Tides Explained….excerpt from Do The Math: Using physics and estimation to assess energy, growth, options—by Tom Murphy December 18, 2011Posted by jaldenh in News & Views.
The Sphere Makes a Good Point
Tides are simply a consequence of putting an extended body in the gravitational field from another body. We exert tides on each other, in fact—though don’t try to use this as an excuse for the bulge that forms around your waist this holiday season!
Some gravity background: since Newton’s time, we have understood gravity to vary as the inverse square of the distance between masses. So gravity scales like 1/r², where r is the distance between sources. Even Einstein’s General Relativity (replaces Newtonian Gravity) respects this relationship, and we have tested that it is accurate to better than a part in ten billion using the lunar orbit. One gnarly consequence of the inverse-square law is that the gravitational force from a spherical body (planet, moon, star, etc.) is exactly the same as if all the mass were located in a point at the center of the body. In other words, the dirt under your feet plays some role in tugging you down. That dirt is very close, so 1/r² is large, but there is not much dirt right under you. Meanwhile, dirt on the other side of the Earth also exerts a pull. There’s more of it (within a given cone angle, for instance), but its pull is much weaker by the same factor. It all evens out to produce an effective gravitational pull toward the center of the Earth, as if all the mass were located there.
As an aside, if the Sun turned into a black hole—keeping its present mass in the process—Earth’s orbit would not change. The Sun is already acting like a gravitational point as far as the Earth is concerned. All that matters is mass and distance to the center, as far as gravity is concerned. Of course in this scenario, I would have to drastically revise my statements about the abundance of solar energy in last week’s post.
What does this have to do with tides? Well, the Moon—siting about 60 Earth-radii away— pulls on the Earth as if from a point. And the extended size of the Earth means that if we say the gravitational pull from the Moon at Earth’s center has strength 1/60², the pull on the near side is 1/59², while the pull at the far side is 1/61². In other words, the Moon’s pull varies by ±3.4% as we cross the Earth. Compared to the average response of the Earth (its center), the side facing the Moon really wants to get closer to the Moon, while the side opposite doesn’t understand what all the fuss is about, and is more sluggish in its attraction to the Moon. The result is an eager bulge on one side and a lethargic bulge on the other side. This is why a location sees two high tides per day, as the Earth rotates under the Moon-pointing bulge (but interaction with continental shelves/coastlines can delay it significantly, so that seeing the Moon high in the sky only means you’re at high tide in the middle of the open ocean).
Meanwhile, the Sun is 23,500 Earth-radii away, so its gravity varies by only ±0.0083% across Earth. But the Sun’s gravity on Earth is about 180 times stronger than that from the Moon, so the absolute force variation from the Sun across the Earth is about 45% as much as it is for the Moon (180×0.000083/0.034). During new and full moon, the Earth, Moon, and Sun are in a line and the bulges add (spring tides). At quarter moon, the high/low from the Sun partly fills in the low/high from the Moon, diminishing the amplitude (neap tides).
For the mathy among you, because tides deal with a difference of force across a small change in distance, tidal force is just the derivative of the underlying gravitational force times the displacement distance. Differentiating 1/r² gives 2/r³, so that the force difference is proportional to 2ΔR/r³, where ΔR is the displacement from the nominal (center) point. For numerical simplicity, I expressed everything above in units of Earth’s radius, soΔR = 1.
For a perfect fluid body (oceanic Earth), the lunar tides would result in a peak-to-trough tidal amplitude of approximately one meter. To get the scale, we realize that the Moon’s mass is one eightieth that of Earth’s and 60 times farther away from the oceans as the Earth center. So lunar gravity is 1/(80×60²) times that of Earth gravity. Since we saw earlier that lunar tides constitute a 3.4% variation of lunar gravity, we end up with the Moon’s gravity varying by ±1.2×10−7 times Earth’s gravity. A small number, yes, but we multiply by the radius of Earth (6378 km) to get the deformation height that establishes potential energy balance. Now we have 0.75 m. Add in solar tides, and we’re close to a meter.
Most of us have seen tides well in excess of a meter. Some special places exceed 10 m, but even run-of-the-mill places like Puget Sound have 4 m tides, and San Diego gets 2–3 m. These are all exhibiting a sloshing that happens in shallow water or geographic restrictions. Take a look at tides in Hilo, Hawaii for contrast. Popping out of deep water in the middle of the Pacific Ocean, Hawaii gets open ocean tides with an amplitude of about—wait for it—a meter.
In the middle of the ocean, water does not have to move very far to raise the tide. Everyone just scrunches a little closer together laterally, forcing some water up. But introduce a continental shelf or an inlet, and we have a pinch point so that lots of water must actually flow past the bottleneck to raise a tide on the other side of the restriction. The flow can get carried away and rise up on the shore more than the tides actually demanded. The sloshing sometimes gets really out of hand when the geography produces a resonant cavity: when the natural fill/drain period is close to the six-hour tidal span. It’s just like how pushing on a playground swing at the resonant frequency produces big motion with little force (kids hate when I push because I like to experiment with out-of-phase pushes).