# Effect of lunar/solar gravity on shape of earth

1. Aug 21, 2011

### daniel6874

In Gamow's short book 'Gravity' he describes the effect of lunar gravity on the earth, in particular the seeming paradox that the oceans, in response to the tug of the moon, swell both in the direction of the moon and on the opposite side as well. He resolves the paradox by explaining that Kepler's laws require that the greater angular velocity of water on the opposite side of the earth send them further out, to satisfy the requirement that greater velocity implies a greater distance from the orbital focus.

My question is this: where does the extra water come from? Is water from the poles drawn down to form the tidal surplus? There is no net addition of water, and water does not "expand." Does this mean that sea level at the poles drops?

Thanks for any enlightenment.

2. Aug 21, 2011

### xts

Water from low tide areas is moved to high tide areas, but only within every single ocean. That is why there are no visible tides on small closed seas, like Baltic. Inertia of that oceanic water flowing in/out shallow waters makes tides to be much higher at the shore than on open ocean.

3. Aug 22, 2011

### chrisbaird

I agree with xts. If your position on earth is facing the moon, you will have high tide, which means that ocean water has been drawn by the moon's gravity to your location. Points on earth that are 90 degrees away from you, measured relative to the line connecting the earth and the moon, will be experiencing low tide. The water was drawn by moon's gravity from those locations to your position. It's not an issue of the poles always loosing water, because the moon does not hover constantly over the equator. The earth's rotation and the moon's orbit means that it is constantly located over different spots on earth, so that every ocean coast constantly cycles through low-tide and high-tide.

4. Aug 22, 2011

### xts

As usually, the main problem is America ;) It almost completely separates oceans. Bering's Straits is narrow and shallow - so no significant amount of water may flow between Pacific and Arctic Ocean. On the South, Drake's Passage is not that narrow, but it still stops any significant flow between Atlantic and Pacific.
But Atlantic is wide enough to make its own pendulum-like waving: as you have high tide at European coast, America is just at low tide. And vice versa. You have even bigger gap between Americas and Japan/Phillipines/Australia.

5. Aug 22, 2011

### Staff: Mentor

Also, I don't know who Gamow is, but it is not correct to attribute the far side bulge to centripetal acceleration. The tidal force is a STATIC force, completely unrelated to any rotation.

6. Aug 22, 2011

### Bill_K

Gamow was a leading physicist who did seminal work on alpha decay and big bang nucleosynthesis.

Laying blame for the tidal bulge on centrifugal force is an unusual way of putting it, but essentially correct. In the rotating frame there are two "forces" acting on the Earth, the centrifugal force and the lunar gravitational attraction. At the Earth's center they are in balance, while at the Earth's far side the gravitational attraction is less and the centrifugal force is more. Both effects combined are responsible for the tidal bulge.

7. Aug 22, 2011

### Staff: Mentor

The centrifugal forces on the near and far side cancel each other out. It is no more correct than adding a term for the degrees of Kevin Bacon to each side of the equation. Sure, the math still works out, but thats just because you haven't actually changed anything.

8. Aug 22, 2011

### K^2

russ, it's a matter of coordinate system choice. You can take a CS that rotates with the planet-moon system and has its center at the center of mass of two bodies. In that CS, the far side is clearly experiencing more centrifugal force. The effective gravitational potential, on the other hand, changes shape, so the gravitational tidal force is reduced. The net effect ends up being exactly the same.

Naturally, taking an inertial frame of reference and getting bulge entirely due to gravitational tidal force is a far less cumbersome way to do it.

9. Aug 22, 2011

### olivermsun

The centripetal accelerations on the opposite sides of the earth do not cancel out, as the gravitational potential is not constant with distance from the orbital focus in the inertial frame.

In terms of Kepler's law, this is the balance between (orbital) angular velocity with and distance.

10. Aug 22, 2011

### Staff: Mentor

You're not hearing me: if the earth and moon were not orbiting each other, there would be no rotating coordinate system and the tidal force would be the same. It is not a matter of picking one coordinate system over another. I understand that the math still works.

Last edited: Aug 22, 2011
11. Aug 22, 2011

### Staff: Mentor

There is no g in the centripetal force equation.

12. Aug 22, 2011

### K^2

I'm well aware of that. Yet, whether or not the force is there is still a matter of coordinate system. That's kind of the deal with any fictitious force, gravity not excluded. To say that one coordinate system is the right one and the other is wrong is silly. Some just happen to be easier to work with. You don't need centrifugal force to explain tidal bulging. The most general explanation would certainly need to avoid it. But it's not wrong when dealing with specific system to say that the bulging is partially caused by centrifugal force. With suitable choice of coordinate systems, it is.

13. Aug 22, 2011

### Staff: Mentor

Let me try this another way: if one learns that the tidal force is a consequence of differential centripetal acceleration, how would they calculate the tidal force in a scenario where there is no rotation?

14. Aug 22, 2011

### K^2

By accounting for ALL fictitious forces. Like I said, gravity is one of them. In one CS, it's just gravity. In another, it's gravity + centrifugal. It's a CS choice.

15. Aug 22, 2011

### Staff: Mentor

So you would invent a rotation if none existed? How would you calculate how fast the universe is rotating in order to select that rotating frame of reference?

16. Aug 22, 2011

### olivermsun

I was talking about the potential due to the sun's gravitational field, which is the centripetal force toward the "orbital focus" in Gamow's words. I was not talking about centripetal force due to the earth's rotation (I don't believe Gamow was, either).

17. Aug 22, 2011

### daniel6874

I was surprised to find additional replies to this question. Russ, to what would you attribute the far-side bulge, then? If the answer to this appears in subsequent discussion, please alert me to the comment number. Thanks.

18. Aug 23, 2011

### K^2

It's a coordinate system transformation. It's not changing laws of physics. I'm not sure what your complaint here is.

19. Aug 23, 2011

### Hurkyl

Staff Emeritus
Agree or disagree:
1. Tidal force from the moon causes a bulge on the near and far sides of the Earth, and flattens the places in-between.
2. Tidal forces from the sun cause a bulge on the sides of the Earth near and far from the Sun, and flattens the places in-between.
3. The Earth's rotation causes a bulge evenly distributed around the equator, and flattens the poles.
4. The Earth's revolution around the Earth-moon center causes a bulge on the far side of the moon and flattens the side nearer to the moon
5. The Earth's revolution around the Sun is insignificant.

I lay 80% odds Russ thinks you are saying "#3 is true, and completely explains the tides", or possibly that you are saying "#1 and #3 are the same thing", or something similar. (my first time reading through the thread, it certainly sounded to me like that was what you were saying)

Last edited: Aug 23, 2011
20. Aug 23, 2011

### K^2

What I'm saying is that choice of center and rate of rotation for a rotating coordinate system are completely arbitrary. For any choice, I'll end up with some centrifugal force that I'll have to add to all massive objects. For some choices of the rotating frame, such as choosing Earth-Moon CoM as center of rotation, centrifugal force is certainly greater on far side, and therefore contributes to tidal bulging.

In general, a rotating frame of reference has its own tidal force associated with it. If you are working in a rotating frame of reference, the CS tidal force must be compounded with tidal force due to metric. (Well, if you transform the metric, that should take care of itself, but I hope you get what I mean.)

21. Aug 23, 2011

### Hurkyl

Staff Emeritus
The primary effect of the gravity of an object is to pull other objects towards it.

Tidal force is the difference in gravitational attraction. If you have an elastic object and you tug (along a line) on the opposite sides with different force, then you stretch the object out along that line. This is the secondary effect of gravity -- to stretch objects. To deform spheres into things like seen at this wiki page.

22. Aug 23, 2011

### Staff: Mentor

I am aware. What I asked was how would you choose one if none was obvious?

For example, if a dumbell shaped object with weights 1000kg, 10km apart is held stationary wrt the background of stars, 30,000km from earth and is dropped, falling straight toward earth, with the bar pointing away from earth, how would you find the rotating frame of reference necessay to calculate the tidal force between the weights?

Last edited: Aug 23, 2011
23. Aug 23, 2011

### Staff: Mentor

Hurkyl, what I think I'm seeing for the near side doesn't sound like any of those:

The tidal force creates a near-side bulge and flattens the far side.

Then #4 would explain the far side bulge.

24. Aug 23, 2011

### willem2

There is really only one way of explaining the tides in a simple way, and that is in a rotating reference frame, where the difference in gravity from the moon is the only cause of the bulges.

If you really want to do it in a rotating reference frame, it can of course be done, but you need to account for:

1. The centrifugal force from the rotating reference frame. $\omega^2 r$ where r is the distance from the barycenter.

2. The acceleration because of the rotation of the earth. Take into account that the earth will rotate a bit slower, because of the reference frame also rotates in the same direction.

3. The coriolis force, because of the rotation of the earth.

4. The differences in gravity from the moon. Since these are equal to what you get in
a non-rotating frame, it follows that the first 3 effects must completely cancel.

25. Aug 23, 2011

### K^2

It. Is. Arbitrary. You can chose ANY coordinate system. In any coordinate system, the results will be the same. In some systems, part of the tidal bulging is due to centrifugal force. In other systems, there is no centrifugal force.

You can choose ANY system you like and do the computations. You'll get the SAME answer.