How Do Lunar Tides Influence Earth's Rotation?

[quasar987]

Symon, Mechanics pp.174: "The effet of lunar tide is gradually to slow down the rotation of the earth. As the angular momentum of the rotating Earth decreases, the angular momentum of the moon must increase."

Why are the lunar tides gradually to slow down the rotation of the earth? (I assume we are talking about the angular momentum of spin of the Earth and the orbital angular momentum of the moon around the earth.)

 

[dextercioby]

I can't think right now of a good explanation,i'll think about it later.But one think i can tell u for sure.You're right about the Earth's spin angular momentum.Earth is a giant boson with spin 3 and as its spin decreases,it will eventually end up in a fermionic state with spin 5/2.If Earth had another natural satellite,it would have been a boson with spin 2,just like a graviton,only bigger,much bigger... :rofl:

Daniel.

PS.I believe Tide (sic) would come up with a reasonable explanation. :tongue2:

 

[russ_watters]

Why are the lunar tides gradually to slow down the rotation of the earth? (I assume we are talking about the angular momentum of spin of the Earth and the orbital angular momentum of the moon around the earth.)

Friction. When a fluid moves inside any container, there is both surface friction and viscous friction associated with the motion. This friction is dissipated as heat (see: volcanos on Io) and since the tides are caused by the moon's gravity and move around the Earth due to the rotation of the earth, the friction dissipates both rotational kinetic energy and the gravitational potential energy of the moon.

 

[quasar987]

How does the fact that there are tides affect the friction of water against the solid Earth more than if there weren't any tides?

 

[Janus]

There are two mechanisms at work that slow the Earth that are caused by the Moon's tides. The first has already been mentioned: Radiated heat loss due to friction. The second is also due to friction but works differently. What happens is that while the Moon tries to align the tidal bulbes with itself, The rotating Earth tends to try the bulge around with it. This tug of war between the Moon and Earth causes the tidal bugle to "lead" the moon a little bit. As the moon tries to pull the bulge back into alignment, the bulge also pulls the moon forward in its orbit. What you end up with is a net transfer of rotational energy from the Earth to the Moon. The Moon as a result climbs into a higher orbit and the Earth slows its spin.

 

[Doc Al]

How does the fact that there are tides affect the friction of water against the solid Earth more than if there weren't any tides?

Think of the tidal bulges as pointing towards the moon while the Earth rotates beneath them. As the Earth rotates, it drags the bulges a bit. As russ explained, this tidal friction dissipates energy, slowing down the Earth's rotation little by little. Also, the tidal bulges are pulled forward a bit by the Earth's rotation, thus changing their gravitational pull on the moon. It works out, as it must, that the moon gains whatever angular momentum the Earth loses.

Note: Janus beat me to it!

 

[kuba]

How does the fact that there are tides affect the friction of water against the solid Earth more than if there weren't any tides?

It doesn't.

Here's what I make of it:

It's just that the gravitational pull of Moon on our oceans is displacing those very oceans, so you have some work (energy) being subtracted from the Earth-Moon system on the astronomical scale and being put into motion of water on Earth. Since the water eventually hits the shoreline, it cannot move past it and it just changes its level (ignoring many important effects for sake of brevity here). But since water has some viscosity, you cannot make it flow (i.e. displace it) in the ocean without dissipating some energy at the very same time.

So essentially Moon's energy in Earth's gravitational field (linked to Moon's angular orbital momentum and so on), is being subtracted from, and the Earth's ocean's are being warmed up by the same amount. A very small, but finite, amount of that energy manages to heat up the stiffer parts of our planet, i.e. crust. I don't know if subcrustal tidal effects have been quantified on Earth or not, but they must be tiny compared to effect of ocean tides.

OTOH, on Io, as has been already mentioned, tides manage to keep that moon in essentially molten state

Cheers, Kuba

 

[GrapesOfWrath]

IIRC, the oceans have little to do with it. The solid Earth tidal displacement is about half that of the surface water--so there is a lot more mantle displaced. The tides are symmetrical, but they are displaced from the centerline--they don't arise/subside instantly. The torque on the nearside bulge is opposed by the torque on the farside bulge, but the nearside is of course closer, so the torque on it is greater.

 

[polyb]

OK guys, then how does this explain the fact that the moon now only revolves once around it's axis in a month (ie it shows the same face to the Earth at all times)? It has been stated to me several times that it was not always so.

Hint: it is the same "mechanism" that is slowing down the earth!

 

[GrapesOfWrath]

OK guys, then how does this explain the fact that the moon now only revolves once around it's axis in a month (ie it shows the same face to the Earth at all times)? It has been stated to me several times that it was not always so.

Hint: it is the same "mechanism" that is slowing down the earth!

Then that explains it, right?

??

 

[pervect]

OK guys, then how does this explain the fact that the moon now only revolves once around it's axis in a month (ie it shows the same face to the Earth at all times)? It has been stated to me several times that it was not always so.

Hint: it is the same "mechanism" that is slowing down the earth!

It's a very similar mechanism. The Earth also raises tides on the moon. They are solid tides, rather than liquid tides, but they are still tides. When the moon was rotating with repsect to the Earth, these tides caused the moon to lose rotational momentum and gain orbital momentum, just as the Earth is doing now.

 

[kuba]

IIRC, the oceans have little to do with it. The solid Earth tidal displacement is about half that of the surface water--so there is a lot more mantle displaced.

Displacement as in linear displacement or volumetric displacement? I find it hard to believe it would be linear displacement. In some regions tides reach a few meters IIRC. I just can't see the seabed displacing a few meters due to tidal effects. I mean if that's true then I'm just ignorant, but on a first look it seems kinda hard to believe.

Could you elaborate?

Cheers, Kuba

 

[GrapesOfWrath]

Displacement as in linear displacement or volumetric displacement? I find it hard to believe it would be linear displacement. In some regions tides reach a few meters IIRC. I just can't see the seabed displacing a few meters due to tidal effects. I mean if that's true then I'm just ignorant, but on a first look it seems kinda hard to believe.

Could you elaborate?

Certainly!

The tidal displacement due to the change in the equipotential surface is a little more than a meter--larger tides are due to coastline configurations that cause the tide to be mulitplied (the so-called tidal wave in the Indian Ocean disaster started out as a displacement of only a couple meters, but as it neared the shore, it slowed and piled up on itself, and produced a much higher wall of water). The solid Earth tide is about half of the ocean tide--if the Earth were not rigid, and was free to respond, then the solid Earth would rise with the tide, and the result would be that we would actually see no difference in water level relative to the shores. "Shore tides" would be a bit greater if there were no solid Earth tide.

 

[Tide]

It looks like you guys have a pretty good handle on this one! :-)

 

[Tigron-X]

I kind of have a different idea on the whole subject. It isn't complete, but I'll share with you guys a basic overview of the idea I’m still researching during my spare time. I agree that the displacement of water on the Earth does affect the rotation of the Earth. And, since the moon affects the Earth and vice versa, it’s needless to say that the moon affects the Earth’s movement through orbit around the Sun. However, I believe we need to acknowledge that there is even a stronger influence on the Earth’s rotation and that is the Sun and the forces acting upon the Sun, as well as the Sun’s reaction to these forces that are unknown…at least to me.

Anyways, the Sun affects the Earth’s core in various ways that cause the particles in the core to slow down or speed up. Depending on the type of acting force on the Earth, the core expands or contrasts.

What we currently know is that the Earth experiences hotter temperatures now than before and the ice caps are melting and have been melting since the Ice Age. From what I’ve come to understand, the cold water sinks to the bottom and begins to travel towards the equator as it’s pushed by fresh melted ice. And, as it reaches the equator the water begins to displace more weight upon the Earth’s surface at various different axis points. (BTW, I don't mean the Earth's mass is changed.) That force eventually pushes on the core and the core reacts by pushing outwards.

Also, that water acts on the tectonic plates and eventually cause the plates to spread apart. However, as the core pushes outwards, the magma also pushes towards the surface which in a matter of speaking binds the plates together as the magma cools. Of course the heat rises and affects our weather patterns. But even more so, this effect pushes tectonic plates into each other which causes earthquakes.

This effect also causes a pressure release on the core, thus slowing down the rotation of the Earth which affects its gravitational pull on the Moon. In this case, it is weakened, and The Moon is allowed to spin slower due to this release in pressure.

Also, this displacement in weight changed the direction of particle movement and particle sequence in the Earth’s core, and that eventually caused a shift in the axis. I also want to mention that the tides allow the Earth to regain an orbital balance around the Sun by shifting land mass.

So in conclusion, the Sun’s solar flares influence the Earth’s orbit and rotation. Now, whether the flares are the only reason for this occurrence or not is currently beyond my understanding.

 

[pervect]

While short to medium term variations in the Earth's rotation do occur as it's mass is redistributed (via a large assortment of mechanisms, including solar heating, glaciers, or anything else that changes the distribution of mass on the earth), these sprts of changes do not change the Earth's angular momentum.

The tidal braking of the Earth's rotation is well known, see for instance

this wikipedia article

and it (as has previoiusly been discussed) occurs because the Earth transfers orbital momentum to the moon via the tidal interactions.

 

[Andrew Mason]

Displacement as in linear displacement or volumetric displacement? I find it hard to believe it would be linear displacement. In some regions tides reach a few meters IIRC. I just can't see the seabed displacing a few meters due to tidal effects. I mean if that's true then I'm just ignorant, but on a first look it seems kinda hard to believe.

This raises an important point. What is the relationship between changes in gravity and the linear displacement that results? If the gravity changes by 1E-8 m/sec^2, what is the amount that the ocean rises? In effect, what is the 'spring constant' for the earth, water (and air)?

The Geosat satellite which measures changes in ocean height due to gravitational effects of the seabed, allowing it to map the seabed, must use such a relationship.

AM

 

[Chronos]

Gravity waves dissipate the energy. In an earth-moon system, the effect is very slight [moon slows down the rotation of the Earth by about 15 micro seconds a year].

 

[GrapesOfWrath]

Gravity waves? How does that work? :)

 

[pervect]

This raises an important point. What is the relationship between changes in gravity and the linear displacement that results? If the gravity changes by 1E-8 m/sec^2, what is the amount that the ocean rises? In effect, what is the 'spring constant' for the earth, water (and air)?


AM

One URL that talks about this is

http://scienceworld.wolfram.com/physics/Tide.html

They give a "rule of thumb" which is

some_constant * tidal acceration * r / g

where g is the gravity at the Earth's surface and r is the Earth's radius

but I"m not sure how much to trust this, it seems to disagree with their main caclulation in the value of some_constant.

I know there is a better URL out there, but I couldn't find it :-(, (I know it exists because once upon a time I read it), so I'm posting this URL on the idea that a partial answer is better than none. In any event, even the calculation on which the above is based is only an upper bound, because it assumes that the Earth's "figure" remains spherical. This relates to the idea of "Love Number", which is also discussed briefly on the same website at a different location.

http://scienceworld.wolfram.com/physics/LoveNumber.html

 

[Andrew Mason]

One URL that talks about this is

http://scienceworld.wolfram.com/physics/Tide.html
http://scienceworld.wolfram.com/physics/LoveNumber.html

Thanks very much for this. I am not sure how this can be a number that is derived from theory. It has to be based on actual data. Is there any experimental verification?

I should think the difference between the radius at the poles and that at the equator would give you the answer. It is about 17km, I believe. If we assume that, but for the Earth's spin, the radius at the pole and at the equator would be the same (apart from the effects of the moon''s gradient), this means that the centripetal force at the equator which reduces g by about .034 m/sec^2., is responsible for a bulge of 17 km. So the 'k' for the Earth should be given by:

$$k/m = 2E-6 sec^{-2}$$

In other words for each reduction in g by 1E-6 m/sec^2, the Earth will 'bulge' a distance of 1E-6/2E-6 = .5 m.

AM

 

[GrapesOfWrath]

In any event, even the calculation on which the above is based is only an upper bound, because it assumes that the Earth's "figure" remains spherical.

I don't think it does assume that. The h's are distinct in the x and z directions (one's perpendicular to the earth/moon line, the other parallel).

On the other hand, that page has numerous errors. We were just discussing it over at the Bad Astronomy Bulletin Board. There seems to be at least four typos in the first ten equations.

Thanks very much for this. I am not sure how this can be a number that is derived from theory. It has to be based on actual data. Is there any experimental verification?

Yes, it is more or less empirically derived.

I should think the difference between the radius at the poles and that at the equator would give you the answer. It is about 17km, I believe. If we assume that, but for the Earth's spin, the radius at the pole and at the equator would be the same (apart from the effects of the moon''s gradient), this means that the centripetal force at the equator which reduces g by about .034 m/sec^2., is responsible for a bulge of 17 km.

The bulge is closer to 21 km, but regardless, the decrease in g at the equator is only half a result of the centrifictional force, the other half is from the fact that the equator is farther from the center.

 

[Andrew Mason]

The bulge is closer to 21 km, but regardless, the decrease in g at the equator is only half a result of the centrifictional force, the other half is from the fact that the equator is farther from the center.

At first I thought, that can't be right - centripetal (which is real, not centri-fictional! - but I like that term) would be much greater. But indeed the increase due to gravitational decrease is even more than that - almost double that due to centripetal force:

$$g_{earth}/g_{sphere}= R_{sphere}^2/R_{earth}^2$$

$$g_{earth}/g_{sphere}= 6353^2/(6353+21)^2 = 0.99342156$$

So:
$$g_{sphere}-g_{earth} = 9.80/.99342 - 9.80 m/sec^2 = .065m/sec^2$$

The centripetal acceleration (which varies as R) is approximately .034 m/sec^2. (w^2R where w = 2pi/(24x60x60-240))

But, ultimately, is it not all due to centripetal acceleration? because it is the centripetal accleration that causes the Earth to expand which causes gravity to reduce which causes further expansion (almost double more). So as a practical matter, the expansion will be .5 m for each (roughly) 2E-6/sec^2. of reduction in g. although 1/3 of that will be due to the centripetal acceleration and 2/3 due to the consequential reduction in gravity from expansion due to centripetal acceleration.

AM