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Why are there high tides twice a day when the earth rotates beneath the moon only once a day? In diagrams it appears the moon's gravity causes the earth's oceans to bulge (creating a high tide) not only on the side toward the moon, but also on the side away from the moon. I've heard some unconvincing explanations for this, including: "the water on the far side is flung away from the earth" (why?); "the moon attracts the earth, and the water on the far side is left behind" (why isn't the water on the far side attracted too?); and "the earth and the moon both revolve around a common point" (I know that, but what does that have to do with the question?). Please help. --Kathleen Hunt, Brookline, Massachusetts

Not to discourage you, Kathleen honey, but this makes 22 questions from you in three months. Think quality, not quantity. This ain't the Lotto, you know.

The explanations you've heard for the tides are technically correct; obviously what's missing is a homely metaphor to bring the point home. Try this one: the earth and moon, which are really dual planets, are like two figure skaters spinning around one another while holding hands. Centrifugal force naturally tends to pull them apart, but their clasped hands (i.e., their mutual gravitational attraction) keep them together. Similarly, centrifugal force tends to fling the ocean outward on the side of the earth away from the moon. On the near side, the water is tugged moonward by lunar gravity.

So why doesn't centrifugal force act on the near side, or gravity on the far side? Because the force of gravity drops off rapidly with distance, and because the side of the earth facing the moon is near the center of common rotation, where centrifugal force is minimal. (Actually, owing to the disparity in size, the center of common rotation is inside the earth, but let's not confuse the issue.) The result is that the oceans bulge on both the near and far sides, and you get high tides twice a day. You following all this? Fine. Hope it holds you for a while.

My high school physics teacher gave us this problem once, but I forget what the answer was. Suppose you've got a bullet in one hand and a pistol in the other, aimed so it's perfectly level. You drop the bullet and fire the pistol at the same time. Which bullet hits the ground first? --L., Indianapolis

Look, I know this is a lot of physics for one day, but this one is so twisted you gotta love it. The average mope reasons like this: the dropped bullet falls only a few feet, whereas the fired bullet travels hundreds of yards. Ergo, the dropped bullet hits the ground first. The average mope with a college education (e.g., a physics teacher) is a little more sophisticated. He figures, hey, the force of the gun propels the fired bullet strictly horizontally. The only downward force is gravity, which acts equally on both bullets. Therefore they both hit the ground at the same time.

Then we have the answer given by those who have achieved spiritual perfection as a result of regular reading of the Straight Dope. This may be summarized as follows: it depends. If the fired bullet travels only a short distance, then yes, both bullets hit the ground at the same time. However, if the fired bullet travels far enough, the earth, being round, curves away from it. (Remember Newton's first law of motion: moving objects tend to travel in a straight line.) Since the fired bullet has farther to fall, it takes longer to hit the earth, so the dropped bullet hits the ground first.

What's more, if the fired bullet travels fast enough (seven miles per second, to be precise--a practical impossibility, but never mind that), it reaches escape velocity and never hits the ground at all. Instead, it goes into orbit around the earth. Amazing, no? Try this one out in your next physics class and you'll kill the whole hour, guaranteed.

Art accompanying story in printed newspaper (not available in this archive): illustration/Cecil Adams.

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