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I am not sure about the derivation in this video although it might be correct? My doubt is with the equation for the acceleration $a$ for the inner phase of the motion where $R$ is the distance from the centre of the Earth and the radius of the Earth is $R_\oplus = 6.37 \times 10^6$ m. $$\ddot{R}(t) = a(t)=-36.36\dfrac{R}{ R_\oplus}$$ During the ...


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Sounds like GPS is best here. The position of the antenna is determined first in coordinates relative to the center of the earth, and then translated onto the ellipsoid and geoid. So finding the position relative to the center of the earth should actually be more accurate than finding its altitude relative to sea level. (Survey-grade can be centimeter ...


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Clarifications There is a difference between a solar flare and a the phenomena that cause things like the 1989 Quebec blackout and/or the 1970s New York blackout (I think it was 1972 but do not recall off hand). The latter phenomena are called coronal mass ejections (CMEs) because they actually involve large amounts (i.e., upwards of billions of tons) of ...


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Just to confirm that CuriousOne is correct as regards direct radiation damage to your body. From Solar Flares Solar flares are gigantic explosions associated with sunspots, caused by the sudden release of energy from “twists” in the sun’s magnetic field. They are intense bursts of radiation that can last for anywhere from minutes to hours. Solar flares ...


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I'm not quite sure how he gets a value of $36.36$ for that constant, because it depends a bit on exactly how he does his approximations and what values he assumes for different constants. But basically it's this: For a spherically symmetric mass distribution like the one considered here, the gravitational acceleration is simply, from Newton's law for the ...


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Yes the Schwarzschild metric describes the spacetime geometry around the Earth, and I describe how to use the geodesic equation to describe objects falling in Earth's gravity in How does "curved space" explain gravitational attraction?. An example of how the Schwarzschild metric describes the Earth's gravitational field is the time dilation of GPS ...


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I agree with Lubos Motl's answer. Geologist indeed consider the magnetic property of molten iron i.e. paramagnetism and ferromagnetism. Iron is a ferromagnetic material. It has a greater degree of magnetism when compared to dia- or paramagnetic material. Now, there is a temperature called Curie's temperature when a ferromagnetic material changes to a ...


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Dynamo Effect : The dynamo effect is a geophysical theory that explains the origin of the Earth's main magnetic field in terms of a self-exciting (or self-sustaining) dynamo. In this dynamo mechanism, fluid motion in the Earth's outer core moves conducting material (liquid iron) across an already existing, weak magnetic field and generates an ...


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The magma has temperature between 700 and 1300 Celsius degrees. The Curie temperature of iron is at 770 degrees Celsius. Above that temperature, iron loses magnetism. Note that right above 770 °C, iron is still solid because the melting point is around 1500 °C. So magma almost never can be magnetic because it's just too hot for that. Incidentally, if it ...


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The Sun pulls on the Earth as well. So both Earth and Moon are "falling towards the Sun" all the time, just as they are moving in almost the same orbit. Earth causes the orbit of the Moon to "wobble" a little bit. If you were simply given the coordinates of the Moon as it moves around the Sun, you would notice there is a deviation from the expected ellipse ...


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Another way is to use fluctuations caused by solar flare coronal masses impacting on the Earth's magnetosphere, which give rise to magnetic storms. These can induce large currents in long conductors such as power grids. However, they are far more destructive than useful. On a more practical note, if you could turn the magnetic field of the Earth into ...


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The surface of the Sun, as you can see from the following picture, is not uniform: But it is impossible to see it with your naked eye. You will need a telescope and specialized solar filters (more information here). Notice also that the picture is not showing visible light, but extreme-ultraviolet light. You will need a special telescope for that. Or you ...


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Short answer, the sun isn't on fire. Flames can flicker with wind or with pockets of fuel that might collect and burn in spurts or bits of water that can steam when the flame touches them, causing movement, or due to small bits of turbulence as the heat expands away from the flame. The visible surface of the sun is much more like a hot iron that has a ...


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I believe that despite the difficulties you'd have getting the atmosphere to rush into a tube, that would be the far easier of the two problems you'd have. The bigger problem is dealing with all that air. Where would it go if not back to earth? Since we're talking about science fiction, I propose the following technology to make it work. Suppose you ...


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The first version of the question asked about just sucking up the atmosphere with a hose, without manipulating gravity. I'll answer that first, since I think it's a very good question that many — including my previous self — get wrong. No manipulation of gravity They did this with a giant vacuum cleaner in the movie Spaceballs. But as knzhou comments, it ...


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The formula is $$v = \varphi\cdot r/t$$ Where v is the velocity, φ the traversed angle in radians, r the distance and t the time, if the object is moving in a transversal direction. If it is approaching or receding you have to split the radial and transversal components with Pythagoras. In your example I get around 1140 ft/sec.


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You sound disappointed that the earth hasn't yet collided with the sun. Applying random formulas to a physical situation is never a good idea. I suspect that the formula you are having issues with is for two bodies in space which are not in orbit around one another, as unfortunately the earth is with the sun. When one body orbits another, the motion of ...


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The answer you refer to is to a question where both bodies are initially motionless. That's not the case for the Earth/Sun system. The Earth orbits the Sun. The gravitational force the Sun exerts on the Earth provides the centripetal force needed to keep the Earth in a stable orbit. In a sense the Earth is constantly free-falling towards the Sun but ...


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As stated in the answer you link, that formula is for 2 bodies starting from rest. The Earth and Sun are not at rest relative to each other, they are in orbit at a relative tangential speed of nearly $30\,{\rm km}\,{\rm s}^{-1}$.


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I think the two big factors would be that the Earth would 'want to' become tidally-locked to the Moon and the Sun. The Moon would win here, which is easy to see because tides are caused more strongly by the Moon than by the Sun. So in due course the Earth would end up tidally-locked to the Moon with a rotation period which would be the same as a lunar ...


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In an isotropic space (which means it's equal in all directions), Noether's theorem tells us that angular momentum is conserved. For the Earth this means that it keeps the angular momentum that it has now. This can be calculated from $$L = I \omega$$ where $I$ is the moment of inertia and $\omega$ the angular speed. You'll see that if $L$ remains constant ...



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