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Below the Schwinger limit photons do not scatter photons, so electromagmetic radiation cannot self compress laterally and cause a red shift. The electric fields due to electromagnetic waves just add linearly.


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Caution: I have not worked this out rigorously, and I'm not entirely convinced that it works (or even that the problem is solvable). Comments appreciated. The answers here all have the disadvantage that they fundamentally require large amounts of mass not in the same non-inertial frame to work properly. But what if there simply was no such mass? In other ...


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A little hard to do, but say.. You build a sun room on the surface of the sun. You tether the axle of a bicycle wheel sitting on the surface of the earth to your sun room on the surface of the sun. As the earth turns, the bicycle wheel turns. It stays in place as it is tethered to the sun. Now - you can extract energy from the turning of the bicycle ...


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Attach magnets, bring wires close, enjoy while it lasts. The changing magnetic field would induce currents in the wires, hence you could extract out energy from the rotating system.


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The fission hypothesis does not really explain why the moon rotates at the same rate as it's orbit of the Earth. The scenario of a external force like another planet or giant comet hiting the Earth and braking off a piece that created the moon would better explain the velocities of rotation. As they say the second hypothesis is not ruled out, but the first ...


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Yes, ultraviolet light photodissociates water to form hydrogen. Both $H$ and $H_2$ are formed. See The photodissociation of water in cometary atmospheres for the relative amounts and a full discussion.


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You can only know that a Foucalt pendulum demonstrates the rotation of the earth by way of astronomical observations - that is, it is by observing the motion of the stars that you can tell how long the day is. That being so, you can determine that the motion of a Foucalt pendulum corresponds to the position of a distant star, that is, it corresponds to a ...


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use a right side prism and place right ahead of eyepiece then it will correct! np


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Yes, indeed there is a correlation between warm regions of the CMB and galaxy clusters. The CMB fluctuations depends on what scale of perturbation that you look at. For large scale perturbations, the opposite to what we expect is true. On the large scales, an overdense region at the time of recombination results in a cold spot in the CMB map. This is ...


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inner radius of habitable zone (AU) $= \sqrt{\frac{L}{1.1}}$ outer radius of habitable zone (AU) $= \sqrt{\frac{L}{0.53}}$ where $L$ is absolute luminosity of the star. See Calculating the Habitable Zone for more information.


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Just because space is curved, it does not mean it is a subspace of a higher dimension. All space is curved, even euclidean space. Flatness is a relation between a curvature of some subspace and the space it is part of. There is the wonderful image of space expanding equally as one might blow up a balloon, but this image might be as easily explained by ...


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Solar noon becomes earlier and then back again more than once a year; it moves 18 minutes earlier from about Feb. 10th until May, then 12 minutes later by July , then back 23 minutes earlier by late October/early November, then forwards again by half an hour by Feb tenth. So the biggest change in Solar noon occurs between early November and mid February by ...


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Surprisingly it's quite easy to answer this because we can use the cosmic microwave background as a reference. The CMB gives us an average inertial frame for the universe so our motion relative to it is the closest we can come to defining the Solar System's motion through space. The CMB is isotropic, but because we are moving relative to it the radiation is ...


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$ L = \int \int {\bf F} \cdot d{\bf s}$ is where you should start, where $F$ is the flux in units of Watts/m$^2$. Blackbody flux is given by $\sigma T^4$ and hence an isotropic flux integrated over a sphere $$ L \int^{2\pi}_{0} \int^{\pi}_{0} \sigma T^4 r^2 \sin \theta\, d\theta\,d\phi = 4\pi r^2 \sigma T^4$$


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Orbits are pretty complicated. Most texts on this deal in terms of predicting positions at a specific time rather than just a simple ellipse, because that model while correct is too basic. As others mentioned position estimation makes it more complex because there are more parameters involved. I'll just pull some stuff for you on the basics. For standard ...


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Extensive googling hasn't helped me to understand how this observed earth rotation angle is used to compute UT1 As explained in: this lecture Earth's angle of rotation = 2π(0.7790572732640 + 1.00273781191135448(Julian UT1date - 2451545.0)) radians. So one observes Earth's angle of rotation and calculates the Julian UT1 date as a decimal date from that ...


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Sound as you hear it is waves of pressure differences in the air, which is interpreted by your ear as sound. So no, you cannot directly hear electromagnetic radiation (EMR). You could, however, take the EMR and convert it into sound waves in the audible range, which you could listen to. This was done in 1990 for Jupiter by Voyager as it passed Jupiter. It ...


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Electromagnetic radiation is light. Like any other star, you can see the radiation from a pulsar if it is not too far away or too dim. Pulsars and other stars do produce very loud sound. We do not hear it because there is vacuum between stars and us. Sound does not travel through a vacuum. Also stars are so very far away that we wouldn't hear it anyway.


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One perspective (heh) involves the following relation among position vectors: $$\vec{r}_{A\rightarrow C} = \vec{r}_{A\rightarrow B} + \vec{r}_{B\rightarrow C}.$$ These position vectors can be for anything; object $A$ could be a house, object $B$ an ant, and object $C$ a leaf on the river. Here's a diagram to help: So if you want to know the position of ...


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Yes, the Sun and our whole Solar System are revolving around the centre of our galaxy, the Milky Way. Milky Way is a spiral galaxy and hence has four major spiral arms and a central buldge. The Sun (and, of course, the rest of our solar system) is located near the Orion arm, between two major arms (Perseus and Sagittarius).


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The surface of the sun is where local plasma cools enough to recombine and go transparent, the photosphere. You would still be deep within the sun's atmosphere, and it would be LOUD. H-bombs are LOUD at the edge of their fireballs.


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helioseismology is what you need to learn about. yes, there are sound waves in Sun


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A good question for anyone who has thought about the big bang,red shift and Hubble etc. There are valid questions to be raised concerning Hubble's red shift leading to the big bang theory. Firstly, space is commonly regarded as a vacuum but this is not the case. Space contains a variety of matter ranging from atoms, to molecules, gases and solids and myriad ...


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Question: How many stars are in the Milky Way Galaxy? Answer: As many as there are! Why this answer? Answer: Even the stars that are close to us are eons older than when their light approximately began to journey toward all directions including Earth. Unknown & thus far unprovable amount of celestial activity continues from when ever it ...


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The Milky Way is receding from the members of the Hydra-Centaurus Supercluster. The Hydra cluster has a red shift of 0.0548. The Centaurus cluster has a red shift of 0.0114. The Norma cluster has a red shift of 0.0157. The local group is and will continue moving away from the Hydra-Centaurus Supercluster.


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There is a reasonable chance that yes, planets can form before the star "ignites" (which I take to mean the fusion of hydrogen into helium, not the very brief phase of deuterium burning which certainly will take place before planets can form). Planets form in a disk of circumstellar material around their parent protostars. The "core-accretion" model of ...


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If you include car lights and similar in what is turned off, then in a big city such as London, which has a radius of about 10 miles, there would be very little light pollution, instantaneously. However, there would still be a lot of atmospheric pollution, and so it wouldn't be possible to see stars as faint as can be seen from, say, the top of a mountain. ...


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No, your eyes would need to adjust to the decrease in light before you would see really well. Also, it is now very difficult to have zero light pollution. Mt Palomar experiences light pollution and it is at least 2 miles from light sources.


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Most of the brightest stars are spectral types B,A,F main sequence stars (50%), but there are also a bunch of O-type main sequence and giant stars (5%) and another big clump of red giants (about 35%) and a few percent are supergiants. There are no white dwarfs, and definitely no neutron stars! I have attached an image which I created by selecting 4992 ...


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Although this is not a direct answer to the question, it may help provide some background to the recent interest. The recent interest in the Mayan calendar, is because it is turning over a high-order digit, although because we have not properly aligned their calendar, it is variously 2005 and 2012 or something. The mayans apparently had a fairly exact ...


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I'm not sure what you are looking to know in your question. There are many problems with specifying a date/time centuries in the future and saying XXXX body will be at RA YYY, DEC ZZZZ. The Mayan calendar was primary designed to predict seasons much like most calendars. The basic elements of the Mayan calendar had little to do with astronomy and much like ...



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