**Question:** I’ve often wondered why the highest tide comes when there’s a full moon. The Sun and Moon are opposite each other at that time, so I would think their gravitational pulls would somewhat negate each other. But when there’s a new moon, and both Sun and Moon are on the same side, it seems their gravity would combine, so that should be when the highest tide happens. Obviously I’m missing something here. — BT, Minoqua, Wisconsin

**Answer:** You’re asking a difficult question BT. I can answer it, but fair warning: on a scale of 1 to 10 in difficulty, the answer is around a 9.572. Your understanding of tides is mostly correct. Tide heights do indeed depend on the relative positions of the Sun and Moon. But the Moon’s affect on tide height dominates that of the Sun by a factor of 33. I’ll explain why below. And (all other factors equal), the full moon and new moon tides are of equal height.

Mariners refer to these highest tides as *spring tides* (or just *springs*). It has nothing to do with the season, but with the fact that the ocean “springs up” to greater heights. Lowest tides are called *neap tides* (*neaps*). The origin of the term “neap” is not known. Best guess: “neap” comes from the Old English “nepflod” which could (perhaps) mean “without power.”

You’re also correct that tides are caused by gravity from the Sun and Moon. But to answer your question, we’ll need the equation for gravity derived by Sir Isaac Newton, and first published in his classic Principia Mathematica in 1687:

F_{g} = G⋅m_{1}⋅m_{2} / r ^{2}

where F_{g} is the force of gravity, G is the gravitational constant, m_{1} and m_{2} are the masses of the two attracting bodies, and r is the distance between the centers of those two bodies. For the purpose of understanding tides, r is the main factor we need to focus on.

Note that r is in the *denominator* of the fraction, which makes F_{g} get smaller as r increases. But r is also squared, which means that a doubling of r causes a 1/4 reduction in the force of gravity. A tripling of r would cause a 1/9 reduction. It’s what mathematicians call an *inverse square* relation. Gravity decreases rapidly with distance.

One counter-intuitive thing the equation tells us is that the gravitational pull from the Sun is only 3% of that from the Moon. Even though the Sun has 27 million times the mass of the Moon, that inverse square term means the Moon’s gravity will dominate since it’s 375 times closer. If you’re a student, look up all the numbers and try to prove that for yourself.

So let’s just imagine the Sun out of the picture for a moment and consider what the Moon does to the Earth. That r^{ 2} term is the key. Gravity is an atom-to-atom force, and Earth is flexible. The Moon pulls on the closest side of Earth harder that it does on the middle, and it pulls harder on the middle than it does on the farthest side. The net effect is to stretch the Earth into an ellipsoid, with tidal bulges both under and opposite the Moon. These are normal high tides, not spring tides.

Water flows between the interconnected oceans and piles up where needed to produce those bulges. As the rotating Earth carries your harbor into either bulge, the tide comes in. More correctly, “you come into the tide.” So that’s why there are two high tides each day, one when the Moon is overhead, and another 12 hours 25 minutes later. That extra 25 minutes is because the Moon orbits in the same direction Earth rotates, and drags its tidal bulge with it. As a result, Earth needs to rotate for slightly more than half a day to catch up to the next tidal bulge.

Even allowing for that extra 25 minutes, it takes all that water some time to move. Coastal topography can speed or slow this process. Also, the Sun’s gravity pulls on the tidal bulge caused by the Moon, speeding it up if it’s approaching, and slowing it when it’s receding. So the actual high tides can precede or lag calculated times by as much as several hours. That’s why all harbors need to keep their own tide charts.

The graphic greatly exaggerates Earth’s distortion. You wouldn’t even notice it at the scale of that image. The world record difference between high and low tides occurs in the Bay of Fundy, where the extreme variation is 16.3 meters (53.5 feet). The average variation is much lower. In the open ocean, it’s only about 0.6 meters (2 feet). But coastal topography can focus the incoming tidal surge to much greater heights.

Now we can put the Sun back into the picture. The Sun produces it’s own tidal bulge, but it’s only 3% as high as the Moon’s. At full moon *or* new moon, the Sun’s tidal bulge aligns with the Moon’s tidal bulge, adding to the total height. That’s a spring tide. A quarter of the way around the globe, where all that water came from, a neap tide is happening.

Next Week in Sky Lights ⇒ Barometric Air Pressure

Hey Dan, thanks for the explanation of the inverse square on the Fg from the sun. I always thought that the pull from the sun and the moon were almost equal (because of the sun’s larger mass). That’s why I thought a new moon tide would be the largest. Thanks Again -B-

Glad I could help straighten you out on that. It was a good question, and though I usually try to avoid equations, I really needed the “inverse square” concept to answer it. FYI, if the Sun had 33 times as much mass, then its NET gravitational pull on Earth would be the same as the Moon’s. But even then, the tidal bulge raised by the Sun would still be less than that of the Moon. Here again, it’s the inverse square law in play. The near side of Earth is only about 0.005% closer to the Sun than Earth’s far side. As a result, the Sun pulls on all parts of the Earth with essentially equal force, and you just don’t get much tidal distortion. For the (much closer) Moon, that distance difference is 1.7%. As a result, the Moon pulls 2.9% harder on the near side than the far side, and you DO get a significant tidal bulge.