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Feng and Gallo have several articles in refereed journals including "Modeling the Newtonian Dynamics for Rotation Curve Analysis of Thin-Disk Galaxies" in Research in Astronomy and Astrophysics, "Galactic Rotation Described by a Thin-Disk Gravitational Model Without Dark Matter" in the Journal of Cosmology, "Deficient Reasoning for Dark Matter in Galaxies" ...

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As noted by Conrad Turner, the approximate shape of the Earth is an oblate spheroid, though it is so close to spherical that you would be pretty hard-pressed to see the difference without precise scientific equipment. This web page has some pretty good information on this topic that you might be interested in. When we talk about this shape, we are generally ...

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The form of the Earth is a approximately (to a very good approximation) an oblate spheroid (difference between the polar diameter and equatorial diameter is ~20km which is ~1/300 of the mean diameter). The shape is distorted by tides (both in the solid body as well as the sea) and by topography. Topographical variation is about +10 to -10km (height of ...

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There are a number of misconceptions with this question. First off, the "gravitational pull" (which I'm interpreting as gravitational force) by the Milky Way on the Earth is seven orders of magnitude smaller than that exerted by the Sun. Secondly, gravitational force is the wrong metric. The Newtonian gravitational force exerted by the Sun on the Moon is ...

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The so called inverse square law makes gravity dominate locally. The Moon orbits Earth, Earth orbits the Sun, and the Sun orbits the Galactic center! Because Earth is about 500 times nearer to the Moon than the Sun is, it is Earth's gravitational field which dominates the orbit of the Moon. And it is not a single thing which attracts mass to orbit the ...

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The tide would be locked in place, roughly, because the Moon always shows the same side to the Earth (tidally locked). But the height of the water at the location facing the Earth and the opposite side should be greater than the mean height, and at the limb, as we see it, the height will be lower than the mean but by half the magnitude of the high tide. ...

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A much easier way to think about this is to consider clouds in the sky. They contain hundreds to thousands of gallons of water molecules but they are very spread out. Same as a hydrogen cloud before it creates a star. When you compact molecules in a tighter space you get a stronger magnetic field in the near vicinity of the object. The amount of molecules in ...

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For bodies with masses smaller that 1 times the Earth mass, surface gravity grows as the square root of the mass. For exoplanets beyond 100 times the Earth mass, surface gravity grows roughly linear with the mass, implying that the radius grows very little. But just between 1 and 100 earth masses, the surface gravity is almost independent of the mass, it ...

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There would be a tide of approximately 28 days due to centrifuge effect,the water being free flowing,about the mid point centre of gravity of the 2 masses. Our tides are monthly but appear daily due to earth 24hr rotation.Variation in sub-marine land masses and ocean baison cross flow have sub tidal effect that can lag or lead the moon phased tide.On a water ...

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In a close approach, Earth would feel tidal forces: the near side to the approach would be affected by the vistor's gravity more than the far side, because the force of gravity decreases with distance from a mass. If you were standing on the far side, though, your effective weight wouldn't increase -- it would decrease! This is because the visitor would be ...

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The Earth wouldn't simply split in half, it would be pulled towards the asteroid as a whole. To simplify matters, let's assume that the mass is equal to twice that of Earth (size alone is not enough, you must also consider density). You can then use Newton's law of gravitational attraction for the calculations: $$F=G \frac{m_1m_2}{r^2}$$ $G$ is a constant, ...

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As a hint: a force directed inward does not change the angular momentum, and therefore does not change the semi-major axis nor the period. It's not clear which angle you're talking about, but knowing that the semi-major axis did not change should let you set the problem up as a simple geometry problem.

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