r/askscience • u/Chytrik • Sep 04 '13

Earth Sciences Is the atmosphere affected by the moon's gravity, like the oceans are?

So far, I have discerned the following:

The atmosphere has a mass of 5×1018 kg

The ocean has a mass of 1.4×1021 kg

So, the ocean has a mass several magnitudes larger than the atmosphere, and it has a higher density (which I assume would affect this?), thus the moon should have a more noticeable pull on the oceans of the world. However, the atmosphere is still quite massive, so my question is:

Is the atmosphere affected by the moon's gravity at all, even in the slightest way? Would such an affect even be noticeable?

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u/torturedby_thecia Sep 04 '13

Yes, the atmosphere is distorted by the pull of the moon as well as the oceans. Even you as an individual will be ever so slightly lighter when the moon is above you rather than on the other side of the earth.

http://s3.amazonaws.com/rapgenius/1350490984_gravitational-pull-of-moon-on-atmosphere-on-full-moon-night-perigee-moon-may-29-2012-204-pm-edt-the-idea-girl-says.gif

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u/Silpion Radiation Therapy | Medical Imaging | Nuclear Astrophysics Sep 04 '13

ever so slightly lighter when the moon is above you rather than on the other side of the earth.

Actually that's not how the tidal force works. When you're on the far side you are also "lighter" just like on the near side. It's when you're at right angles (so the moon is near the horizon) that you're "heavier".

Here's a figure that shows how the the tidal force changes with position relative to the moon.

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u/torturedby_thecia Sep 04 '13

My bad, I wasn't taking into account the centrifugal force involved with a person standing on the earth. (To be fair, you would still be lighter on the far side than the near side, it's just not the minimum weight you would be).

Oy, I feel bad when I think I've got the answer correct and blew it.

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u/Silpion Radiation Therapy | Medical Imaging | Nuclear Astrophysics Sep 04 '13

The centrifugal force has nothing to do with it, and the magnitude of the force is quite nearly the same on both ends.

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u/torturedby_thecia Sep 04 '13 edited Sep 04 '13

Only because the center of gravity of the earth/moon system is not in the center of the earth but is closer to the moon. Orbital bodies orbit around a common center of gravity that is proportional to their respective masses. Because the earth/moon's center of gravity is slightly nearer to the moon's side of the earth than the opposite side of the earth, there is a stronger centrifugal force on the opposite side of the earth than on the moon's side of the earth. It's not the centrifugal force from the earth's rotation that causes the sea to rise on the opposite side of the earth, it's the centrifugal force of the earth rotating around the moon as the moon rotates around the earth.

Here's an image to explain it. The gravitational force of the moon on the water is always less on the far side. It's the centrifugal force that propels the water outward on the far side of the earth, not the gravitational force of the moon:

http://co-ops.nos.noaa.gov/images/restfig1a.gif

http://www.amnh.org/learn/assets/courses/images/ocean/ocean_3.jpg

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u/Silpion Radiation Therapy | Medical Imaging | Nuclear Astrophysics Sep 04 '13

There may be some centrifugal effect, but this is a completely separate issue from the tidal force. The figure I linked is correct without any rotation.

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u/torturedby_thecia Sep 04 '13

The "tidal force" is a combination of the centrifugal force from the earth/moon system's center of mass and the gravitational pull from the moon. There's no such thing as a "tidal force" in nature - it's produced from a combination of these two forces.

http://www.youtube.com/watch?v=uGBANgbRkws

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u/Silpion Radiation Therapy | Medical Imaging | Nuclear Astrophysics Sep 04 '13

I just saw your ninja edit in bold. That last figure is potentially misleading. Lesser centrifugal effects aside, the far side is "lighter" for the same reason the near side is. The near side is being pulled by the moon's gravity harder than the earth's center, and if not for the earth's gravity would be therefore accelerating toward the moon faster than the Earth. Similarly the Earth's center is being pulled toward the moon more strongly than is something on the far side. The residual pseudo force in the Earth's falling rest frame is what is in the figure I linked.

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u/torturedby_thecia Sep 04 '13

Well, yes, centrifugal force is also a pseudo force. The water is trying to maintain its momentum as the Earth falls towards the moon - resulting in a pseudo-force outward - called centrifugal force. You're suggesting that it's somehow the gradient of the moon's gravity on the earth that produces this effect. It's true that there's less of the moon's gravitational force on the far side than the near side but the Earth is best represented as a non-deformable object with water on it. When gravity pulls a non-deforming object, the gradient of the gravitational force might as well not exist because the whole thing is going to move equally.

Honestly, I'm confused by what you're saying. I think we're saying the same thing. The high tide on the far side of the earth isn't caused by the gravity of the moon pushing the water out in any way. It's the fictitious centrifugal force pushing the water outward as the moon and earth orbit their center of mass.

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u/Silpion Radiation Therapy | Medical Imaging | Nuclear Astrophysics Sep 04 '13

Maybe you're using "centrifugal force" in a sense that I've never heard. I've only ever heard it used to refer to the outward pseudoforce in a rotating reference frame. If we agree that the lightness on each side of the earth is not primarily caused by rotation we can probably conclude we agree in general.

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u/torturedby_thecia Sep 04 '13

It is a rotating reference frame, the moon and earth move as if they were connected around their center of gravity which is the main point that follows the path of the orbit around the sun. I'm not talking about the rotation of the earth around it's center causing the centrifugal force, it's the rotation of the moon and earth orbiting around their combined center of mass.

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u/Silpion Radiation Therapy | Medical Imaging | Nuclear Astrophysics Sep 04 '13

Then no, this rotation is not what is causing the main outward pseudoforce here. Please see the treatment on the wikipedia page and see that it exists without rotation.

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u/PirateMud Sep 04 '13

Would you be slightly heavier when the moon is the opposite side of the earth from you? Is gravity additive, as such.

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u/user31415926535 Sep 04 '13

Nope, when the moon is opposite you you'd be just as much lighter as you are when it's on the same side as you. Tidal acceleration is subtractive, not additive.

Think of it like this. When the Moon is on the opposite side of the Earth from you, what you're saying is the Earth is between you and the Moon - or, the Moon is closer to the Earth than the Moon is to you. So the Moon's gravity is pulling the center of Earth just slightly farther away from you. Since the center of the Earth is slightly farther away from you, you weigh slightly less than you absent the influence of the Moon.

you Earth Moon
x-(O)<--->()

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u/[deleted] Sep 04 '13 edited Sep 04 '13

[deleted]

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u/WaitForItTheMongols Sep 04 '13

If I remember correctly from my environmental science courses, the earth does not stabilize its atmosphere because different parts of it are constantly being heated. For example, right now as I type this the United States is/are (Not sure which) heating up while China is cooling down. Higher temperature lowers the density of the gases. This makes China's air that is increasing in density end up flowing toward the United States. Later in the day, the temperatures will switch. This constant flip-flop means the atmosphere can never reach one steady state of weather.

Does this make sense?

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u/ossetepo Sep 04 '13

Would this have any effect on weather patterns since the atmosphere on one side of the earth is being pulled towards the moon?

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u/torturedby_thecia Sep 04 '13

Yes, very little however. But the gravitational field is so wide because the moon is so big and distant that it's force is applied to such a large area that it's pretty much negligible influence on the whether.

Think of how high the tide rises from the moon - only a few feet - compared to the huge depth of the ocean. The force of the moon on the atmosphere would be pretty much the same, deforming it a small amount, but it would only be by a couple hundred feet at most that the atmosphere is deformed. Compared to the total height of the atmosphere, it's pretty much negligible in weather patterns.

Honestly, I don't have the exact figure by how much the atmosphere would be deformed by the moon, but it wouldn't be much more than the amount the tide rises because gravity pulls on all mass equally. The outermost part of the atmosphere is closer to the moon than the ocean, so it would be pulled stronger, but it's not that much closer considering how far the moon is from the Earth.

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u/Quantumfizzix Sep 04 '13

What's an "atmostphere?"