Now is the time on Do the Math when we scan the energy landscape for viable alternatives to fossil fuels. In this post, we’ll look at tidal power, which is virtually inexhaustible on relevant timescales, is less intermittent than solar/wind (although still variable), and uses old-hat technology to make electricity. For this exercise, we mainly care about the scale at which the alternatives can contribute, leaving practical and economic considerations sitting in the cold for a bit (spoiler alert: most are hard and expensive).
Previously, we looked at solar and wind, finding that solar can satisfy our current demand without batting an eyelash, and that wind can be a serious contributor, although apparently incapable of carrying the load on its own. Thus we put solar in the “abundant” box and wind in the “useful” box. There’s an empty box labeled “waste of time.” Any guesses where I’m going to put tidal power? Don’t get upset yet.
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.