Understanding the Moon's Role in Tides

[chmate]

Hi all! I was just wondering if anyone can explain me why the tides (created by moon effect) are created on opposite sides.

(image is on attachment)

Thank you.

 

[sylas]

Hi all! I was just wondering if anyone can explain me why the tides (created by moon effect) are created on opposite sides.

(image is on attachment)

Thank you.

In simple terms... the bit closest to the moon gets a bit more pull than the rest of the Earth, and so there is a hump towards the Moon.

The bit furthest away from the moon gets a bit less pull than the rest of the Earth, and so there is a hump away from the Moon.

Cheers -- Sylas

 

[granpa]

gravity and centrifugal force

 

[russ_watters]

...centrifugal force

Since that is an insignificantly small fraction of what is going on (not sure how small, but certainly less than 1%), it is best not to bring it up when someone is looking for the broad overview of tides.


chmate, I tend to word it slightly differently than cylas: The hump nearest the moon is caused by that ocean being pulled away from the earth. The hump furthest from the moon is caused by the Earth being pulled away from that ocean.

 

[D H]

gravity and centrifugal force

What centrifugal force? In what rotating frame? Centrifugal forces only arise in rotating reference frames. You should be able to explain this phenomenon from any reference frame, and there is no centrifugal force in a non-rotating frame. Some phenomena (e.g., hurricanes) are much easier to explain in terms of a rotating frame than a non-rotating frame. This is not one of them.

IMO, the easiest frame to use as a basis for this explanation is an Earth-centered inertial frame. This is an accelerating frame, so a fictitious inertial force arises from the gravitational acceleration of the Earth toward the Moon. The apparent gravitational force exerted by the Moon at some point on the surface of the Earth is the gravitational acceleration toward the Moon at this point less the acceleration of the Earth as a whole toward the Moon. This apparent force is

• Directed toward the Moon (and away from the Earth) on the side of the Earth facing the Moon,
• Directed away from the Moon (and away from the Earth) on the side of the Earth opposite the Moon, with nearly the same magnitude as the apparent force at the sub-Moon point, and
• Directed toward the Earth (but at half strength compared to the sub-Moon point) at the points where the Moon is on the horizon.

The result: Bulges. Looking at just the sub-Moon and anti-Moon points omits what is happening elsewhere. The squeeze in the middle is just as important as those outward forces at the nodes.

 

[granpa]

gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit). there is an excess gravitational force on the moon side of the Earth and an excess centrifugal force on the opposite side of the earth.

at the center of the Earth the 2 forces are balanced.

 

[granpa]

http://www.scienceforums.net/forum/showthread.php?t=36766

http://www.coas.oregonstate.edu/research/po/research/tide/index.html

 

[D H]

gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit). there is an excess gravitational force on the moon side of the Earth and an excess centrifugal force on the opposite side of the earth.

at the center of the Earth the 2 forces are balanced.

In what rotating frame? Whenever you invoke centrifugal force you need to specify what your frame of reference is. There are no centrifugal forces in a non-rotating frame.

 

[granpa]

gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit/rotate).

 

[DaveC426913]

The hump nearest the moon is caused by that ocean being pulled away from the earth. The hump furthest from the moon is caused by the Earth being pulled away from that ocean.

This is the explanation I have found to be the most clarifying.

 

[D H]

gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit/rotate).

The Earth and Moon don't orbit in the rotating frame used to explain tidal effects from the perspective of centrifugal force. They are stationary in such a frame.

 

[DaveC426913]

The Earth and Moon don't orbit in the rotating frame used to explain tidal effects from the perspective of centrifugal force. They are stationary in such a frame.

I don't know if you're beating him up excessively.

Inasmuch as centrifugal force can explain to the layperson why water stays in a swung bucket on a string, surely this is also an acceptable answer for the tides.

Not that I think it's the best explanation...

 

[D H]

If you do the math properly, the Earth's ocean tides can be explained from the perspective of a Mars-centered, Mars-fixed reference frame. Just because it can be done doesn't mean it is the best choice. Explaining the tides from the perspective of a rotating frame is not the best (i.e., easiest) choice, either. If you pick a frame rotating with the Earth and Moon and with origin at the Earth-Moon barycenter, the sub-Moon point is a lot closer to the origin than is its antipodal point. If you pick a Moon centered or Earth centered frame, you have the fictitious inertial force plus the fictitious centrifugal force to deal with. Any way you cut it, the math is messy. There is no reason to invoke a rotating frame when a non-rotating frame gives a clearer and cleaner explanation.

 

[Vanadium 50]

I don't know if you're beating him up excessively.

I think he is not beating him up excessively. If anything, not enough.

The answer provided was simply "gravity and centrifugal force". Now, I think we can all agree that "gravity" by itself is a poor explanation. While gravity is the cause, the word "gravity" alone is not descriptive. So therefore, it is natural to conclude "and centrifugal force" means that centrifugal force plays a major role in the explanation. While it may be possible to twist things so this is technically correct, as an explanation, it's poor.

Russ' answer is, in my mind, the clearest. The moon's gravity pulls the near-side ocean away from the Earth and the Earth away from the far-side ocean.

 

[pixel01]

I know some textbooks explain tides by centrifugal force with mass center of the two bodies.
But i myself agree with " The moon's gravity pulls the near-side ocean away from the Earth and the Earth away from the far-side ocean"

 

[russ_watters]

gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit). there is an excess gravitational force on the moon side of the Earth and an excess centrifugal force on the opposite side of the earth.

at the center of the Earth the 2 forces are balanced.

That effect is so small as to be unimportant. Perhaps you could actually calculate its magnitude and compare it to the magnitude of the tidal force...

http://www.scienceforums.net/forum/s...ad.php?t=36766

The fact that someone in another forum said it means exactly the same as when someone (you) in this forum said it: they got it wrong. Not sure what the point of the other link was.

 

[russ_watters]

I don't know if you're beating him up excessively.

Inasmuch as centrifugal force can explain to the layperson why water stays in a swung bucket on a string, surely this is also an acceptable answer for the tides.

Not that I think it's the best explanation...

No, it is not an acceptable explanation. In the swinging bucket, if the bucket isn't swinging, the water falls to the ground. In the tides example, if the Earth and moon are not in orbit around each other, the tides are still there. What he is describing is not tidal force. It is a different effect entirely, and a miniscule one at that.

 

[russ_watters]

I know some textbooks explain tides by centrifugal force with mass center of the two bodies.

If they do, they are wrong - as I said before, the tides exist even in a non-rotating system. What granpa is describing is not tidal force. Are there terms in the tidal force equation for location of the barycenter and period of rotation?

Since it is still too early in the morning for math, for now just consider this problem:
The Earh has no moon, but a moon-sized asteroid is on a direct collision course with earth. Are there two tidal bulges or one?

 

[yenchin]

Here is what one of my profs has to say:

http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

 

[LURCH]

No, it is not an acceptable explanation. In the swinging bucket, if the bucket isn't swinging, the water falls to the ground. In the tides example, if the Earth and moon are not in orbit around each other, the tides are still there. What he is describing is not tidal force. It is a different effect entirely, and a miniscule one at that.

So, if the Earth and Moon were not rotating around a common center of gravity, and were stationary relative to one another, the oceans would still bulge out on the far side? What would cause this bulge?

 

[sylas]

So, if the Earth and Moon were not rotating around a common center of gravity, and were stationary relative to one another, the oceans would still bulge out on the far side? What would cause this bulge?

The bulge occurs when the Earth, and its oceans, are all in freefall. If you actively hold the Earth stationary somehow, it's no longer the same situation.

 

[DaveC426913]

So, if the Earth and Moon were not rotating around a common center of gravity, and were stationary relative to one another, the oceans would still bulge out on the far side? What would cause this bulge?

The key is, if they were not rotating around each other, they would not be stationary relative to each other. They would accelerate toward each other. What would happen to the oceans in this case?

 

[sylas]

You'd still get the bulge, on both sides. Briefly. Then the ocean would be vaporized in the collision.

 

[D H]

I know some textbooks explain tides by centrifugal force with mass center of the two bodies.

If they do, they are wrong - as I said before, the tides exist even in a non-rotating system.

Russ is exactly right. The tides exist, so one had better be able to explain them from any frame of reference. Some choices are good in the sense that the explanation is clear and concise. Other choices, such as a frame with origin at the center of Mars and rotating with Mars' rotation rate (1 revolution per 1.026 days), are not so good in the sense that the explanation is anything but clear and concise.

Here is what one of my profs has to say:
http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

From the link, a bit out of order,

Some argue in the same way as above, but in order to get an outward pointing force at the far side, they consider a centrifugal force caused by the Earth's rotation around the Earth-Moon center of mass. This is not correct.​

Hmm. That's just what Russ said.

This explanation seems to be popular among oceanographers. All the sites below use a centrifugal force.​

It appears oceanographers have been infected by a bad meme.

They argue that the centrifugal force is the same at every point of the Earth, and that it must be equal to the gravitational pull at the center of the Earth. I find it too much of a “coincidence” that the centrifugal force happens to equal the gravitational pull at the center (or the center of mass).​

This is an incredible hand-wave that is an aberration of the concept of centrifugal force. More below.

Some people, including Aslamazov and Varlamov in the book “The Wonders of Physics” use a centrifugal force, but they let the force depend on the distance from the center of rotation, which is clearly wrong. If we ignore the rotation of the Earth, then all points on the Earth describes circles with the same radius, but different centers, in the course of the month.​

I disagree with the author of that website here. The centrifugal force on some object of mass m located at some point r from the origin of a rotating reference frame is defined as $m{\boldsymbol{\omega}}\times{\boldsymbol r}$. If one insists on using a rotating frame, one should at least do it right.

The underlying assumption needed to arrive at the conclusion that "all points on the Earth describe circles with the same radius" is that the Earth is not rotating with respect to inertial space. This in turn means that the Earth is rotating with respect to the rotating Earth-Moon frame, and hence one has to consider Coriolis force as well as centrifugal force in explaining the tides. To make things worse, the authors of the sites erroneously use the term "centrifugal force" to describe what is really the sum of the centrifugal and Coriolis forces. Why do that when all one has to do is invoke the concept of tidal gravity and be done with it?

But i myself agree with " The moon's gravity pulls the near-side ocean away from the Earth and the Earth away from the far-side ocean"

Personally I like tidal gravity (or gravity gradient) as the explanation. The explanation pops out clear and concise from the math, and this also explains why the squeeze at the middle. Words don't do as much for me as does mathematics.

 

[russ_watters]

So, if the Earth and Moon were not rotating around a common center of gravity, and were stationary relative to one another, the oceans would still bulge out on the far side? What would cause this bulge?

Sorry, I'm not sure if I said it(i was thinking it) but the Earth and moon can't be held stationary with respect to each other: if a hypothetical support was constructed to keep them there, the regular gravitational force would swamp the tidal force and all of the water would rush to the side of the Earth that the moon is on, so it isn't good for visualizing the issue. So consider instead the scenario I described before:

The Earth has no moon, but a moon-sized asteroid is on a direct collision course with earth. Are there two tidal bulges or one?

There are two tidal bulges.

 

[russ_watters]

You'd still get the bulge, on both sides. Briefly. Then the ocean would be vaporized in the collision.

Briefly, but not too briefly - if the object were traveling at 30,000 km/hr (reasonable for an object on a collision course with us), it would take hours for it to get here from a distance of the moon's orbit (it would accelerate during that time of course, so i don't want to try to calculate how long exactly). Plenty of time to observe and measure the tidal bulges as they both grow prior to the collision.

 

[russ_watters]

Here is what one of my profs has to say:

http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

From the link:

In fact, a simple computation shows that the centrifugal force caused by the Earth's rotation around the Earth-Moon center of mass is tiny compared to the gravitational differential.

Which is what I said and I honestly intend to do the calculation if I become less lazy in the future... I'm curious to know, because this is a common misunderstanding and it has come up before.

 

[Count Iblis]

I know some textbooks explain tides by centrifugal force with mass center of the two bodies.
But i myself agree with " The moon's gravity pulls the near-side ocean away from the Earth and the Earth away from the far-side ocean"

I'm quite sure textbooks don't call this "centrifugal force", rather they will work in the frame which is co-moving with the Earth's center of mass. In this reference frame, the Earth's surface is stationary (we don't need to consider the Earth's rotation around its axis) because the Earth is (to good approximation) a rigid body. The gravitational force plus the fictitious force yileds the acceleration in this frame of reference.

On the side of the Moon, the sum of the gravitational force and the fictitious force points in the direction of the Moon, on the opposite side it points away from the Moon.

 

[D H]

I'm quite sure textbooks don't call this "centrifugal force", rather they will work in the frame which is co-moving with the Earth's center of mass.

Unfortunately, you are wrong. For example, see
http://books.google.com/books?id=80...=X&oi=book_result&ct=result&resnum=1#PPA53,M1

In this reference frame, the Earth's surface is stationary (we don't need to consider the Earth's rotation around its axis) because the Earth is (to good approximation) a rigid body. The gravitational force plus the fictitious force yileds the acceleration in this frame of reference.

On the side of the Moon, the sum of the gravitational force and the fictitious force points in the direction of the Moon, on the opposite side it points away from the Moon.

Simple. Why invoke a rotating frame, and why invoke any frame but an Earth-centered frame? The goal, after all, is to compute the acceleration relative to the Earth.

 

[Count Iblis]

But the explanation given in thet book is correct, the only thing is that he calls the fictitious force the "centrifugal force". Now, the book is written by Andreĭ Andreevich Varlamov, Alekseĭ, so it could be a mistranslation.

 

[pixel01]

In reality, the two tide bulges are symmetrical (or nearly). If proved by centrifugal force and center of mass, the two bulges are not.

 

[D H]

But the explanation given in thet book is correct, the only thing is that he calls the fictitious force the "centrifugal force."

From the book,

Consider now the revolving frame where both the Earth and the Moon state at rest. Inasmuch as the reference frame is noninertial every mass element experiences not only the force of gravity but a centrifugal force as well. ...​

That is definitely a centrifugal force. Unfortunately, the authors (I gather this is a physics for Russian poets text) did not do any math. If they had, it would have been wrong because the explanation in the book is, simply put, wrong.

This physics for Russian poets text is wrong. The teaching of how tides arise in American oceanography classes also invokes centrifugal force and in doing so is wrong.

Here is what one of my profs has to say:
http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

This link provides several links to how oceanographers envision the tides as arising. For example, lecture notes from E&ES 106 at Wesleyan University in 1999 (not your average Whatsamatta U.). Starting at slide 20,
http://soconnell.web.wesleyan.edu/ees106/lecture_notes/lecture-tides/HTML%20Presentation%20folder/sld020.htm,

Tides are differential gravitational forces, the difference between the gravitational forces exerted on two neighboring particles by a third more distant body.​

This is exactly correct. Unfortunately, the subsequent slides do not develop this concept. They instead incorrectly invoke the concept of centrifugal force.

The same explanation is http://web.vims.edu/physical/research/TCTutorial/origin.htm" (NOAA).

The problem with invoking the concept of centrifugal force is that doing so will inherently give the wrong answer for the simple reason that the apparent force in a rotating frame is equal to the force in a non-rotating frame at the frame origin only. Asking what the forces are from the perspective of a non-accelerating frame (e.g., the Earth-Moon barycenter, ignoring the Sun) is equally wrong because the ultimate goal is to determine the acceleration with respect to the center of the Earth.

There are at least a couple of ways to arrive at the correct answer when reasoning from the basis of a rotating frame or a non-Earth centered frame. The right way is to properly develop the equations of motion in the selected frame and then reframe the answer in terms of an Earth-centered inertial frame. The equations of motion as properly developed in the selected frame will include a differential gravity term plus a whole lot of other terms such as centrifugal and coriolis acceleration. Reframing the answer in terms of an Earth-centered frame will remove all that extraneous stuff, leaving only differential gravity. The answer will be correct, but why bother.

These centrifugal force explanations did not use this route. The promulgators instead mix frames, ignore coriolis force, mislabel some hand-wavy thing as the centrifugal force, and do not recast the answer in terms of an Earth-centered frame. They do come up with the right answer, but that is only because they know what the right answer is ahead of time.

 

[granpa]

http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

3.
Some people, including Aslamazov and Varlamov in the book “The Wonders of Physics” use a centrifugal force, but they let the force depend on the distance from the center of rotation, which is clearly wrong. If we ignore the rotation of the Earth, then all points on the Earth describes circles with the same radius, but different centers, in the course of the month.

why would we 'ignore the rotation of the earth'?

 

[granpa]

as the Earth spins on its axis it produces an outward centrifugal force but over millions of years the whole Earth itself (not just the water) has bulged at the equator thereby cancelling out this force. most of the rotation can therefore be safely ignored. but not all of it. the fraction of its rotation equal to 1 rotation per month does not produce a regular equatorial bulge. it produces a bulge on the side facing the moon and a bulge on the opposite side away from the moon. in between there is no bulge.

therefore most of the Earth's spin can be ignored but the once per month rotation can not be ignored.

hence to a first approximation when calculating tidal force we can treat the Earth as though it rotated once per month. and the centrifugal force (due to the Earth and moon orbiting each other) is proportional to distance from the earth-moon center of mass (around which they orbit). the centrifugal force at the center of the Earth (and roughly speaking at the poles) is equal to the gravitational force. to a first approximation the gravitational force can be considered to be a constant.

 

[D H]

why would we 'ignore the rotation of the earth'?

Rotation of an object means angular motion about an axis that passes through an axis internal to the object in question. The Earth rotates once per sidereal day. If you want to attack the issue of tides from the perspective of a rotating Earth, have at it. We ignore the daily rotation of the Earth to get a first-order approximation of what makes the tides arise.

That said an Earth-centered, Earth-fixed frame is the proper reference frame for describing the tides as the oceans are more-or-less rotating with the Earth. A real-world model of the tides is incredibly messy and is also quite ad-hoc (37 frequency components that vary from place to place over the world).

as the Earth spins on its axis it produces an outward centrifugal force but over millions of years the whole Earth itself (not just the water) has bulged at the equator thereby cancelling out this force.

Wrong. The shape of the Earth is, ignoring mountains and valleys, an equipotential surface, not a constant force surface. Mean sea level is an equipotential surface.

most of the rotation can therefore be safely ignored. but not all of it. the fraction of its rotation equal to 1 rotation per month does not produce a regular equatorial bulge. it produces a bulge on the side facing the moon and a bulge on the opposite side away from the moon.

Wrong again. This bulge has nothing to do with orbital motion. As Russ mentioned earlier, an asteroid headed straight toward the Earth with zero angular momentum would still raise tides -- and on both sides of the Earth. The tides result from the gradient in the gravitational force, not from centrifugal force.

in between there is no bulge.

In between there is the opposite of a bulge. Tidal forces are directed inward at the points where the Moon is on the horizon.

hence to a first approximation when calculating tidal force we can treat the Earth as though it rotated once per month. and the centrifugal force (due to the Earth and moon orbiting each other) is proportional to distance from the earth-moon center of mass (around which they orbit). the centrifugal force at the center of the Earth (and roughly speaking at the poles) is equal to the gravitational force. to a first approximation the gravitational force can be considered to be a constant.

This is nonsense.

Do the math.