Given that Earth has a much stronger gravitational pull than the Moon, how does the Moon have any influence on Earth's oceans?
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14$\begingroup$ To the downvoters: Just because a question appears to be naive does not mean that the question is bad and is worthy of a downvote. $\endgroup$– David HammenAug 15, 2019 at 15:21
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1$\begingroup$ I guess the downvotes are because this question shows no attempt to find an answer to this problem. $\endgroup$– JiKAug 16, 2019 at 23:09
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1$\begingroup$ Sorry. I'm new here and I guess I thought that a question was a legitimate way of finding an answer. $\endgroup$– SteveAug 17, 2019 at 8:52
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$\begingroup$ Welcome, Steve. What JiK is getting at is that on Stack Exchange sites people are expected to do some prior research before posting a question, as explained in How to Ask. So it's legitimate to downvote questions which show no prior research. $\endgroup$– PM 2RingAug 17, 2019 at 10:28
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1$\begingroup$ Given that you earn a salary from your job, how does the Christmas cheque from your great aunt have any affect on your bank balance? $\endgroup$– David RicherbyAug 17, 2019 at 12:46
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2 Answers
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Everything in the universe has a gravitational influence on everything else in the universe. It isn't a question of the strongest gravitational pull winning out and all the others doing nothing.
The Earth is the strongest pull on the oceans, but the Moon and the Sun both have easily measurable effect in addition to the Earth's. Other bodies (Venus, Jupiter, a small asteroid in another galaxy,....) all have much smaller effects which will be hard or impossible to detect amidst the noise due to waves and so on.
answered Aug 15, 2019 at 12:23
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The following diagram from the wikipedia article on the tidal force shows the tidal force that results from a moon.
Note that the tidal force is directed away from the center of the planet when the moon (satellite) is directly overhead or underfoot but is directed toward the center of the planet when the moon is on the horizon. You are right that these very small influences. The tiny changes in the vertical component of the tidal force from the Moon on the Earth have very little effect on the oceans.
What does matter are those places where the angle between the line segment from the center of the planet to the moon and the line segment from the center of the planet to a point on the surface is approximately 45° or 135°. The tidal force is purely horizontal in those places. Minuscule as that tidal forcing is, this horizontal component of the tidal forcing function is unopposed by gravitation the Earth itself. This horizontal forcing makes the waters "want" to flow sideways.
The direction of this flow changes constantly due to the Earth's rotation. The Coriolis effect comes into play precisely because the Earth is rotating. The shapes of the oceanic basins and continental margins also come into play. The end result is a set of amphidromic systems, each of which involves large scale oceanic waves that rotate about amphidromic points.
answered Aug 15, 2019 at 15:15
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3$\begingroup$ "Minuscule as that tidal forcing is": how minuscule is it? How does its magnitude compare to that of the coriolis force? $\endgroup$– phoogAug 16, 2019 at 13:42
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$\begingroup$ The maximum vertical acceleration occurs when the Moon is directly overhead or underfoot and is about a tenth of a microg (or about 100 nanog). The maximum horizontal is smaller yet, about 3/4 of that. So very small indeed. The Coriolis acceleration is also rather small. The Coriolis effect tends to turn the shallow waves generated by the tidal forces sideways. The horizontal component of the Coriolis effect at the equator is null. (Once again, it's only the horizontal component that matters with regard to a liquid.) It's quite significant near the poles. (continued) $\endgroup$ Aug 16, 2019 at 23:11
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1$\begingroup$ This is why equatorial equatorial amphidromic systems are much larger than are polar ones. This tension between the tidal forcing functions and the Coriolis effects are the primary reason that the so-called tidal bulges do not and cannot exist. $\endgroup$ Aug 16, 2019 at 23:17