<p>I can give this a shot.  Anyone smarter than me is welcome to offer corrections.</p>

<p>The first issue, based on your question is how much closer you move the Moon.  (I'm assuming that you're not playing with other variations like the Earth's rotation speed or the Moon's mass).</p>

<p>If you want Tides to be 100 times greater, you have to move the Moon 10 times closer.  The tidal force is equal to the square of the distance (Radius to Radius, not surface to surface).   Putting the Moon 10 times closer would make it enormous in the sky and full moons would be bright enough to read by and the Tidal force would be 100 times greater.   </p>

<p>The Tidal Force is basically a bulge in the Earth that stays pointed at the Moon.  The land bulges too, not just the oceans.</p>

<p>Optimally, tides require about 1/4 of the circumference of the Earth to have full effect.  That's why lakes basically don't have tides.  Anything smaller than an ocean doesn't have tides to speak of.  Now if you make the tieds 10 times greater, the ratio ocean to large lake would still be the same, but large lake and sea tides (like the Mediterranean) would become much more apparent.  </p>

<p>Think of tides as raising the sea level of the entire ocean.  Of-course, it's not really like that, but locally, that's essentially what happens.</p>

<p>If the current tide is 50 CM (above average sea level), then imagine, in that area, what it would be like if sea level rose 50 meters.   In-land it wouldn't be an issue, but most coastal cities couldn't survive that.   100 times the tides would be the end of virtually all coastal living.</p>

<p>See chart.</p>

<p><a href="https://i.sstatic.net/zMw8S.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zMw8S.jpg" alt="enter image description here"></a></p>

<p><a href="https://en.wikipedia.org/wiki/Tide" rel="nofollow noreferrer">Source</a>.</p>

<p>As far as nearly geostationary, that would have 2 effects.   The slower movement would create more leveling off, less variation that you see in the chart above, where one part of the Earth can get tides that are amplified can get 2 or 3 times the tides than average, and the Tides would happen more slowly, as you get two high tides every time the Moon is overhead.  If the Moon took 6 months between rises, you'd get a high tide every 3 months, but you'd still have the 50 foot tide problem.   The geostationary orbit distance of the Moon is pretty close to the first number given, about 36,000 km. </p>

<p>Currently Tides vary quite a bit, but that's because the Sun and Moon both interact with tides, so you get pretty significant tidal variation.  You also get variation as the Moon's distance to the Earth changes.   If the Moon was 10 times closer, the Sun-tides wouldn't change and would be arguably ignorable (less than 1/2 of 1% variation), where, currently the position of the sun to the Moon is important in tidal calculation.  </p>

<p>You'd need to take into account the Moon's orbital period and the earth's rotation to estimate how quickly the tides move around the Earth.   Currently the Moon takes 29 days to orbit the Earth, so it's primarily the Earth's rotation (once every 24 hours) that governs the tides.   If you bring the Moon into 36,000 or 37,000 km, then the orbital speed is important because every few thousand KM would make a measurable difference. </p>

<p>My answer is pretty general.  I can try to touch on some specifics if I missed anything.  You could also ask in Astronomy stack exchange.  Tides are perhaps closer to Astronomy than Earth Science.  </p>

<blockquote>
  <p>Shouldn't the tidal forces go as the inverse cube of distance, not the
  inverse square? Given that and that orbit radius is the only change</p>
</blockquote>

<p>It's definitely the inverse square law.  I've done enough reading up on tidal forces to be sure of that.   The formula also clearly shows it as a product of inverse square too.  See <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/tide.html" rel="nofollow noreferrer">here</a> or just google "Tidal force formula". - Google's a handy shortcut for looking up formulas.</p>

<p>But I get what you're saying.  Thinking about it gets a little complicated.   This <a href="http://milesmathis.com/tide.html" rel="nofollow noreferrer">website</a> gives a better explanation than I'm likely to.   As a rule of thumb, the tidal force is equal to the size of the object in the sky (distance from the center of the Earth, so near by objects don't count), times the density.   The Sun and Moon are about the same size in the sky, the Moon is more dense, so the Moon causes greater tides.</p>

<blockquote>
  <p>Shouldn't such a large change be partially damped by the resulting
  increase in basin volume as low-lying areas flood?</p>
</blockquote>

<p>I hadn't thought of that, and probably, but I don't think that would affect it a great deal.  When a tidal bulge happens, the entire ocean (from the point of view of land) lifts up.  Oceans are very large, so unless the land is very flat for a very long distance, I don't think there's going to be a huge difference.   Lets say the tidal bulge peaks at 50 meters.  The tidal "wavelength" if you will is 10,000 km from peak to troth.  5,000 km from 50 meters to zero.  While it's a sign wave, not a linear drop, if we measure it linearly, which would be an exageration, the tidal bulge drops 1 meter every 100 km of distance.  Land usually rises quite a bit faster than that along the coastline almost everywhere.  There are some low-lands that would permanently become lakes, but mostly, I'd say at least 90%-95%, you can calculate how far the tide would reach based on meters above sea level.  If the land stays flat for a long time, you could adjust downwards.  That's my best guess anyway.   I'm not really an expert.</p>

<p>The bigger issue is the shape of the coastline.  Some coastlines focus the tide and effectively increase it.  The map above gives a pretty good estimate of that.  Basically, when the coastline is convex, the tide is focused and increases and when the coastline is concave, it decreases.  The exception being the gulf of Mexico, which has very low tides, despite being a very convex body of water, but the tides are blocked by Florida and many islands.   If you slow the orbit of the Moon down significantly, like you said, a moonrise every 6 months, then I think that effect would be largely reduced due to the much slower moving tide. It would have a lot more time to even out. </p>

<blockquote>
  <p>Could oceanic boundary conditions limit tides non-linearly? Might the
  increased tidal mass increase response time and flatten the bulge?</p>
</blockquote>

<p>Not crystal clear what you're asking here.  </p>

CC BY-SA 3.0
2016-04-30T20:40:06.787Z
Shouldn't the tidal forces go as the inverse *cube* of distance, not the inverse *square*? Given that and that orbit radius is the only change, I agree that the Newtonian tidal elongation will respond linearly with tidal force. However, it is not clear to me that that would translate into a linear response of actual coastal tidal amplitudes. Shouldn't such a large change be partially damped by the resulting increase in basin volume as low-lying areas flood? Could oceanic boundary conditions limit tides non-linearly? Might the increased tidal mass increase response time and flatten the bulge?
Geoffrey
CC BY-SA 3.0
2016-05-01T00:23:57.993Z
It's definitely inverse square.  I'll explain above & give an example or two.
userLTK
CC BY-SA 3.0
2016-05-01T00:47:40.910Z
It's definitely inverse cube.
David Hammen
CC BY-SA 3.0
2016-05-01T00:55:13.087Z
@DavidHammen  I welcome corrections, but if you want the tidal force from the moon to be 100 times greater, you need to move the moon 10 times closer.   The formula is clear.   Though I understand that the inverse cube plays a role, the effect that moving the moon closer has on the tide is inverse square to the distance.   http://hyperphysics.phy-astr.gsu.edu/hbase/tide.html
userLTK
CC BY-SA 3.0
2016-05-01T01:13:34.787Z
@userLTK -- Read that page carefully. If tidal forces followed an inverse square relationship, the Sun would be the dominant driver of the tides, by a [factor of about 180](http://www.wolframalpha.com/input/?i=((sun+mass)%2F(1+AU)%5E2)%2F((moon+mass)%2F(385000+km)%5E2)).  That is not the case. It's the Moon, not the Sun, that drives the tides, by a [factor of over two](http://www.wolframalpha.com/input/?i=((sun+mass)%2F(1+AU)%5E3)%2F((moon+mass)%2F(385000+km)%5E3))
David Hammen
CC BY-SA 3.0
2016-05-01T01:54:51.420Z
Thanks for the responses. The [tidal force wiki](https://en.wikipedia.org/wiki/Tidal_force) has a pretty clear derivation of the inverse cube form. And it should be that way, right? Since gravity is an inverse square law and the tidal force is based on the *difference* between the gravitational force at two places in a field. What I meant with the last part is that we do not observe a >1m tide - it's more like 50cm as you say - so clearly there are limits imposed on the model. Might those limits be non-linear (i.e. reduce greater tides by more than lesser ones).
Geoffrey
CC BY-SA 3.0
2016-05-01T02:00:08.967Z
And a larger tidal bulge will have more mass and therefore more inertia, lagging farther behind the passage of the moon unless the moon were slowed down significantly (like in geostationary orbit). So shouldn't that lag cause the bulge to be sheared somewhat, broadening and flattening it? Couldn't that affect tidal height?
Geoffrey
CC BY-SA 3.0
2016-05-12T12:04:38.547Z
@DavidHammen  apologies to you & Geoffrey for getting that wrong.  What I remember, from Tidal explanations is that the size of the object in the sky is roughly the strength of the tide, so if the Moon was 10 times closer, the size would be 100 times larger, but I may have gotten that estimate wrong.  I need to do more research there.
userLTK
CC BY-SA 3.0
2016-05-12T12:14:59.003Z
@Geoffrey Inertia doesn't really make sense to me.  The tidal bulge points towards the Moon and due to the rotation of the earth, is only 3 degrees ahead of the moon by this website:  https://www.lhup.edu/~dsimanek/scenario/tides.htm  If the Moon was closer to a geostationary orbit, that 3 degrees would be even smaller, and tides would move slower but the bulge should be roughly the same shape I would think.
userLTK
CC BY-SA 3.0
2016-05-12T17:12:47.773Z
@userLTK: What tidal bulge? http://physics.stackexchange.com/a/121858/52112
David Hammen
CC BY-SA 3.0
2016-05-12T22:35:36.350Z
@DavidHammen  fair enough and excellent answer there.  (Maybe I should stop answering questions altogether) . . . but in response to Geoffrey's question, the rate the Moon orbits in relation to the Earth's rotation and the Tidal (er, shape), I think my answer was correct, a slower moving moon in the sky wouldn't create much change in the Tides' lift.  Only in the rate it moves relative to the land.  Slower moving tides would reduce the effect the shape of the coastline has on local tides, but not on the lift overall.
userLTK
CC BY-SA 3.0
2016-04-30T16:54:00.853Z
2016-05-01T00:52:54.667Z
2016-05-01T00:52:54.667Z
userLTK
<p>Starting from scratch and building a "toy model" is unlikely to be successful here, because while you could certainly model Newtonian tides, much of what happens on the continental shelves is tied very closely into the shapes of coastlines and bathymetry, the interaction between the inertia of the tidal wave and friction on the seabed, and so forth - a lot of non-linear effects, some of which are not fully understood.</p>

<p>My suggestion would be to start with an existing, validated, global ocean model and tweak its input parameters. For example, to simulate a change in the gravitational pull of the moon you could change the amplitudes of the lunar tidal constituents that drive the model (and perhaps also their periods, if you wanted to work out the appropriate orbital mechanics!).</p>

<p>Be aware that running models of this sort tends to be computationally expensive, obviously depending on the resolution - but it should be within the realms of what is feasible on a home computer, if you don't mind things taking a week or two to run (I assume, since this is a hobby project, that you don't have access to HPC resources for it). I'm not sure whether anybody makes a simple but validated model openly available - if anybody knows of one, perhaps they could comment. If not, you might have to build your own using an open source code, which would be substantially more work.</p>

CC BY-SA 3.0
2016-05-02T22:32:52.797Z
Thanks for the advice. I feared that this would be the case. At least now I'm feeling more confident that I haven't missed anything.
Geoffrey
CC BY-SA 3.0
2016-05-03T02:08:01.597Z
The physicist's need for a spherical cow is, in this case, just wrong.
David Hammen
CC BY-SA 3.0
2016-05-02T12:26:20.807Z
2016-05-02T12:26:20.807Z
Semidiurnal Simon
<p>Are there any publicly available programs or source code for modelling global (or local) changes in sea level which include parameters for gravitational tidal effects?</p>

<p>I have a physics background, and recently I started playing with large-scale fluid dynamics as a side project. The oceans and their boundary conditions are incredibly complicated, and I've gotten pretty far out of my depth at this point. I don't have the time or motivation to completely invent the wheel on this, so ideally, I'm looking for a program that would allow me to vary lots of parameters - including the tidal potential of the moon - and output topographical data for sea levels at different locations in the ocean or along the coast. However, I'd be happy with just an open source program that I could modify with additional parameters.</p>

<p>I'm looking into this question predominantly from a hobbyist's "toy model" perspective. In particular, I'm curious about how oceanic tides would change with tidal forces, say, 100x greater than the moon's or with an almost tidally locked moon (e.g. six months between moonrise and moonset). How much land would be flooded with each high-tide and how much reclaimed with each low-tide? If anyone has an answer to that question, I'd be interested in hearing it as well.</p>

<oceanography><models><sea-level><tides><ocean-models>
Modelling sea levels with gravitational parameters
Geoffrey
# Modelling sea levels with gravitational parameters

(Asked by: username_0 on 2016-04-30)

Are there any publicly available programs or source code for modelling global (or local) changes in sea level which include parameters for gravitational tidal effects?

I have a physics background, and recently I started playing with large-scale fluid dynamics as a side project. The oceans and their boundary conditions are incredibly complicated, and I've gotten pretty far out of my depth at this point. I don't have the time or motivation to completely invent the wheel on this, so ideally, I'm looking for a program that would allow me to vary lots of parameters - including the tidal potential of the moon - and output topographical data for sea levels at different locations in the ocean or along the coast. However, I'd be happy with just an open source program that I could modify with additional parameters.

I'm looking into this question predominantly from a hobbyist's "toy model" perspective. In particular, I'm curious about how oceanic tides would change with tidal forces, say, 100x greater than the moon's or with an almost tidally locked moon (e.g. six months between moonrise and moonset). How much land would be flooded with each high-tide and how much reclaimed with each low-tide? If anyone has an answer to that question, I'd be interested in hearing it as well.

**Tags:** oceanography, models, sea-level, tides, ocean-models


---

## 2 answers

### Answer by username_3 on 2016-05-02. Score: 4 | Accepted answer

Starting from scratch and building a "toy model" is unlikely to be successful here, because while you could certainly model Newtonian tides, much of what happens on the continental shelves is tied very closely into the shapes of coastlines and bathymetry, the interaction between the inertia of the tidal wave and friction on the seabed, and so forth - a lot of non-linear effects, some of which are not fully understood.

My suggestion would be to start with an existing, validated, global ocean model and tweak its input parameters. For example, to simulate a change in the gravitational pull of the moon you could change the amplitudes of the lunar tidal constituents that drive the model (and perhaps also their periods, if you wanted to work out the appropriate orbital mechanics!).

Be aware that running models of this sort tends to be computationally expensive, obviously depending on the resolution - but it should be within the realms of what is feasible on a home computer, if you don't mind things taking a week or two to run (I assume, since this is a hobby project, that you don't have access to HPC resources for it). I'm not sure whether anybody makes a simple but validated model openly available - if anybody knows of one, perhaps they could comment. If not, you might have to build your own using an open source code, which would be substantially more work.

> **Comments:**
> **username_0**: Thanks for the advice. I feared that this would be the case. At least now I'm feeling more confident that I haven't missed anything.  
> **username_2**: The physicist's need for a spherical cow is, in this case, just wrong.  

---

### Answer by username_1 on 2016-04-30. Score: 1

I can give this a shot. Anyone smarter than me is welcome to offer corrections.

The first issue, based on your question is how much closer you move the Moon. (I'm assuming that you're not playing with other variations like the Earth's rotation speed or the Moon's mass).

If you want Tides to be 100 times greater, you have to move the Moon 10 times closer. The tidal force is equal to the square of the distance (Radius to Radius, not surface to surface). Putting the Moon 10 times closer would make it enormous in the sky and full moons would be bright enough to read by and the Tidal force would be 100 times greater.

The Tidal Force is basically a bulge in the Earth that stays pointed at the Moon. The land bulges too, not just the oceans.

Optimally, tides require about 1/4 of the circumference of the Earth to have full effect. That's why lakes basically don't have tides. Anything smaller than an ocean doesn't have tides to speak of. Now if you make the tieds 10 times greater, the ratio ocean to large lake would still be the same, but large lake and sea tides (like the Mediterranean) would become much more apparent.

Think of tides as raising the sea level of the entire ocean. Of-course, it's not really like that, but locally, that's essentially what happens.

If the current tide is 50 CM (above average sea level), then imagine, in that area, what it would be like if sea level rose 50 meters. In-land it wouldn't be an issue, but most coastal cities couldn't survive that. 100 times the tides would be the end of virtually all coastal living.

See chart.

[![enter image description here](https://i.sstatic.net/zMw8S.jpg)](https://i.sstatic.net/zMw8S.jpg)

[Source](https://en.wikipedia.org/wiki/Tide).

As far as nearly geostationary, that would have 2 effects. The slower movement would create more leveling off, less variation that you see in the chart above, where one part of the Earth can get tides that are amplified can get 2 or 3 times the tides than average, and the Tides would happen more slowly, as you get two high tides every time the Moon is overhead. If the Moon took 6 months between rises, you'd get a high tide every 3 months, but you'd still have the 50 foot tide problem. The geostationary orbit distance of the Moon is pretty close to the first number given, about 36,000 km.

Currently Tides vary quite a bit, but that's because the Sun and Moon both interact with tides, so you get pretty significant tidal variation. You also get variation as the Moon's distance to the Earth changes. If the Moon was 10 times closer, the Sun-tides wouldn't change and would be arguably ignorable (less than 1/2 of 1% variation), where, currently the position of the sun to the Moon is important in tidal calculation.

You'd need to take into account the Moon's orbital period and the earth's rotation to estimate how quickly the tides move around the Earth. Currently the Moon takes 29 days to orbit the Earth, so it's primarily the Earth's rotation (once every 24 hours) that governs the tides. If you bring the Moon into 36,000 or 37,000 km, then the orbital speed is important because every few thousand KM would make a measurable difference.

My answer is pretty general. I can try to touch on some specifics if I missed anything. You could also ask in Astronomy stack exchange. Tides are perhaps closer to Astronomy than Earth Science.

> Shouldn't the tidal forces go as the inverse cube of distance, not the
> inverse square? Given that and that orbit radius is the only change

It's definitely the inverse square law. I've done enough reading up on tidal forces to be sure of that. The formula also clearly shows it as a product of inverse square too. See [here](http://hyperphysics.phy-astr.gsu.edu/hbase/tide.html) or just google "Tidal force formula". - Google's a handy shortcut for looking up formulas.

But I get what you're saying. Thinking about it gets a little complicated. This [website](http://milesmathis.com/tide.html) gives a better explanation than I'm likely to. As a rule of thumb, the tidal force is equal to the size of the object in the sky (distance from the center of the Earth, so near by objects don't count), times the density. The Sun and Moon are about the same size in the sky, the Moon is more dense, so the Moon causes greater tides.

> Shouldn't such a large change be partially damped by the resulting
> increase in basin volume as low-lying areas flood?

I hadn't thought of that, and probably, but I don't think that would affect it a great deal. When a tidal bulge happens, the entire ocean (from the point of view of land) lifts up. Oceans are very large, so unless the land is very flat for a very long distance, I don't think there's going to be a huge difference. Lets say the tidal bulge peaks at 50 meters. The tidal "wavelength" if you will is 10,000 km from peak to troth. 5,000 km from 50 meters to zero. While it's a sign wave, not a linear drop, if we measure it linearly, which would be an exageration, the tidal bulge drops 1 meter every 100 km of distance. Land usually rises quite a bit faster than that along the coastline almost everywhere. There are some low-lands that would permanently become lakes, but mostly, I'd say at least 90%-95%, you can calculate how far the tide would reach based on meters above sea level. If the land stays flat for a long time, you could adjust downwards. That's my best guess anyway. I'm not really an expert.

The bigger issue is the shape of the coastline. Some coastlines focus the tide and effectively increase it. The map above gives a pretty good estimate of that. Basically, when the coastline is convex, the tide is focused and increases and when the coastline is concave, it decreases. The exception being the gulf of Mexico, which has very low tides, despite being a very convex body of water, but the tides are blocked by Florida and many islands. If you slow the orbit of the Moon down significantly, like you said, a moonrise every 6 months, then I think that effect would be largely reduced due to the much slower moving tide. It would have a lot more time to even out.

> Could oceanic boundary conditions limit tides non-linearly? Might the
> increased tidal mass increase response time and flatten the bulge?

Not crystal clear what you're asking here.

> **Comments:**
> **username_0**: Shouldn't the tidal forces go as the inverse *cube* of distance, not the inverse *square*? Given that and that orbit radius is the only change, I agree that the Newtonian tidal elongation will respond linearly with tidal force. However, it is not clear to me that that would translate into a linear response of actual coastal tidal amplitudes. Shouldn't such a large change be partially damped by the resulting increase in basin volume as low-lying areas flood? Could oceanic boundary conditions limit tides non-linearly? Might the increased tidal mass increase response time and flatten the bulge?  
> **username_1**: It's definitely inverse square.  I'll explain above & give an example or two.  
> **username_2**: It's definitely inverse cube.  
> **username_1**: @username_2  I welcome corrections, but if you want the tidal force from the moon to be 100 times greater, you need to move the moon 10 times closer.   The formula is clear.   Though I understand that the inverse cube plays a role, the effect that moving the moon closer has on the tide is inverse square to the distance.   http://hyperphysics.phy-astr.gsu.edu/hbase/tide.html  
> **username_2**: @username_1 -- Read that page carefully. If tidal forces followed an inverse square relationship, the Sun would be the dominant driver of the tides, by a [factor of about 180](http://www.wolframalpha.com/input/?i=((sun+mass)%2F(1+AU)%5E2)%2F((moon+mass)%2F(385000+km)%5E2)).  That is not the case. It's the Moon, not the Sun, that drives the tides, by a [factor of over two](http://www.wolframalpha.com/input/?i=((sun+mass)%2F(1+AU)%5E3)%2F((moon+mass)%2F(385000+km)%5E3))  
> **username_0**: Thanks for the responses. The [tidal force wiki](https://en.wikipedia.org/wiki/Tidal_force) has a pretty clear derivation of the inverse cube form. And it should be that way, right? Since gravity is an inverse square law and the tidal force is based on the *difference* between the gravitational force at two places in a field. What I meant with the last part is that we do not observe a >1m tide - it's more like 50cm as you say - so clearly there are limits imposed on the model. Might those limits be non-linear (i.e. reduce greater tides by more than lesser ones).  
> **username_0**: And a larger tidal bulge will have more mass and therefore more inertia, lagging farther behind the passage of the moon unless the moon were slowed down significantly (like in geostationary orbit). So shouldn't that lag cause the bulge to be sheared somewhat, broadening and flattening it? Couldn't that affect tidal height?  
> **username_1**: @username_2  apologies to you & username_0 for getting that wrong.  What I remember, from Tidal explanations is that the size of the object in the sky is roughly the strength of the tide, so if the Moon was 10 times closer, the size would be 100 times larger, but I may have gotten that estimate wrong.  I need to do more research there.  
> **username_1**: @username_0 Inertia doesn't really make sense to me.  The tidal bulge points towards the Moon and due to the rotation of the earth, is only 3 degrees ahead of the moon by this website:  https://www.lhup.edu/~dsimanek/scenario/tides.htm  If the Moon was closer to a geostationary orbit, that 3 degrees would be even smaller, and tides would move slower but the bulge should be roughly the same shape I would think.  
> **username_2**: @username_1: What tidal bulge? http://physics.stackexchange.com/a/121858/52112  
> **username_1**: @username_2  fair enough and excellent answer there.  (Maybe I should stop answering questions altogether) . . . but in response to username_0's question, the rate the Moon orbits in relation to the Earth's rotation and the Tidal (er, shape), I think my answer was correct, a slower moving moon in the sky wouldn't create much change in the Tides' lift.  Only in the rate it moves relative to the land.  Slower moving tides would reduce the effect the shape of the coastline has on local tides, but not on the lift overall.  

---

