Eli5: how does the moon affect low and hightide of the ocean which is pretty heavy but does not cause other things which are lighter to start floating?

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Eli5: how does the moon affect low and hightide of the ocean which is pretty heavy but does not cause other things which are lighter to start floating?

In: Physics

6 Answers

Anonymous 0 Comments

Google has all the answers, so here’s what I found. This is not my answer:

Everything on earth feels the gravity of the moon. Because the force varies with the square of the distance, it is relatively tiny on any tiny portion of the earth – like your balloon. The moon is approximately 60 earth radii away from Earth. But a tiny force applied everywhere is huge. As some have pointed out, the force is proportional to the mass of both objects. When one of said objects is all the free-standing water in the oceans of the world, that is a tremendous combined force and has tremendous effects. Because the water is fluid and relatively free to move, it reacts in aggregate to the attraction from the moon, and follows it. If we didn’t have large land masses, the bulge in the ocean that is the tide would simply follow the moon around the world without interruption. Instead, it sloshes back and forth in the big pools that are the ocean basins.

Your tethered balloon (do we assume that you meant helium-filled, and therefore floating? or did you mean air-filled and therefore hanging from its string? — I will arbitrarily assume that you meant helium-filled) does indeed feel a pull from the moon, no matter where the moon is in its orbit. And by the way, the moon is overhead of a given patch of earth during the night, only for part of the month. Another part of the month, the moon is visible in the sky during daytime. During the other two quarters of its orbit, the visible moon straddles night and morning, or afternoon and night. So the direction of pull is constantly changing throughout the day and night, and throughout the month.

BUT, due to the tiny mass of the balloon (less, even, than the same volume of air molecules), OTHER effects are much more detectable. For example, you specified that it was tethered to your roof. That means it is exposed to every outdoor breeze. That effect would completely drown out any observation you could make of the moon’s gravitational pull on that balloon.

Even if you got strangely lucky and the air was dead calm for as long as you cared to perform your measurements, the balloon and the land and buildings around it are still exposed to the sun. The balloon would expand in the sun’s warmth and would pull upward more strongly while it was being heated, which would swamp the minuscule attraction of the distant moon on such a low-mass object as that balloon. Then, the sun would set and the region would grow colder, and the balloon would shrink and sag. This change, too, would overwhelm any signal you could detect from lunar gravity. Just the heat of your body, standing near the balloon would be enough to set up convection currents in the air near the balloon.

[Source for the original author](https://www.quora.com/Why-doesnt-the-gravity-of-the-Moon-affect-anything-else-on-Earth-besides-the-tides-If-it-is-strong-enough-to-affect-oceans-shouldnt-it-pull-other-things-too-For-example-if-I-tie-a-balloon-on-the-roof-would-it-float-a-bit-higher-at-night)

Anonymous 0 Comments

Small object are attracted by moon but the affect is too small so the movement isn’t visible, there isn’t gravitational force between them. However due to the combined mass of water in ocean enough gravitational force is created that can be visible.

That’s why small water bodies like rivers, lakes don’t have tides.

Anonymous 0 Comments

I suppose if you had a few day old party balloon that was nearly neutral you might have some luck getting it to float at the time of a high tide. The effect the moon has is very very small, much less than one percent on measured weight. The reason the tide is observable is there is so stinking much ocean that a tiny fraction of it is still a lot of water.

Anonymous 0 Comments

Gravity pulls on all mass the same. A kilogram will experience the same force if it’s the same distance away no matter what it is a part of. A quadrillion kilograms of ocean will therefore feel a quadrillion times the force that a kilogram dumbbell will feel.

The moon pulls ever so slightly harder on one side of Earth than the other. It’s a difference in force of (by my math) something like 0.1%. This tiny difference causes a little bit of tidal force on every atom of Earth. This is slowly causing us to spin less. Since water is free to move, that small amount of force adds up and can move huge amounts of water.

Anonymous 0 Comments

The moon does not “suck up” the water. It’s the centrifugal force of the earth causing that. The moon only disturbs it. If the moons gravitation really would rise the water, you would only have one high-tide wave every 24 hours, not two of them every 12 hours. The second high-tide is *opposite* of the moon.

Anonymous 0 Comments

In essence, the top layer of water on the ocean is squeegeed by the force of gravity. It isn’t *lifted* like you were taught in school, that is asinine.

[https://rmpbs.pbslearningmedia.org/resource/what-physics-teachers-pbs-space-time/what-physics-teachers-pbs-space-time/](https://rmpbs.pbslearningmedia.org/resource/what-physics-teachers-pbs-space-time/what-physics-teachers-pbs-space-time/)

[https://www.youtube.com/watch?v=pwChk4S99i4](https://www.youtube.com/watch?v=pwChk4S99i4)

This (both the same but one presented by PBS instead of just youtube) explanation is probably the best one I have seen. There is a tidal force operating on any and all water, but you don’t notice it because the you just don’t have enough water for you to see it happening. Get enough water, like an ocean or a great lake, and the tidal effect is evident.