# Tag Info

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Half. The escape velocity for an object at a distance $D$ from an object of mass $M$ is $\sqrt{2GM/D}$. The circular orbital velocity (the Moon is on an orbit that's close enough to circular that I'll just assume this) at the same distance is $\sqrt{GM/D}$. Setting the escape velocity from the Earth with it's new reduced mass $M_{\rm new}$ equal to the ...

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Assuming that we are making a certain amount of the earth's mass disappear instantly somehow, doing this will decrease the escape velocity of the earth. The moon is already moving at its orbital velocity in an orbit centered on the current center of mass of the earth-moon system. If we took away so much mass (instantaneously, somehow) that the escape ...

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No. One way of knowing this is by measuring the solar constant (which is not a constant in the sense that physicists usually mean) which is the total power from the Sun at 1 AU. There have been good measurements of this for a while: for the last three sunspot cycles the variation has been around a tenth of a percent (Wikipedia is a good starting point and ...

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What may be confusing you is that the passage you quote about the Space Shuttle is talking about speed relative to a fixed frame of reference: one fixed relative to the distant stars. On the other hand, when you think about aircraft flying through the air (or people walking along the ground), you think about a co-rotating frame of reference: fixed relative ...

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There is a slight effect on the lift requirement for an aircraft. A "stationary" object on the equator is actually traveling in a circle, at one earth-circumference per sidereal day velocity. That lessens its apparent weight because it is accelerating toward the center of the earth (centripetal acceleration). The effective gravitational constant (g minus ...

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Actually it's not like that when Earth is rotating all the objects in it's gravitational influence will also move with the same angular velocity $w$ so even if you are at certain height you are still stuck with your initial point but you can contradict it simply by using another external force which can cancel out your gravitational pull or you can go up to ...

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This actually contains two questions. 1) The sun seems brighter (more dazzling) if there is more scattering in the atmosphere. The sun would actually look very small to us in the sky if there were no atmosphere (it's the same angular size as the moon) and most of the brightness seen in the direction of the sun is from small deflection rayleigh scattering. ...

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The simplest answer is that the Earth is even MORE massive. Don't let Earth's lack of stature relative to the Jovian systems fool you. The gas giants are truly massive but are not nor do they have physical land and water at their surface. This may seem laughable on its face as what could be more massive than a gaseous body in a liquid state? Nay, veerily....

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Phenomena is called Cold Welding The reason for this unexpected behavior is that when the atoms in contact are all of the same kind, there is no way for the atoms to “know” that they are in different pieces of copper. When there are other atoms, in the oxides and greases and more complicated thin surface layers of contaminants in between, the ...

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It all depends on how clean the surface of the metal are. If there are oxides and impurities on the surfaces they will not bond together. However if the surfaces are ultra clean with no gas molecules from the atmosphere adhering to them, which on Earth would require them to be an ultra high vacuum chamber, then the surfaces would join together having made ...

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I would intuitively expect that the positive particles will go one way, and the negative particles (electrons) will go the opposite way, as per Lorentz force. The Lorentz force is acting on the particles in the standard way you would expect, but the magnetic field close to the Earth (i.e., within ~2-4 $R_{E}$) near the geographic equator is roughly ...

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@CuriousOne is approximately right. Others have thought of it, the url below is a paper from one who concludes with slightly more careful calculations (but still relatively straightforward) that the most one could get in terms of a continuous power from it is 50 MW, and typically a lot less. 50 MW is 5% of a large power plant. http://www.electrostatics.org/...

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First of all, let me correct a misconception. According to your question, your understanding of the structure of the Earth is that there's a solid crust of 5 to 50 km and below it there is liquid. This is not correct. The Earth is mostly solid. It's solid almost all the way down: you can drill down 3000 km without seeing any liquid at all. The liquid you do ...

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Lots of these answers expect you to know a lot about the way gravity works. From these, I've managed to come to the answer, and in this post I'll try to dumb it down to my level. To start with, I'd like you to imagine a tray, or an empty biscuit tin - something with a flat base and rims. Now imagine you pour water into this tray (feel free to actually pour ...

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I think you have started from a picture of the tides making water go up and down that is a natural picture to have, but it is wrong: they don't. If that picture were right, you ought to be able to measure the tides just by digging a deep well. What happens is that the influence of the moon (and sun) changes the shape of what is meant by "level" - ...

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Tides are the resultant of attraction from earth (major effect), moon (secondary) and sun. Other planets'influence can be neglected. The position of the moon (and of the sun) cause a slight variation in equipotential surfaces on the earth, and the water as a liquid naturally tends to follow an equipotential surface. Of course those variations are by ...

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incompressible: not able to be compressed. compressible: In thermodynamics and fluid mechanics, compressibility is a measure of the relative volume change of a fluid or solid as a response to a pressure (or mean stress) change. So water does not change its volume. The same volume of water can take many shapes. When no other gravitational force except ...

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