\ \\ \text{This leads to} \\ \\ \text{I have no idea how to solve this. Wolfram Alpha}^1\text{ could help, but I don't} \\ \text{have a subscription. Something out there will help you.} \\ \\ \text{When you have that, substitute in for $A$, $B$ and $C$ and plug it all in to } \\ \\ \frac{\partial\zeta}{\partial t}+\frac{1}{a\cos\varphi}\left[\frac{\partial}{\partial\lambda}(uD)+\frac{\partial}{\partial\varphi(vD\cos\varphi)} \right]=0 \\ \\ \text{Solve for } \zeta \text{ and enjoy.} \\ \end{array} } \endtoggle$$


1 You can find a potential starting point here.

Note:

This is just to let you know what the math is. There are other answers that cover everything else, so I've decided to leave this as is, because I couldn't add anything else.

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    $\begingroup$ Oddly enough WAlpha seems to choke on your system of DEs, but it's a simple first-order system; the solution is $u = \frac{1}{C}\bigl([C u(0) + B]\cos(C t) + [C v(0) - A]\sin(C t) - B\bigr)$ and $v = \frac{1}{C}\bigl([C v(0) - A]\cos(C t) - [C u(0) + B]\sin(C t) + A\bigr)$. $\endgroup$ – David Z Apr 19 '15 at 6:19
  • $\begingroup$ @DavidZ Thanks; I'll try to work it out. I was quite surprised that Wolfram Alpha had difficulties when I first plugged it in. $\endgroup$ – HDE 226868 Apr 19 '15 at 13:01
  • $\begingroup$ Laplace's equations describe a tidal bulge in an ocean that covers the entire planet to a uniform depth. They're theoretically interesting, but on the actual Earth there are all kinds of things getting in the way, and the waters of the oceans end up moving in directions quite different from the Moon. $\endgroup$ – David K Apr 19 '15 at 21:56
  • $\begingroup$ @DavidK I duly noted that. It's the best approximation I could find. $\endgroup$ – HDE 226868 Apr 19 '15 at 21:57
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It would not only disrupt shipping but everything else needed to run a modern civilization. If the Moon were as close to the Earth when it formed, about ten times closer, the tides would be about a thousand times higher. The Earth's crust and the magma beneath it would then also experience significant tides. You would then probably have permanent flood basalt eruptions at the plate boundaries at a scale of the one that led to the formation of the Siberian Traps.

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It would not discourage oceanic shipping, but it would make the times it can be done more selective.

If you have a port that is 20m deep and full at high tide, with our Moon you should still have at least 10m left at low tide (depending on the tides in that area). That's plenty to operate big ships in.

If your Moon is closer and your tides bigger, your 20m harbor might be empty at low tide. This means ship captains need to plan more accurately: they need to be in and out of a harbor this shallow within a couple of hours. Delays in voyages would lead to big delays at the destination waiting for the next tide.

The alternative is that harbors are built bigger, which costs a lot more in both money and time.

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Moon illusion, i.e., a harvest moon:

enter image description here

Size is just as important as distance. Our Moon is outside the Earth's Roche limit and we are losing it at about 4 cm a year. Shortly after it formed, it would have appeared 15 times larger than today:

The moon [was] so close that rock and magma [were] tidal. The lunar pull [was] 4000 times greater than today [...] In the sea, every wave was a tsunami.

What If We Had No Moon? -YouTube, Discovery Channel (short answer: we wouldn't be here)

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    $\begingroup$ The strength of the tide is directly proportional to the mass of the moon, however, it is related to $ \frac {1}{r^3} $ of the distance between the bodies. Based on this, I'd say distance is far more important than mass. Also, it would be possible to have a large but low mass moon made of less dense materials. $\endgroup$ – Jim2B Apr 19 '15 at 4:09
  • $\begingroup$ @Jim2B The strength of the gravitational field is related to $1/r^2$. Where does the extra factor of $1/r$ come from? $\endgroup$ – David Richerby Apr 19 '15 at 15:48
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    $\begingroup$ @DavidRicherby A uniform gravitational field would not create tides; it is the change in the Moon's field from one side of the Earth to the other that causes tidal forces. So we need to look not at $1/r^2$, but at $\frac{d}{dr} (1/r^2)$, which is proportional to $1/r^3.$ $\endgroup$ – David K Apr 19 '15 at 20:17

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