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JamesF

What causes the tides?

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Ignoring the tidal bulge on the side of the Earth opposite the Moon and the effects of the Sun etc. for the moment, we all know it's mostly the Moon's gravity attracting the water in the oceans, right?

Well, hmmm.  I was lying in bed the other night contemplating life, the universe and everything when it struck me that I wasn't really sure that's a particularly good explanation.  (Look, just don't ask me to explain what goes on in my head, ok?  I'm not entirely sure it's all under my control in the first place.)  My thoughts followed thusly:

The Moon's gravity exerts a force on an ocean facing it, but it's nowhere near the force exerted on the water by the Earth, so the Moon can't actually move the water towards it.  It just means that the net force on the water in the direction of the Earth is reduced.  It won't be dragged "up" towards the Moon because of the Moon's gravitational pull therefore.  It will stay where it is.  So what actually causes the "bulge" of water that occurs?

Is it that due to the spinning of the Earth the water actually wants to obey Newton's first law and disappear off into space all the time, but is prevented from doing so by the Earth's gravity, though not prevented quite so much when the Moon counters some of the Earth's gravitational attraction thereby allowing it to move further from the centre of the Earth before Earth's gravity "overcomes" it?

Perhaps that's not a very good explanation either.  Assuming it's basically the correct idea perhaps a clearer explanation would be to show how all the forces involved are acting.  So it wouldn't be the Moon's gravity attracting the water, but that force changing the equilibrium of all of the others?

It leads me to another question as well: Does a given mass of water weigh the same at sea level at all points on the Earth, or does it vary depending on the position of the Moon?

Is there some other explanation?  Or should I just be taking more Valium?

James

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Or

Could do the free open Moon course. It covered tides, and the earth itself bulges not just the water

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It's all vectors. The water is exerted a lot by the Earth and a little by the Moon. Doesn't matter if the gravitational force of the Moon and the Sun is much less than that of the Earth, the resulting vector points a little less to the center of the Earth as a result. Kepler was right, Galileo was wrong.

Edit: I hate to reference Wikipedia but it actually explains this very well: https://en.wikipedia.org/wiki/Tidal_force

 

Edited by Waddensky
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3 minutes ago, happy-kat said:

Could do the free open Moon course. It covered tides, and the earth itself bulges not just the water

Is that a Lancaster Uni course?  My son is hoping for an offer from them to study Natural Sciences, if Cambridge won't have him.

James

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I think you are overthinking it.

However you explain it

- Moon pulling on water

- Moon making Earth pull less on the water

- Waver just following straight path in bent space time

you are essentially describing the same phenomena.

Perhaps, third option would be "the most precise" explanation - as far as our understanding goes at this point.

There is more to the whole story - low/high tide change every 12h rather than 24h - this means there is "counter bulge" on the opposite side. That one is due to system having certain springiness in it and it is just oscillation in sync with the motion of the Moon (similar to what happens when marching on the bridge and wind blowing over structures - ah, yes, resonant frequency is the term).

As for water moving away from earth - there is spinning but also there is pressure. Water is not hugely compressible but it does compress somewhat and there are enormous pressures in depths. If we were to suddenly turn off gravity - all that ocean water would bounce of the surface of the earth into space.

Mass of the water is not the same on sea level - because a) - sea level is not the same height over surface of the earth, and b) surface of the earth is not same distance from the center of the Earth

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The key part which is being missed is that it's the non-uniformity in the field which is important.

If we imagine the Earth as a rigid sphere then:

  • the water closest to the Moon "feels" a stronger acceleration towards the Moon than the Earth does
  • the Earth "feels" a stronger acceleration towards the Moon than the water furthest from the Moon does

the net result is that you get a tidal bulge at each side.

It's got nothing to do with the springiness of water or pressure or anything like that. You see tidal forces in action in merging galaxies, where stars are (effectively) only interacting gravitationally. You still see two bulges.

However, there is a whole other interesting topic which is due to the "springiness" of the system: tidal locking. It's no coincidence that one side of the moon always faces us.

 

If you're in a uniform gravitational field there is no tidal bulging in the direction of the field.

 

This graphic from the wikipedia may help visualise:

1024px-Tidal-forces.svg.png

Graphic of tidal forces. The top picture shows the gravity field of a body to the right, the lower shows their residual once the field at the centre of the sphere is subtracted; this is the tidal force.

Edited by randomic
included caption for graphic
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39 minutes ago, happy-kat said:

Thanks for that.  In fact, double thanks, as doing some of these courses seems like an ideal way to spend some time whilst I don't have much work.

James

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As a side note: counterintuitively, this means that you not only weigh less when the moon is directly overhead but also when the moon is directly underneath.

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52 minutes ago, vlaiv said:

I think you are overthinking it.

However you explain it

- Moon pulling on water

- Moon making Earth pull less on the water

- Waver just following straight path in bent space time

you are essentially describing the same phenomena.

Perhaps, third option would be "the most precise" explanation - as far as our understanding goes at this point.

I agree that it's the same phenomenon, but I got to the point where I wasn't happy with the simple explanation.  It didn't seem to account for what happens in a way that felt sufficiently "thorough".  Factoring in the behaviour of the water due to the rotation of the planet for me provides a greater depth of understanding.  In fact I might suggest that now I understand it (at some level), whereas before yesterday I would merely be repeating what I had been told was the reason.

James

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46 minutes ago, randomic said:

The key part which is being missed is that it's the non-uniformity in the field which is important.

Very interesting point, and indeed I think you are right, however, I have observation to make - I'm not sure if I'm right about this as it just sprang to my mind:

Sun is much more massive and is much further, in fact, it accounts for only 46% of the influence of the Moon according to this article:

https://oceanservice.noaa.gov/education/tutorial_tides/tides02_cause.html

summarized in this image:

tide02_480.gif

However, if we look at three points - one in the Center of the earth and one on each side - those distances are much larger in comparison to distance to the Moon than are with respect to distance to the Sun. Not sure if I put that correctly, here is another attempt - radius of earth in comparison to distance to Moon is much larger than in comparison to distance to the Sun.

This means that tides due to the Sun should be much smaller although Sun exerts about half of the pull that of the Moon, right?

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8 minutes ago, vlaiv said:

However, if we look at three points - one in the Center of the earth and one on each side - those distances are much larger in comparison to distance to the Moon than are with respect to distance to the Sun. Not sure if I put that correctly, here is another attempt - radius of earth in comparison to distance to Moon is much larger than in comparison to distance to the Sun.

It shouldn't be too hard to run the numbers. We'll keep things Newtonian for simplicity. So a = GM/r^2

G is Gravitational constant 6.67e-11
Radius of Earth, re, is 6,371km
Mass of Sun is 2e30 kg
Distance to Sun from Earth centre, rs, is 149,785,000km (going off the above diagram)
Mass of Moon is 7.3e22 kg
Distance to Moon from Earth centre rm, is 384,835km

a at Earth centre due to Sun = 6.67e-11 x 2e30 / (1.5e11 + 0)^2 = 0.0059289 m/s^2
a at Earth close edge due to Sun = 6.67e-11 x 2e30 / (1.5e11 - 6.371e6)^2 = 0.0059294 m/s^2
The difference between these two is the tidal force due to the Sun = 0.0000005 m/s^2

a at Earth centre due to Moon = 6.67e-11 x 7.3e22 / (3.8e8 + 0)^2 = 0.000033719 m/s^2
a at Earth close edge due to Moon = 6.67e-11 x 7.3e22 / (3.8e8 - 6.371e6)^2 = 0.000034879 m/s^2
The difference between these two is the tidal force due to the Moon = 0.00000116 m/s^2

So the Sun imparts a tidal force around 43% of the Moon's tidal force. It's a bit different from the 46% in the diagram above because I did a lot of rounding.

To me, it's astonishing that the Sun imparts such a significant tidal force given how far away it is. It really helps appreciate just how ridiculously massive the Sun is compared to everything else in the solar system.

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The way I was taught to think about this was that the earths ocean, being a liquid, want to have an equipotential surface. That is the gravitational force (Newtonian model) is equal everywhere on the surface. 

If not the net force causes the water to flow towards one. Neglecting the continents etc. this is what you get.

Regards Andrew 

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All I can say is that my Nan told me the afternoon at Bembridge IOW when I was a kid was a great way of getting winkles for my toothless grandad.

They were probably trying to get me drowned .

M

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22 hours ago, randomic said:

As a side note: counterintuitively, this means that you not only weigh less when the moon is directly overhead but also when the moon is directly underneath.

But weight isn't a tidal effect.

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I feel the need to return to this because my brain just wouldn't let it go and I spent part of another night lying in bed trying to properly understand what's going on.  (I hope I'm not the only person who does this sort of thing.  Not that there's much I can do about it if I am.)

First I tried to decide, if the Moon and Earth were stationary relative to each other, would there be a tidal bulge and if so, how would its height compare with the tides we have in the "real world"?  I couldn't make up my mind about that as presumably they'd just crash into each other after a while.

Ignoring the effect of the Moon for a moment, I came to believe that there may be no force on the water causing it to move away from the planet.  It's just attempting to continue in a straight line at a constant speed given the inertia it has gained from the Earth's rotation (though I'm not certain about this for liquids, and perhaps particularly for polar liquids).  So presumably the height of a body of water is given by some sort of equilibrium being reached between the water continuing to move in a straight line and the force on it due to the Earth's gravity?

If that is the case then presumably the effect of the Moon is to allow that equilibrium point to move away from the surface of the Earth because it reduces the total force on the water in the direction of the centre of the planet?

It also occurred to me to wonder: do the seas and oceans "slosh towards" the land on their western edges as the Earth rotates to the east?  For example given two land masses of equal height above the centre of the Earth with a large body of water between them, would the water level be higher when measured against the land to the west than the east?

James

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13 minutes ago, JamesF said:

First I tried to decide, if the Moon and Earth were stationary relative to each other, would there be a tidal bulge

Yep, the bulge is caused by the non-uniform gravitational field across the Earth. We see the bulge in the oceans because the Earth is (relatively) rigid and water is not. Check my posts in this thread for more information.

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11 minutes ago, randomic said:

Yep, the bulge is caused by the non-uniform gravitational field across the Earth. We see the bulge in the oceans because the Earth is (relatively) rigid and water is not. Check my posts in this thread for more information.

So you'd say that only gravity is required to explain the effect, in which case given my stationary Earth and Moon example presumably we'd get similar tidal bulges, at least for a while?

James

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Try this link for a mathematical description Tides

Regards Andrew 

Edited by andrew s
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That will keep me quite for a while :)  Thank you, Andrew.

James

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I made a diagram which might help visualise. These points are not interacting so in the right-hand case they would drift apart indefinitely. Obviously, the oceans are gravitationally bound to the Earth so they don't just leave 😂

The red ellipses are not supposed to be an accurate representation of the tidal bulge shape, rather just a visual aid.

tqtxQuK.png

P.S. The end of the caption on the right hand diagram should say "the outer two move away from it."

Edited by randomic
P.S.

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3 hours ago, randomic said:

I made a diagram which might help visualise. These points are not interacting so in the right-hand case they would drift apart indefinitely. Obviously, the oceans are gravitationally bound to the Earth so they don't just leave

Thanks.  There's still something here that isn't clear to me yet though, and I'm not sure what it is.  I may just be being dense, over-complicating things or modelling the entire thing in my head in the wrong way.

If I think about a molecule of water at, say, "average" sea level in the middle of the sea with the Moon directly above it, so it's actually sitting on a line drawn between the centre of the Moon and the centre of the Earth, there's a force on the molecule towards the centre of the Earth due to the Earth's gravity.  There's also a much, much smaller force on the molecule towards the Moon due to the Moon's gravity.  The net force is therefore always towards the Earth.  So how can that molecule of water move towards the Moon?

It's also just occurred to me to wonder if there are "atmospheric tides" as well as sea tides :)

James

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7 hours ago, JamesF said:

Thanks.  There's still something here that isn't clear to me yet though, and I'm not sure what it is.  I may just be being dense, over-complicating things or modelling the entire thing in my head in the wrong way.

If I think about a molecule of water at, say, "average" sea level in the middle of the sea with the Moon directly above it, so it's actually sitting on a line drawn between the centre of the Moon and the centre of the Earth, there's a force on the molecule towards the centre of the Earth due to the Earth's gravity.  There's also a much, much smaller force on the molecule towards the Moon due to the Moon's gravity.  The net force is therefore always towards the Earth.  So how can that molecule of water move towards the Moon?

It's also just occurred to me to wonder if there are "atmospheric tides" as well as sea tides :)

James

I think you are missing that the average pull of gravity is balanced out but the strength of the materials of the solid earth and liquid water. This is essentially the EM force. 

What matters is the small pererbations from the average. These are shown in fig 5 of the link I gave. Section 3 gives a hand waving description. Just try reading the paper and forget the maths that's what I do first time through.

Regards Andrew 

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