The tides are due to differential gravitional force. If instead of a big body in orbit, we had a radial line of little particles in orbit- the close ones would go faster, the further ones slower, and eventually they would form a spiral, and after a few orbits, a big mess.
But, the local body’s gravity keeps everything together. So, everything orbits the same speed, even though some parts are closer than others. On earth, the water, because it is liquid, can react where rock can’t as easily; so it flows. The water closer to the moon is going too slow for it’s orbital distance, so it wants to fall inward (gravity). The water opposite the moon is going too fast for it’s distance, and wants to fly away (centrifugal force). In each case, you have a force that slightly cancels earth gravity to allow the water to rise, a whole ten feet or so.
“But wait!” you say. “The moon orbits the earth.” Well, actually it doesn’t. The earth and moon orbit their common centre of gravity, which is about 2000 miles from the centre of the earth. To complicate matters, there is a similar effect between earth and sun which can exaggerate the tides during full and new moon, when the sun, earth, and moon are in line.
The key here is “differential gravity”, how fast does gravity fall off with distance? the stronger the rate of fall-off, the higher the tide. (But, in bays like Fundy, you have the “slosh” effect where the shape and size of bay can create exaggerated effects - I read somewhere actual tide effect on open seas is about 3 or 4 feet).
Of course, I live nowhere near an ocean and cannot validate this against real observations.
I presume this is in reference to Cecil’s column: Why are there two high tides a day when the moon is only overhead once a day?
Since this a reference to Cecil’s column and not to the Mailbag, I am moving it to the forum COMMENTS ON CECIL’S COLUMNS.
Welcome to the SDMB, and thank you for posting your comment.
Please include a link to Cecil’s column if it’s on the straight dope web site.
To include a link, it can be as simple as including the web page location in your post (make sure there is a space before and after the text of the URL).
Cecil’s column can be found on-line at the link mentioned by my colleague and dear friend CKDextHavn.
The column (including Slug Signorino’s illustration) can also be found on pages 199-201 of Cecil Adams’ book
“Triumph of the Straight Dope”.
I’m afraid Cecil has the right idea, but is slightly wrong about the physics. “The Flying Circus of Physics” (with answers) is a great source for correct explanations of seemingly odd physics.
Cecil is right in that a tidal force results from changes in a force with position. Unfortunately it is not the change the magnitude of the moon’s gravitational force, but rather the direction that causes two tides. The moon’s gravity pulls everything on the earth toward a line connecting the center of the earth with the center of the moon. The moon also pulls along the line, but that doesn’t result in the tides, from what I’ve read. The highest lunar tides are on the line, at both “ends” of the earth. The lowest low tides are 90 degrees off of the line. (The sun also contributes to tides, but that just confuses the issue.) The water at low tide is pulled toward that line more strongly than the water at high tide, because the water at high tide is already on the line. (Actually, the tides slightly lag this line, but I have a job and can’t go into that here.) The upshot is that the water at the low tides exerts pressure that supports the high tides. This is somewhat like what happens when you squeeze a balloon - balloon is narrower where it is squeezed, and extends to where it is “squoze”.
There is a lot of confusion about tides, so I wrote a web page describing the situation. Check out: http://www.badastronomy.com/bad/misc/tides.html
I’m afraid The Bad Astronomer’s explanation also relies on change in magnitude of the moon’s gravitational force along the line connecting the centers of the earth and moon. That explanation assumes that the water on the moon side is pulled away from the earth. Given that the ocean still touches the ocean floor, this could only happen if the density of the water were to change. Since water is incompressible, that can’t be right.
The only way to raise the level of water is to add water. That is why, in mathematical terms, you rely on the components of the differential of the force that are tangential to the earth’s surface, not the radial component. Those components are the ones that drive the flow that creates the ocean tides. Note that my earlier explanation was not quite correct either, as the perpendicular to the line connecting the earth and moon is radial at low tides, but I was trying to point this out in a short space.
If this all seems confusing, relax. If I remember correctly, Newton also made some mistakes on tides in his Principia.
Again, check out “The Flying Circus of Physics with Answers” by Jearl Walker, 1977 (or at least my edition is 1977, but I’m old).
As far as they go, I don’t see anything seriously wrong with Cecil’s or the Bad Astronomer’s explanations.
A complete answer would involve the calculation of the equipotential surfaces surrounding the earth, taking into account the rotation of the earth and moon, and the earth and moon gravity. The surface of the ocean is an equipotential surface. As the equipotential surface changes (as the earth rotates under the moon), the water moves to re-establish equilibrium with the equipotential surface. It “flows downhill.”
By a weird coincidence, I read Jearl Walker’s explanation for tides just yesterday, and I was puzzled by it. My edition is also 1977, but I used to have the edition without answers. His explanation is in the answer to 4.52: “The tides are not due to the moon or sun pulling the water radially outward from the earth. Instead the bulges are caused by the horizontal components of the gravitational forces from the moon or sun collecting the water in the bulges. Since the horizontal components are less below the moon’s or sun’s position in the sky and on the opposite side of the earth, the bulges collect there.”
As Tim points out, there is a problem with this explanation: at the points farthest away from the moon-earth line (approx. low tide, ignoring the tidal lag), the horizontal components are either zero (in the reference frame moving with the earth) or directed towards the moon (in an inertial reference frame). So, what causes the water to “leave” those points and move to the back of the earth? I think Jearl’s explanation needs a little work (he’s only human–see his answer to 4.103, where he apparently calls himself a liar). On a long thin rod oriented towards and orbiting a planet, there would be a “tidal” stretching of the rod–and Jearl’s explanation seems to miss this point.
Tim Allison said:
You then go on to state that the change in tide level is caused by the water moving rather than being stretched. But I can’t see how that disagrees with the answer that says the water is pulled away from the earth. It’s pulled, it’s incompressible, and it’s a fluid - of course the water that’s next to it is pulled in behind.
As RM said, the water is seeking an equipotential.
You are close, if I am right, but out of phase! (Sorry, it’s a lunar thing.)
The horizontal (to the Earth’s surface) are zero at the point under the moon, and the antipodal point. They are maximal at low tide, that is why the water leaves low tide and goes toward high tide. (Or tries to, it is the effort that generates the pressure that maintains the water column that is high tide.)
The radial component is maximized under the moon, and at the antipodal point, but the water can’t travel radially from the center of the earth. This what Jearl Walker talks about. If the radial component could pull the ocean towards the moon, the water would be less dense under the moon. This implies that water is compressible, which it isn’t. This is where Cecil, and the Bad Astronomer, have their explanation wrong. They both show, or discuss, the radial component of the tidal force.
To some extent, this is more an arguement of how to heuristically explain the existence of two tides, than of why there are two tides. We all agree that the ocean lies on the equipotential surface of the earth-moon-sun system. It seems to me one could argue that the radial component could pull water towards the moon. (Or, push it away on the antipodal point, assuming the sign of the tidal force changes direction at the Earth’s center.) The horizontal components would then move water in from the sides to keep the density of the ocean constant. I suppose the only way to resolve which heuristic picture is most correct, is to do the math for the shape of the equipotential with only the radial component, only the horizontal components, and then with both.
You are close, if I am right, but out of phase! (Sorry, it’s a lunar thing.)
The horizontal (to the Earth’s surface) are zero at the point under the moon, and the antipodal point. They are maximal at low tide, that is why the water leaves low tide and goes toward high tide. (Or tries to, it is the effort that generates the pressure that maintains the water column that is high tide.)
The radial component is maximized under the moon, and at the antipodal point, but the water can’t travel radially from the center of the earth. This what Jearl Walker talks about. If the radial component could pull the ocean towards the moon, the water would be less dense under the moon. This implies that water is compressible, which it isn’t. This is where Cecil, and the Bad Astronomer, have their explanation wrong. They both show, or discuss, the radial component of the tidal force.
To some extent, this is more an arguement of how to heuristically explain the existence of two tides, than of why there are two tides. We all agree that the ocean lies on the equipotential surface of the earth-moon-sun system. It seems to me one could argue that the radial component could pull water towards the moon. (Or, push it away on the antipodal point, assuming the sign of the tidal force changes direction at the Earth’s center.) The horizontal components would then move water in from the sides to keep the density of the ocean constant. I suppose the only way to resolve which heuristic picture is most correct, is to do the math for the shape of the equipotential with only the radial component, only the horizontal components, and then with both.
Both is good. I’ve done that, as an exercise in geophysics. Jearl misses the mark because he ignores the radial components–but the tidal force is most often presented as a radial force. The Bad Astronomer’s diagrams work pretty well for presenting that. The tidal force depends upon the inverse cube of distance because it is the derivative of the gravitational force, which is an inverse square.
That’s why the sun’s gravitational force on the earth is 200 times as great as the moon’s, but since it is 400 times farther away, it’s tidal force is half (200/400) that of the moon.
Tim, you wrote:
Your argument seems to rely on water being incompressible. How would the tides be qualitatively different if water were compressible? It seems to me (admittedly after only a minute or two of thought) that there would still be two high and two low tides per day.
I’ve thought about this for a while, and I’m not even sure that that follows. Why would it necessarily be less dense?
And you expect an answer four years later?
It’s almost semantics whether the tides are described as due to an upward force towards and away from the Moon, or due to an inward force radially towards the line intersecting the Earth and Moon’s centers, or due to both.
The difference in force at the Earth’s surface due to the Moon relative to the force at the center of the Earth is proportional to 2 * X * Xhat - Y * Yhat - Z * Zhat, where Xhat, etc. are unit vectors, X, Y, Z specifies a point on the Earth’s surface, and the Moon is on the X axis at Xmoon >> X. The Earth obviously adds a large force towards the Earth’s center, -F * Rhat. The total force is then
-F*Rhat + Fmoon * ( 2 * X * Xhat - Y * Yhat - Z * Zhat )
but with a small change in F, I could just as easily describe the total force as
-F1*Rhat + Fmoon * 3 * X * Xhat
or
-F2*Rhat - Fmoon * ( 3 * Y * Yhat + 3 * Z * Zhat )
This assumes a spherical Earth, ignores the tidal lag, blah, blah, blah.
Maybe with the new fees, they can afford to buy some vectors.