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thunderchicken
2004-Jun-17, 11:01 PM
Hi. This is my first post! First off, I would like to say that I know next to nothing about astronomy, so please bear with me if I am WAY off base!

I would like to know if my reasoning on the creation of tides is correct (I have searched the BA page and found lots of info on tides, but I remain unsure if I grasp the basic concepts).

FACTS??

The centrifugal forces created by the earth rotating on its axis are of no consequence when examining the differential, tide creating forces.
The earth and the moon revolve around a common centre of mass called the barycenter. This point is located about three quarters of the way to earth’s surface, on the side nearest the moon. The “centrifugal” forces created by THIS rotation can be used in an analysis of the forces that create tides.

MY REASONING:

POINT A - Side of earth farthest from the moon:

At this point I would see a large ‘centrifugal force’ away from the barycenter, and a small gravitational force from the moon acting in the opposite direction.

&lt; -------------Fc + Fg--- > = &lt; ----------

POINT B – Centre of earth:

Because the centre of the earth is not located on the barycenter, I would expect to see a ‘centrifugal force’ (smaller than at Point A) away from the barycenter. I would also expect to see a gravitational force (larger than at Point A) from the moon acting in the opposite direction. Because the earth and moon do not fly off from one another, these forces should be equal and opposite.

&lt; -----Fc + Fg----- > = balance

POINT C – Barycentre:

At the barycenter I would not expect to see any ‘centrifugal force’ at all. I would, however, expect to see a gravitational force larger than at Point B acting toward the moon.

Fg------ >

POINT D – Side of earth nearest the moon:

As this point is nearest to the moon, I would expect to find the largest looner gravitational pull of all. This point is also on the opposite side of the barycentre from points A and B. Just as at these points, I would expect to see a ‘centrifugal force’ directed away from the barycentre. However, this time, that force would be toward the moon.

Fc-- > + Fg-------- > = ---------- >

Therefore, I end up with this:

&lt; ---------- @ POINT A 0 @ POINT B @ POINT D---------- >

The two opposite net forces create two tidal bulges, one on the near side and one on the far side of the earth.
[/b]

ToSeek
2004-Jun-18, 12:35 AM
I don't think you do. The centrifugal effects are minor in comparison with the variations in lunar gravity. Phil has a good explanation here. (http://www.badastronomy.com/bad/misc/tides.html)

Anyhow, welcome to the BABB!

JustAGuy
2004-Jun-18, 01:59 AM
thunderchicken? Any relation to MaxFun?

stu
2004-Jun-18, 05:34 AM
To add a link (with pictures!), I will suggest one of my own pages: http://home.cwru.edu/~sjr16/earth_moon.html . If that's a bit less than you want, there's also http://home.cwru.edu/~sjr16/advanced/earth_moon.html which has lovely equations, too!

Note that on both pages, the info on tides is towards the bottom.

Ricimer
2004-Jun-18, 05:36 AM
basically it works like this:

The moon's gravity pulls on three different regions of the earth, the near side (a) the middle (b) and the far side (c).

Since gravity gets weaker with distance, it pulls harder on A than B and C.

As such the material there is drawn up and away from the further sections, creating a bulge.

The center (B) is pulled harder than C (the far side). And so it is pulled away fromt he far side. This means a part of the far side is left behind. This look, to us on earth, as another bulge, but opposite the side of the moon.

The rotation of the earth acts to increase the bulges uniformly, and more at the equator than the poles.

The Earth-moon revolution (the cause of your centrifugal forces) acts to mitigate (partially) the bulge on A, and increase (slightly) the bulge on C. This effect is minimal however.

Eroica
2004-Jun-20, 03:32 PM
The center (B) is pulled harder than C (the far side). And so it is pulled away from the far side. This means a part of the far side is left behind. ...
I used to think as much, but there is a problem with this view: it seems to imply that the Earth is "lurching" towards the Moon, leaving the tidal bulge on the far side behind. But the distance between the Earth and Moon need not change. If the two bodies were orbiting the barycentre in circular orbits, there would still be tides - but no question of anything being "left behind."

stu
2004-Jun-20, 05:33 PM
But the distance between the Earth and Moon need not change. If the two bodies were orbiting the barycentre in circular orbits, there would still be tides - but no question of anything being "left behind."

But there is a tidal bulge formed on Earth and Moon. Obviously there's the one in the water on Earth, but the land also does bulge. I think it's by a couple feet, but this was from a class nearly 3 years ago. The bulge on the moon, I believe, is more like 25 feet.

And - at least in my opinion - an even more interesting consequence is that the the bulge actually trails the position of the Earth-Moon line, causing the moon to be pulled slightly back in its orbit. As a result, the Earth-Moon distance actually increases by 3.8 cm per year.

Ricimer
2004-Jun-20, 06:17 PM
There's still the differential gravity, and that descript puts to words what the math says (though you're right, it is a bit misleading).

But basically, the bulge opposite the moon is caused because the center and near sides are pulled harder.

Stu: It's actually pulled forward in it's orbit.

Note: The displacement of the bulges is due to the earth's rotation dragging the water (and the bulge in the crust too!) forward a bit.

thunderchicken
2004-Jun-21, 09:07 PM
Hi, thanks for the replies to my post!

In my original post, I was reasoning (wrongly??) that the earth and moon have an outward momentum that is equal, but opposite to their gravitational attraction, thus keeping them in a stable orbit around the barycentre. I guess this would have meant that the centrifugal effects would be just as large, or larger than the gravitational effects. If that is in fact where I went wrong, I would like to try another line of reasoning that I hope is correct???

“The centrifugal effects are minor in comparison with the variations in lunar gravity”. This is because the moon is more or less ‘free-falling’ toward the earth with enough horizontal component of motion to avoid an impact. This would mean that the gravitational pull of the earth on the moon is far larger than any outward momentum (if the moon were to lose it’s horizontal component of motion, it would fall to the earth)?? This is why we can generally ignore ‘centrifugal forces’ and consider only the variations in lunar gravity over the diameter of the earth?

If I am right up to this point, I still do not understand what Phil is saying in his explanation: “…..Now, we measure the gravity of the Earth relative to the center of the Earth; everywhere on the Earth, the center is "down". In a sense, we see the center of the Earth as "at rest". It is mathematically correct to then subtract the force of the Moon on the center of the Earth from the force felt on the near and far sides.”

Am I a deeply confused individual, or am I starting to understand this whole tide thing?
:o

Irishman
2004-Jun-21, 10:12 PM
stu, I take issue with this:

Earth has a certain velocity as it orbits the sun, and this causes an outward force. Just as the moon's gravity pulls the water and creates a bulge towards it, the outward force also creates a bulge, but on the side opposite the moon. Thus there are two high tides on the planet on opposite sides at once.

Your "outward force" appears to mean inertia. Gravity is applying an inward force to pull the Earth's motion into an orbit. The outward force is the inertia. But the inertia always pulls outward from the Sun, while the Moon circles the Earth. So sometimes the outward force pulls toward the Moon and sometimes it pulls away. Therefore, inertia cannot be the cause of the opposite tidal bulge that is always present.

The opposite tidal bulge is caused by the Earth getting pulled more that the water on that side, because the water is farther away.

I used to think as much, but there is a problem with this view: it seems to imply that the Earth is "lurching" towards the Moon, leaving the tidal bulge on the far side behind. But the distance between the Earth and Moon need not change. If the two bodies were orbiting the barycentre in circular orbits, there would still be tides - but no question of anything being "left behind."

But "lurching" towards the moon is correct. The Earth lurches toward where the Moon is, but the Moon moves, so the Earth lurches toward where the Moon is now, but the Moon moves again. That perpetual lurch is the pivot of Earth about the barycenter.

thunderchicken, your biggest problem with your original explanation is trying to use a centrifugal explanation. Physicists and engineers get trained to think of inertial coordinate systems and told "centrifugal force is not a real force", so they get conditioned not to look at it that way.

It is probably cleaner to look at it the way it is being explained by the BA and others. The Moon pulls on nearer parts harder. The near water is pulled hardest, and able to move more, so it bulges toward the Moon. The dirt also bulges, but not as much because it is more rigid. As you move away from the Moon (through the planet), the pull lessens. At the far side, the water is pulled least of all, and is less rigid so stays more bulged away. This way you don't have to consider any rotational factors.

The rotational factor that becomes of interest is the Earth's rotation pulling the bulges slightly out of alignment, the near-moon bulge getting slightly ahead of the moon, thus slowing the Moon down, thus causing the Moon to drift outward. But no arguing over centripetal and centrifugal required.

stu
2004-Jun-22, 02:10 AM
stu, I take issue with this:

Earth has a certain velocity as it orbits the sun, and this causes an outward force. Just as the moon's gravity pulls the water and creates a bulge towards it, the outward force also creates a bulge, but on the side opposite the moon. Thus there are two high tides on the planet on opposite sides at once.

Your "outward force" appears to mean inertia. Gravity is applying an inward force to pull the Earth's motion into an orbit. The outward force is the inertia. But the inertia always pulls outward from the Sun, while the Moon circles the Earth. So sometimes the outward force pulls toward the Moon and sometimes it pulls away. Therefore, inertia cannot be the cause of the opposite tidal bulge that is always present.

The opposite tidal bulge is caused by the Earth getting pulled more that the water on that side, because the water is farther away.

My bad. Thanks for finding the mistake! (And I'll fix it as soon as I get some free time ... :cry: )

milli360
2004-Jun-22, 09:47 AM
This is why we can generally ignore ‘centrifugal forces’ and consider only the variations in lunar gravity over the diameter of the earth?
Gravity varies over the Earth, as the distance varies from the sun or moon varies, and this results in the differences that we know as tides. Centrifugal force (whether calculated from the center, or the barycenter) is essentially constant over the Earth, in direction and magnitude, so there are no differences that can cause tides.

thunderchicken
2004-Jun-22, 11:38 PM
I think I finally understand. Thanks.

Just a question: If the distance from the earth to the barycentre were far greater than it is, would we then see centrifugal effects coming into play?

Also, I found an explanation of tides in the Microsoft Encarta encyclopedia that uses differential centrifugal forces, balanced with differential gravitational forces as the cause of tides. Has anyone else seen this? Is it wrong?
8-[

milli360
2004-Jun-23, 04:10 AM
Just a question: If the distance from the earth to the barycentre were far greater than it is, would we then see centrifugal effects coming into play?

No, not really. The centrifictional force is constant across the mass of the Earth, essentially. It doesn't induce a difference, hence no part of the tide. Same for the moon, which is very far from the Earth/moon barycenter, although the moon's synchronous rotation means the tides are not as changing anyway.

Also, I found an explanation of tides in the Microsoft Encarta encyclopedia that uses differential centrifugal forces, balanced with differential gravitational forces as the cause of tides. Has anyone else seen this? Is it wrong?
Haven't seen it, but if they say that there is a difference in the centrifictional force from side to side, chances are it's wrong.

Ricimer
2004-Jun-23, 04:19 PM
“The centrifugal effects are minor in comparison with the variations in lunar gravity”. This is because the moon is more or less ‘free-falling’ toward the earth with enough horizontal component of motion to avoid an impact. This would mean that the gravitational pull of the earth on the moon is far larger than any outward momentum (if the moon were to lose it’s horizontal component of motion, it would fall to the earth)?? This is why we can generally ignore ‘centrifugal forces’ and consider only the variations in lunar gravity over the diameter of the earth? Sounds good to me.

[qoute]If I am right up to this point, I still do not understand what Phil is saying in his explanation: “…..Now, we measure the gravity of the Earth relative to the center of the Earth; everywhere on the Earth, the center is "down". In a sense, we see the center of the Earth as "at rest". It is mathematically correct to then subtract the force of the Moon on the center of the Earth from the force felt on the near and far sides.”

Am I a deeply confused individual, or am I starting to understand this whole tide thing?
:o[/quote]

Okay, what he's saying here is everyone on earth generally considers the center of the earth to be, well, the center of our system. Now, the earth's center is normally being pulled by the moon. However, if we treat it as the reference point, the "zero" point or whatever, the part that doesn't change (it's arbitrary, btw) just like we use the ground for the reference point on a lot of other things (like altitude).

We can reference any point, but some are more meaningful than others. Here it makes more sense to use the center of the earth, as opposed to...china.

Now, if we want it to be our reference point, it shouldn't move. The force there, should be defined as zero. But there is a force there! Ack! ....wait. Since it is our reference point, we can subtract the force exerted there, from the force exerted everywhere else.

Example: Height, again. Sure, a plane is travelng 1 mile above the earth's surface. But it's ~12,001 miles from the core. Since nobody cares about the distance to the core (you crash when you hit the surface, not the center) everybody just subtracts (or doesn't even consider) that 12,000 mile portion.

So there's a force acting on the earths center, it's stronger than on the far side, weaker than on the near side (relative to moon). So if we subtract it, we get a negative force on the far side (2-5=-3...) which means....it's pointing in the opposite of a positive force. We also get a positive force on the near side...

So we end up with two forces at the surface, one pointing left, the other right (or whichever, just so long as thier opposites). Due to symmetry (or actual math I don't want to do) I also posite that they are ~the same strength and orientation.

Viola! We get the tides.

milli360
2004-Jun-23, 04:39 PM
Viola! We get the tides.
Music to my ears

Ricimer
2004-Jun-23, 04:45 PM
....grrrr.... mispelling.

Ah well. Briings a smile to my face, so it stays.

P.s. Is it to late to say I meant to do that?

milli360
2004-Jun-23, 05:00 PM
P.s. Is it to late to say I meant to do that?
Not at all. You probably got it from Travis (http://www.badastronomy.com/phpBB/viewtopic.php?p=274678#274678). You may even have got it from me (http://www.badastronomy.com/phpBB/viewtopic.php?p=14211#14211)!

I editted my last post to add a PS, and then I took it out when I saw you'd responded. Here it is:

PS: Do not, under any circumstances, trust the tides webpage explanation (http://co-ops.nos.noaa.gov/restles1.html) hosted by our National Oceanographic and Atmospheric Organization. It's an embarassment:

At the surface of the earth, the earth's force of gravitational attraction acts in a direction inward toward its center of mass, and thus holds the ocean water confined to this surface. However, the gravitational forces of the moon and sun also act externally upon the earth's ocean waters. These external forces are exerted as tide-producing, or so-called "tractive" forces. Their effects are superimposed upon the earth's gravitational force and act to draw the ocean waters to positions on the earth's surface directly beneath these respective celestial bodies (i.e., towards the "sublunar" and "subsolar" points).

High tides are produced in the ocean waters by the "heaping" action resulting from the horizontal flow of water toward two regions of the earth representing positions of maximum attraction of combined lunar and solar gravitational forces. Low tides are created by a compensating maximum withdrawal of water from regions around the earth midway between these two humps. The alternation of high and low tides is caused by the daily (or diurnal) rotation of the earth with respect to these two tidal humps and two tidal depressions.

ToSeek
2004-Jun-23, 05:48 PM
It's an embarrassment

Not to mention virtually incomprehensible. Sounds like something Nancy would try to pass off as Zetatalk.

Ricimer
2004-Jun-23, 06:06 PM
well, it's technically right.

But poorly written.

And doesn't give a clue as to the origin of the second bulge.

milli360
2004-Jun-23, 06:07 PM
It's an embarrassment

Not to mention virtually incomprehensible. Sounds like something Nancy would try to pass off as Zetatalk.
I think (http://boards.straightdope.com/sdmb/showthread.php?s=&amp;threadid=159592&amp;perpage=40&amp;pagen umber=2#post2982783) it was really wrong years ago, and they tried to patch it without completely cleaning it up.

milli360
2004-Jun-23, 06:09 PM
well, it's technically right.
The two tidal bulges are the sublunar and subsolar points? That's not true at all.

And doesn't give a clue as to the origin of the second bulge.
No, it says there are two bulges, one below the sun and one below the moon.

Ricimer
2004-Jun-23, 06:56 PM
hmmm....well then...

You're right, that's a fundamental mistake.

I just assumed they'd integrated the solar tides the way it's usually done. You know, neap and spring tides.

Well, then...yeah, that's close, but no cigar. Acceptable from an individual, but not from an organizaiton in the field.

Can I complain to them somehow?

milli360
2004-Jun-23, 07:39 PM
Can I complain to them somehow?
I complained years ago--that was when they started to patch it up, I think.

Check out the rest of the comments.