# Why two tides?

1. Dec 8, 2008

### tgt

What is the explanation for the tide on the surface of the earth opposite the moon?

2. Dec 8, 2008

### Danger

It's due to the inverse-square law of gravity. The strongest pull of the moon is upon the ocean closest to it. The second strongest pull is upon the Earth as a whole. Less so is the pull upon the opposite ocean.
The gravity of the moon therefore pulls the closest ocean away from the Earth, and then pulls the Earth away from the farthest ocean.
That's a gross oversimplification, but the best that I can contribute.

3. Dec 8, 2008

### Georgepowell

The moon does not just simply spin around the earth. The moon and the earth spin around their common center of mass. i.e. you can treat the moon and the earth as one object spinning around its own center of mass.

What this leads to, is that the part of the earth further away from the earth is like the outer side of a ball being swung round your head, and so the ball elongates in the direction of the point it is spinning from.

Of course the moons gravity still pulls the opposite side of the moon closer to it, but it is not as strong as the pull on the closer side of the earth.

4. Dec 8, 2008

Staff Emeritus
TGT, Danger's description is the correct one.

5. Dec 8, 2008

### Georgepowell

So is mine wrong?

He said that the earth is being pulled away from the ocean, but the earth is not getting any closer to the moon. I am aware that the earth is accelerating towards the moon and not getting any closer, but as the ocean on the other side of the earth is also staying the same distance away from the moon I dont see how his description works.

His explanation would only work if the earth was always accelerating towards the moon at a faster rate than the ocean was. Mine works because both the ocean and the earth would naturaly carry on in a straight line at a tangent to its orbit, and the ocean at the other side of the earth is trying to do this the same amount that the earth is, but the gravitational pull of the moon has more effect on the earth because it is closer. And therefore the ocean "bulges" on that side.

Correct me if im wrong, but tell me why as well.

Thanks

6. Dec 8, 2008

### Ian

George,
Imagine that you are swinging on a trapeze.
In this 'you' are the earth and the 'moon' is the trapeze pivot. If you do a 'loop the loop' as you swing, your hands holding the bar will feel a force toward the pivot - this force represents the higher tide on the side of the earth facing the moon.
Your shoes however, will try to fly off your feet - this the force represents the tide on the far side of the earth from the moon.
Like the others said, the key to understanding the tides is to realise that the earth and moon rotate as one body about a single point.

7. Dec 8, 2008

### Georgepowell

Yes, so we agree. You simply came up with another analogy (very similar to my analogy of swinging a ball around your head).

I don't know if you read my first post, or if you knew that it was me that posted it. I was the one who mentioned the earth and moon spinning round their barycenter (centre of mass of the two planets). And I was asking why I was wrong... But by the looks of it you agree with me as well.

Danger said: "...and then pulls the Earth away from the farthest ocean" which I said I thought was wrong, or at least significantly incomplete.

I am willing to accept I am wrong, but I need some convincing arguments :tongue:.

8. Dec 8, 2008

### Danger

That caveat was acknowledged in my last sentence, so I certainly take no offense at your mention of it. I merely answered with my best understanding of the situation as it was explained to me previously. Since I'm not a scientist, I always defer to those who actually know what they're talking about.
Now I'm just going to lie back and learn from the ensuing exchange.

9. Dec 8, 2008

### tgt

From the discussion it seems that on the far side of the earth (compared to the moon), the centrifugal force resulting from the rotation of earth and moon outweighs the gravitational pull from the moon, resulting in a net force away from the earth hence the second tidal wave.

10. Dec 8, 2008

### tgt

This pull on the Earth away from the farthest ocean must be the centrifugal force on the earth since the earth is not moving toward the moon as pointed out by george.

In this way both george and danger are both correct?

11. Dec 8, 2008

### D H

Staff Emeritus
I'll make a few observations first and then proceed with some math.

First point: The tides at some point in general are not aligned with the Earth-Moon orientation. The tides at some point have various frequency components. The magnitudes and times at which these components peak varies from place to place on the Earth. The dominant frequency at most places is the M2 tidal, with a period of 12.421 hours (half of a lunar day). A plot:

http://svs.gsfc.nasa.gov/stories/topex/images/TidalPatterns_hires.tif

The plot has a bunch of white curves emanating from various nodes where the M2 tidal effect is zero. The M2 tidal contribution will peak at the same time at any point along one of those white curves.

Second point: Some places on the Earth (e.g., most of the Gulf of Mexico and the Caribbean) essentially have but one tide per day. The M2 tidal is very small throughout the Gulf and the Caribbean.

The ocean tides are a complex interaction between the driving tidal force and friction between the oceans and the solid Earth. Rather than dwelling on these complexities, I'll finish with the gravitational forces that drive the tides. The easiest way to look at things is from the perspective of an Earth-centered frame. As this is a non-inertial frame, pseudo-force terms are needed to give the appearance that physics follows Newton's laws. In particular, the Earth is accelerating toward the Moon and the Sun. Worrying about the Moon only, the Earth as a whole is accelerating toward the Moon:

$$\boldsymbol{a}_{e\to m,\,\textrm{inertial}} = -\,\frac {GM_m}{||\boldsymbol{r}_{e\to m}||^3}\boldsymbol{r}_{e\to m}$$

The acceleration at some point pon the surface of the Earth differs slightly from this:

$$\boldsymbol{a}_{p\to m,\,\textrm{inertial}} = -\,\frac {GM_m}{||\boldsymbol{r}_{p\to m}||^3}\boldsymbol{r}_{p\to m}$$

The acceleration toward the Moon at that point as perceived in an Earth-centered frame is the difference between these accelerations:

$$\boldsymbol{a}_{p\to m,\,\textrm{Earth-centered}} = -\,GM_m\left( \frac {1}{||\boldsymbol{r}_{p\to m}||^3}\boldsymbol{r}_{p\to m} - \frac {1}{||\boldsymbol{r}_{e\to m}||^3}\boldsymbol{r}_{e\to m}\right)$$

The acceleration at the point on the Earth's surface directly between with the Earth's and Moon's centers of mass is directed toward the Moon with magnitude equal to

\aligned & GM_m\left(\frac {1}{(r_{e\to m}-r_e)^2}-\frac {1}{r_{e\to m}^2}\right) \\ &\approx 2GM_m\,\frac {r_e}{r_{e\to m}^3} \endaligned

The tidal acceleration at the point on the opposite side of the Earth has nearly the same magnitude and is directed away from the Moon. The driving force has two nodes, and hence the two tides per day.

Last edited by a moderator: Apr 24, 2017
12. Dec 8, 2008

### D H

Staff Emeritus
That's not a good explanation. Think of it this way: The sun and earth orbit (don't use spin here; that means something different) about their common center of mass, too, and the Sun-Earth center of mass is located *a lot* further from the Earth's center of mass than is the Earth-Moon center of mass.

The tidal forces result from the gradient of the gravitational acceleration, and this is to first order an inverse-cube effect. Even though the Sun is much more massive than is the Moon, the Moon has a much greater tidal influence than does the Sun because the Sun is so much further from the Earth than is the Sun.

13. Dec 8, 2008

### Georgepowell

Both the sun and the moon exert a gravitational force on the earth, but the difference from one side of the earth to the other is much greater for the moon's pull than it is for the sun's pull. If I have this right, then I already understood this.

What this explains is why the water is pulled towards the moon less, it is still pulled, just less than on the closer side of the earth. This alone therefore does not explain the tide on the far of the earth; Just because a force is smaller it doesn't make the ocean move further away from the moon.

The position of the earth-moon centre of mass means that this water is being swung outwards.

Are you saying that if their was no Centrifugal force (I say this with caution, I know it is not a force, just an illusion) then the tide on the other side of the earth would still be there? If so, what pushes it?

14. Dec 8, 2008

### Staff: Mentor

Yes! An object in freefall directly toward anotherh object (as opposed to being in orbit) feels tidal forces. If you model this as three objects connected via two springs and drop this system from a stationary point above a massive object (and the objects are oriented along that line), both springs will stretch as the three objects are pulled apart.

People often talk about orbits as situations where centrifugal force equals gravitational force, but this is a little bit of a misleading way of looking at it and tidal forces are an example of why. To be honest, I'm not sure if doing the math on your answer would produce the correct result, but even if it does, it doesn't give a full treatment to the concept of tidal force.

15. Dec 8, 2008

### D H

Staff Emeritus
Sure it does. The relevant gravitational acceleration is that relative to the Earth. The relative acceleration toward some third body at the subbody point at its antipodes are

$$a_{\textrm{tidal}} = GM_{\textrm{body}}\left(\frac {1}{(R_{\textrm{body}}\mp R_\textrm{e})^2}-\frac {1}{R_{\textrm{body}^2}}\right)$$

where GMbody is the body's gravitational parameter, Rbody is the distance from the center of the Earth to the center of the body, and Re is the radius of the Earth. (Note: For non-aligned points the full vector form must be used. I am using the fact that the Earth and body centers of mass, and the subbody point and its antipodes are collinear.)

Expanding the above,
$$a_{\textrm{tidal}} &= \pm \frac{GM_{\textrm{body}}R_{\textrm{e}}}{R_{\textrm{body}}^3} \left(2\pm3\frac{R_{\textrm{e}}}{R_{\textrm{body}}}+\cdots\right) \approx \pm 2\frac{GM_{\textrm{body}}R_{\textrm{e}}}{R_{\textrm{body}}^3}$$

The tidal acceleration is directed toward the third body at the subbody point and away from the third body at the antipodes, and the accelerations are nearly equal to one another (the omitted term is $O((R_{\textrm{e}}/R_{\textrm{body}})^4)$).

16. Dec 8, 2008

Staff Emeritus
DH has done a good job explaining how tides work. I'd like to take a different approach, and show why centrifugal force doesn't: the earth is traveling 40x faster around the sun than the moon travels around the earth - and the speed at which the earth travels around the earth-moon system is smaller still. Centrifugal force goes as the square of velocity, so solar tides would be at least 1600x larger than lunar tides. This isn't what we see, so that theory can't be the explanation.

17. Dec 9, 2008

### Georgepowell

I think that we both have the same theory behind the tide, just I might be explaining it differently...

As i have said, the centrifugal force is not a real one, just a tendadncy for objects to stay moving in a straight line. And in your analogy where the three objects on springs are in free-fall, the object on the further side has a tendancy to 'stay in the same place' whilst the other parts are falling towards the masive object. So your explanation is really just another way of explaining the centrifugal force.

18. Dec 9, 2008

### D H

Staff Emeritus
George, your barycenter argument just doesn't work. See post #16.

You will get something closer to the gravity gradient model if instead of looking at the Earth orbiting the Earth-Moon barycenter, you look at it in terms of the Earth orbiting the Moon. It still won't be right, however.

This model also will not be able to explain the tidal force on point on the Earth's surface whose distance to the Moon's center of mass is equal to the distance between the Earth's and Moon's centers of mass. Here, the magnitude of the tidal force is about 1/2 that of the force at the sub-Moon point and at its antipode. While the force at the sub-Moon point and at its antipode are directed away from the Earth, the tidal force at these equidistant points (a small circle on the Earth) is directed inward.

This small circle of equidistant points is very close to the great circle equidistant between the sub-Moon point and its antipode. One way to look at the tidal forces is to imagine the Earth as a rubber ball with strings attached to the sub-Moon point and its antipode. The tidal forces pull on the strings and squeeze on the middle.

19. Dec 9, 2008

### Staff: Mentor

In my example, there is no rotational component, so no centrifugal force.

20. Dec 9, 2008

### Georgepowell

OK, lets get this straight; I don't like arguing with someone that I agree with. Firstly, I think your explanation is completely valid, and is virtually the same explanation as mine:

In the following (correct) analogy:

If you where standing on the middle object (the earth) you would see the object moving away from you, and it would look like something was pushing it away. The same thing happens when that object is in perpetual free-fall (orbit), and it is this exact same effect that we call the centripetal force, or at-least a form of it. Each part of our planet is subject to this centripetal force, and each part of it is 'trying' to escape of at a tangent. But because of the "gradiant" of the strength of the moon's gravity, the further side of the earth, and its oceans, are more effected by it, the earth as a whole is less effected, and even less effected is the ocean closer to the moon.

I think you are all correct, and I know that the gradient of the moons gravitational strength on the earth is part of the explanation of the tides. But it is the tendency for the parts of the earth to keep on going in a straight line that causes the bulge:

If the barycentre of the earth-moon was directly in the centre of the earth, then there would be no tide on the far side of the earth, because the earth would not be orbiting around anything, and would not be subject the the water at the far side of the earth would not have the initial velocity to bulge out away from the barycentre.

I hope this ends our discussion.

Thanks - George

edit: This agrees with me as well: http://oceanlink.island.net/oinfo/tides/tides.html

Last edited: Dec 9, 2008