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Consider an arbitrary point P and a point close to it, let's say A. $\underline{g}_P = -\frac{GM}{r^2}$ Whereas $\underline{g}_A = -\frac{GM}{(r+a)^2}$ Taylor expansion on $\underline{g}_A = -\frac{GM}{(r)^2}(1-\frac{2a}{r} +...)$ Finding the difference between $\underline{g}_P$ and $\underline{g}_A$ yields $\frac{2aGM}{r^3}$ note that this answer ...


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You probably got voted down cause this can easily be google searched, but the simplest way to explain it is that a tide happens because the lunar tug on one side of the ocean is measurably more than on the other side of the ocean and as the earth rotates the tidal "bump" follows the moon so you get 2 high tides and 2 low tides a day. A tide is effectively ...


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Analogous to the tides of Earths oceans, do the Moon and Sun cause our atmosphere to bulge in what could be described as a low and high tide? The answer is yes, if you generalize beyond gravitation. Sunlight heats the atmosphere, and this causes atmospheric tides. The two dominant effects are absorption of visible and near infrared sunlight by water ...


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The differential force of gravity on the atmosphere works the same as it does for the rest of the earth (the oceans etc). However, moving the equipotential surface by a few m will be almost undetectable on the atmosphere, since the density of the atmosphere decreases so gradually – over many km. Contrast this with the surface of the ocean, which ...


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I'm going to answer the question of your title, and also address the curious statement that "tidal gravity=real gravity". Let's begin with your statement: Tidal gravity, by my understanding, is the difference in gravity between two points. You're very much on the right track here. When people talk of "tidal effects" and "tidal gravity" when not ...


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If the moon is on the opposite side of the planet while the sun is over our heads, I would be under the impression the moon pulling the atmosphere towards it would allow warmer days or cooler nights as the atmosphere is thinner during this time.


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Yes, though the effect is greater on earth-bound or terrestrial tracking stations than on spaceborne or orbiting satellites. See Table 2.9 ("Perturbing accelerations acting on a GNSS satellite") and sec. 10.1.2 ("Site Displacement Modeling: Solid Earth Tides, Pole Tides, and Permanent Tides") in the BERNESE Software Manual


2

There is a lot more to it than just astronomy. For example, the tide times inside Boston Harbor are significantly different from those on the southeast coast of Cape Cod. It is true that the primary force behind tides is the position of the Moon, but the macro tidal bulges take a long time to propagate around/across oceans, and then the shoreline shape ...


1

(In this post I use bold face to denote triples of real numbers. That is, points in 3D space, with ${\bf x}=(x,y,z)$) With a moon mass $m_m$ position ${\bf x}_m$, the vector force on your surface element mass $dm$ position $\bf x$, is $$G (dm) m_m\frac{ {\bf x}_m-{\bf x}}{\| {\bf x}_m-{\bf x}\|^3}$$ so it accelerates as the following if you ignore the ...


1

The effect of earth's gravity is not disregarded. It is what keeps the earth in one piece and approximately spherical. This is the dominant effect in the the earth's vicinity. Without it the sun and the moon would most likely stretch and tear it apart, as would earth's own rotation. Tides are a tiny perturbation on top of this effect caused primarily by the ...


1

Let me summarize the different accelerations here: Acceleration of Earth's gravity. This is very important, it points "down" in your coordinates no matter where you are on the Earth, and it keeps the oceans on the planet, so that they roughly describe a round distribution around the Earth. Acceleration of the Moon's gravity on the center of the Earth. This ...


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But isn't there a difference in Earth's gravity between antipodal points of the earth (opposite directions)? I don't follow your statement above, but my sincere apologies if I have misunderstood your question and I will be happy to delete this answer if it does not work out for you. I think you may be mixing up the Earth's tidal force on the Moon, ...


3

I agree with Walter, they don't. However, in addition to the common envelope drag and mass exchange a very important feature of their evolution is the loss of angular momentum through a magnetised wind. The loss of angular momentum from the orbit also leads to orbital shrinkage and closer contact, until presumabaly at some point they truly merge. This ...


3

Short answer: They don't. Contact binaries are a possible evolutionary phase of close binary systems, where both stars fill and/or overflow their Roche limit, i.e. 'kiss' each other at the L$_1$ point. As a consequence, they lose matter to a common envelope. This exerts a drag onto the stars and leads to further shrinking of the binary, which eventually ...



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