In the explanation of tides on earth there seem to be different versions for the second water bulge on the side opposite to the moon, while everybody seems to agree that the bulge on the moon side is due to the gravitational pull of the moon, I found the following explanations for the second bulge:

A) Centrifugal force due to the rotation of earth around the center of mass of earth and moon.
B) Earth is pulled away from the water (as in this Feynman lecture).
C) Smaller gravitational force of the moon due to the inverse square law as compared to the low tides in the middle and the first bulge closest to the moon.

My question is basically are all of these correct? Do they describe the same effect (in different frames of reference) or do you need to take into account all of them for a full description?


2 Answers 2


The tides (rise and fall of water in the sea) don't actually work like that. (A rare example of Isaac Newton getting it completely wrong.) It's more like the periodic forces act like the sloshing of water in a bucket when you carry it. The forces drive resonant oscillations in the water that actually circle around ocean basins (due to the Coriolis effect). It is high tide on one part of the coast when it is low tide on another quite close nearby, and the wave revolves around a central point.

Have a look at the map of amphidromic points for more details.

But I think your question is really about the idealised situation for tidal forces where the force is upwards (pointing away from the moon) on the opposite side of the world to the moon.

I don't think any of the explanations are very good. I think the best way to think of the tidal force is as the shape of the gravitational potential's curvature around the observer. The gravitational potential of the moon is shaped like a trumpet bell. If you are sat somewhere on the inner slope, you have a force on you pulling you towards the centre, which we can approximate by drawing a tangent plane to the potential surface. The tangent plane gives a linear approximation to the shape of the potential, and the force vector tells us only about this part. It is like the first term in a Taylor series approximating a function - it just gives you the linear component.

If you are only interested in a point particle, this is all you need. But for extended bodies you need more information. So we subtract this tangent plane from the trumpet-shaped potential to see what the next component is. We find this next-order correction is below the plane in directions towards and away from the centre of the potential well, and above the plane for directions at right angles. A body is stretched along the radial direction, and squeezed along tangential directions, taken as a correction to the pure radial force vector. This is the tidal force.

The best of the offered explanations is C), that it is because of the fall-off of gravity due to the inverse square. But that doesn't explain why the tidal force squeezes in tangential directions. The centrifugal force explanation is definitely incorrect - tidal forces apply even when there is no rotation. In my view, the best way to explain it is to draw the tangent plane to the potential, and subtract.

  • 1
    $\begingroup$ yowza on the taylor series argument! $\endgroup$ Mar 23, 2023 at 18:37
  • $\begingroup$ "The centrifugal force explanation is definitely incorrect - tidal forces apply even when there is no rotation." If there's no rotation, then there's no tides, although you may have been through off by the use of the word "rotation" rather than "revolution". If the Earth and Moon don't revolve around the Earth/Moon barycenter, then they crash into each other, and there are no tidal forces. $\endgroup$ Mar 24, 2023 at 3:22

With apologies for the hastily-drawn, MS Paint pictures:

Let's say you have four stars arranged in the shape of a circle.

enter image description here

Now add the gravitational force due to a distant object directly below this setup. We can calculate the forces acting on each of these stars using $F = GMm/r^2$. We get something that looks like this:

enter image description here

The bottom star experiences the largest force. The two stars on the side experience a force that does not point straight down, but rather at the gravitating object directly.

Because the forces & their directions are different, after a short while, the original circle is no longer a circle, but looks like this:

enter image description here

Now we'll add a circle to this diagram.

enter image description here

The circle corresponds to the surface of the Earth, while the stars correspond to the local water level. Note the star closest to the gravitating object (the Moon) is well displaced above the circle. This implies high tide. Similarly, note the topmost star is also well displaced above the circle, which also implies high tide. That is why there is a second bulge.

(Of course for the Earth things are slightly more complicated, because there are two gravitating bodies - the Moon and the Sun - that both affect the tides. Also, this picture of tides itself is idealized, per the answer in Does Earth really have two high-tide bulges on opposite sides?)

I'd say none of the three explanations you give satisfactorily explain tides, but ultimately whatever makes sense to you is the best explanation (unless you need it for an exam, in which case whatever makes sense to your examiner is the best explanation).


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