New answers tagged

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This results in the dotted line (net force) drawn, which speeds the moon in its orbit. This actually causes it to move to a larger radius (per the comment by Logan R Kearsley). The problem with this statement as it stands is that if the moon were to be moved to an orbit with a larger radius and the same shape, it would actually slow down, since for ...


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Gravity acts on all matter, not just water (it just so happens that water flows with less resistance than rock) which is why we get noticeable water tides but not very noticeable earth tides. However, if you were to bring a very large gravitating body too close to earth, you would find that the earth isn't quite as solid as it feels. The answer to your ...


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First have a look at my answer to Black holes and positive/negative-energy particles for some background on Hawking radiation. The pairs of virtual particles analogy is just an analogy and not what actually happens. In fact virtual particles don't really exist in the way that real particles do - see this article by Matt Strassler for more on this. But your ...


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The gravitation of the stars in our galaxy keep the solar system in it. I'm not sure whether that's important for earth or for life on earth though, but it makes for nicer night skies. Gamma ray bursts, if close enough and (im)properly oriented, affect earth. A gamma ray event from a "soft gamma repeater", SGR 1900+14, is known to have affected earth's ...


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The distant stars are also responsible for cosmic rays, which in turn can affect the Earth's weather


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Machs principle, that inertia is caused by the distribution of distant stars was a principle that Einstein tried to incorporate into GR, but failed. However Barbour, quite recently incorporated an aspect of Machs principle into his theorising of time: ephemeris time An ephemeris gives the position of celestial bodies, and duration is deduced in terms of ...


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The stars in our galactic neighbourhood do have a dynamical, gravitational effect on the inner workings of the solar system: They built the Oort cloud The Oort cloud is a roughly spherical cloud of icy bodies that is thought to act as a reservoir of long-period comets (and which we speculate exists to explain said comets' existence). These icy bodies ...


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The other answers talk about some of the effects. This is a complementary answer that attempts to put a number to the force behind one of the effects - gravitational attraction. Proxima Centuri is the closest star to our solar system. It is about 4 × 1016 m away and has a mass of 2.45 × 1029 kg. The mass of Earth is about 5.97 × 1024 kg. Plugging these ...


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The amount of deflection of light (the bending of the null geodesic) passing a star is a deflection angle of 4m/R where m is the mass of the star and R is the radius of the star. The mass of the sun is 2 x 1030 kgs, where as the mass of the Earth is 6 x 1024 kilograms. The radius of the earth is 6 x 106 meters and the radius of the sun is 7 x 108 meters. ...


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The issue with massive waves on a 1 meter deep ocean is that the waves cannot propagate fast enough on a planetary size object. We get fast moving shallow tsunami waves in the open ocean over a thousand meters deep. The tsunami piles up when the wave slows down due to contact with a shallow shoreline. Hundred meter high waves could never propagate fast ...


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Gravitationally, there is little immediate effect on earth on a daily basis, though over very long periods of time, stars that pass near enough to the sun could disrupt the orbits of Oort cloud objects and send them towards the sun (and earth or other planets in our Solar System). Culturally, stars have a very big impact on our species. Religion, art, ...


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A lot (to put it mildly) of elements are created in stars and supernovae. These elements then travel through space until they fall to Earth (or, to be exact, some microscopic portion of them reach us). Earth itself wouldn't exist if stars hadn't generated elements which then clumped into dust, into minerals, and so on until a big ball of matter started to ...


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I don't think that light from the stars other than Sun is of much practical use nowadays except for the classic navigation, where it's essential of course. I guess any effect comes from the limitless reach of the gravitational force, which drops with the square of the distance but grows linearly with the mass exerting the force. A star most obviously ...


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This is how you do the calculation. The elapsed time on an observer's clock is called the proper time, $\tau$, and it is calculated by integrating the metric: $$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - \frac{dr^2}{1-\frac{2GM}{c^2r}} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 $$ In this case we'll assume all motion is radial so $d\theta = ...


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I'll reduce your question to its simplest expression: "What is mass?" And give you my best, simplest answer:"It is a measurement of how much an entity opposes acceleration or deceleration". I believe that in the end it all comes to that...


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Gravity and dark energy would have to be one & the same, due to the fact that gravity is spacetime and space time is basically the mother board or rather the rawness or better, the naked form of the universe. If einstein is correct about spacetime curvature than both are just different movements of the same entity, sort of like, how a river moves ...


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This is an answer to question 3 relating to simple motors and dynamos. A simple motor and a simple generator are one and the same thing When a current flows through the coil of a motor the coil rotates in a magnetic field and so the coil acts as a generator with the induced electromotive force (emf from Faraday’s Law) in the opposite direction to the ...


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It is not so clear what you mean by "infinite", but maybe you mean to ask why gravity has infinite range. It comes from the fact that the force of gravity decreases as $1/r^2$, thus slow enough for it to be felt by objects at the other end of the Universe. The same property is seen in the electrostatic Coulomb's force that decreases as $1/r^2$ as well. This ...


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Expansion A gas is a collection of atoms or molecules that are constantly moving and colliding with each other. Credit for this great visualization goes to Greg L at the English language Wikipedia. If the bounding box were to disappear, the particles would escape in the direction they were previously travelling - which for lots and lots of particles ...


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The state of a substance depends very much on its temperature. There can be a change of state due to expansion that is how gas liquefiers work. A gas is compressed and then is allowed to expands rapidly. In expanding the gas does work against the surroundings and in separating the gas molecules so the average kinetic energy of its molecules drops ie it ...


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Gluons could form glueballs (theoretically) because they interact in the QCD model that is part of the standard model (there are three gluon vertices, for example). QED lacks such entities because there are no three (or four) photon vertices. Since there is no theory of quantum gravity, there is no answer to your question. @JohnRennie (see above comment) ...


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There are $20$ numbers which are of maximum importance in physics. Well, $20$ is probably not going to do it these days (maybe you were reading something written before the neutrino oscillation discovery of the late $1990$'s). Before the Super Kamiokande experiments in the $1990$'s, we all thought that neutrinos were massless and didn't mix with each other, ...


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Fermi transport will be the equations that describe how each polarisation component changes direction, as the ray passes through the gravitational field (or rather, curved spacetime). So for circular polarisation to become eliptical, you need to check whether othogonal components cease to be orthogonal (when Fermi-Walker transported along an arbitrary ...


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Let's simplify the scenario slightly - imagine an object at the distance of the Moon, that has had its angular momentum slowed sufficiently that it will approach the Earth to a distance less than the radius of the Earth plus the radius of the Moon. Distance Moon-Earth ~ 400,000 km Radius Moon ~ 1,700 km Radius Earth ~ 6,300 km Mass Moon ...


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The energy of the collision is equal to the kinetic energy $K$ of the two bodys just before they collide. The mechanical energy is conserved during the whole process, so $$ V_1+K_1=V_2+K_2 $$ where $V$ is the gravitational potential energy of the bodies and $K$ their kinetic energy. Subscripts $1$ and $2$ refer to the initial time and the time of the ...


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Yes and no. Remember in special relativity whenever someone asks a question, they always are told to draw a spacetime diagram. The same thing happens in general relativity. If you want to see what is possible, consider drawing a Carter-Penrose diagram. For a black hole you can draw the event of a test particle crossing the event horizon. The past light cone ...


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Why is the gravitational constant.. constant? We don't actually know that it is. Check out the Dirac Large Numbers hypothesis: "According to Dirac's hypothesis, the apparent equivalence of these ratios might not be a mere coincidence but instead could imply a cosmology with these unusual features: The strength of gravity, as represented by the ...


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In a comment you say (I fixed a few words in this quote) "If it was a neutron star an atom would lose its electrons and protons before becoming a part of that star"... And in the question you say, "Are photons absorbed by atoms compressed out by gravity". This reminds me of Feynman's father. If you Google "feynman father photon", you should find the story ...


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Notice the photons are reduced around the smaller one. Is that happening before the photon sphere? The photon sphere by definition does not send any photons in our direction, as it is a spherical region of space where gravity is strong enough that photons are forced to travel in orbits. So the photons seen come from the region before, ...


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Since most the universe is empty space The short answer to this surprise is to see that the density $\rho = \frac M V = \frac M {\frac 4 3 {{\frac{2 G M}{c^2}}^{3}} \pi} = \frac{positive-constant} {M^2}$ decreases quickly as the mass increases. Now let's play with the values of size and mass that one can find on Wikipedia and other publications. ...


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Without being able to manipulate gravity. We are manipulating gravity all the time, except on earth, labs and constructions do not allow timing gravitational effects, which is why newtonian gravitational theory which has instantaneous effects is so successful. How do we know that gravity is restricted to the speed of light? or gravitational ...


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Gravity acts as a source term in the equations, and it is a source term on the energy and momentum equations. The mass conservation equation is not modified by gravity. So, looking at the momentum equation with gravity, we have: $$ \frac{\partial \rho u_i}{\partial t} + \frac{\partial \rho u_i u_j}{\partial x_j} = -\frac{\partial p}{\partial x_i} + \mu ...


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"Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is." It seems like the night sky isn't so very big, so it should be easy to observe objects, right? Wrong. Once you use telescopes, the night sky becomes huge and if you don't know what you are searching for, you'll only find out about it by coincidence. The ...


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All we can tell (assuming of course that the conclusions of the CalTech team are correct) is that there is a large mass in a distant orbit around the Sun. The mass could in principle be anything, but some things are more likely than others. It seems very plausible that the mass could be a planet that got ejected from an orbit nearer the Sun because: we ...


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In principle, any body with the mass of the proposed planet would have the same gravitational effect as the planet. Therefore, it would explain the orbit of those other bodies equally well. We know of a lot of planets in orbit around stars (and we have theories about how they form). However, I don't think we've ever seen or theorized black holes in orbit ...


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Think of a star as a big globe with ideal gas inside. Gravity acts as a force compressing the globe, the more it compress, the more energy goes to thermal part since $$dE = TdS-PdV$$ so a shrinking volume decreases $PdV$ term ($P$ is negative in this case, otherwise the system would be expanding), and for $dE=0$ because no reactions are occurring and no ...


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Both gravity and electrostatic forces depend on distance ($r$) like $1/r^2$. So changing the separation between 2 atoms changes both forces equally. So whichever force is stronger initially (at any distance) will always be stronger. To determine which is stronger consider the ratio of gravitational to electric force. $$ F_g/F_e = 4\pi \epsilon_0 G ...


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The formula for acceleration at a given height above the Earth would be: \begin{equation} a = -gR^2/r^2 \end{equation} where R is the radius of the earth and r is the radius you are at. This is because gravitation varies with the square of the distance between two objects. I'm afraid you no need calculus to be able to derrive escape velocity due to the fact ...


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I believe what he's pointing out is that energetic particles have sufficient kinetic energy to counter the pull of gravity, but as the star cools, the net kinetic energy decreases until the particles cannot go "up," or away from the centroid of mass. At that point, the star's mass collapses inward. The total mass is probably less than before, since the ...


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It would take less than 8 mins. It depends on elasticity of space time fabric. Consider Put a marble on a cloth and then observe how much it descends and curves cloth. Now, suddenly remove marble The time taken by cloth to regain its original position, so that it ends point feel no curvature, clearly depends on elasticity of fabric and amount of depth it had ...


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You don't need acceleration. Just calculate change in potential energy with altitude and make sure your initial Kinetic energy is sufficient that your final velocity is greater than zero. You cannot use your formula without integration. As for actual formula it would be $$g= GM/r^2$$ where M is mass of earth and r is varying distance.


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It's hard to see how gravitational repulsion between matter and antimatter would do any of those things, (1) because gravity is weak, and (2) because matter and antimatter are intermixed in the early universe, so the matter-antimatter repulsion would be competing with matter-matter attraction and antimatter-antimatter attraction.


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Yes, it "Is possible to generate energy from gravity". Gravity affects the orbit of the moon. The moon affects the level of tide waters. The movement & mass of tide waters can be captured & converted into electrical energy using energy harvesting technologies. e.g.: https://www.youtube.com/watch?v=7jsOerwz4Z8


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As per userTLK's comments, there are two ways we harness gravitational potential energy: Hydro systems are systems where water is given gravitational potential energy by being evaporated by the sun's radiant energy. That gravitational potential energy then converts to kinetic energy - the falling of the water - which is then converted to electricity by the ...


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Have you learned about gravitational potential? The force of gravity follows an inverse square law, $F=\frac{GM_1M_2}{r^2}$; the potential, which is the integral $\int F dr$, goes as $-\frac{GM_1M_2}{r}$. As you go from "infinity" (where the potential is zero) to some smaller radius, some potential energy is released and converted to kinetic energy, $\frac12 ...


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According to this problem set from Goddard Spaceflight Center in the upper Triassic the day was $23.5$ hours. The rotation speed was then $\frac {24}{23.5} \approx 1.021$ times higher. The equatorial speed of the earth is now $465.1$ m/sec. The centripetal acceleration at the equator is now $\frac {465.1^2}{6378100}\approx 0.034$ m/sec, which reduces $g$ ...


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Since I am not aware of any aspect of gravitational theory which corresponds to magnetism, I do not see how gravitational theory can account for the effects of motion. Gravitomagnetism is in fact a known and measured phenomenon. It emerges from Einstein's general relativity, rather than Newtonian gravity (which, as has been noted in other answers, is ...


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I used to count dofs in terms of spinor representation. There are two spinor representation of SO(3,1) (1/2,0) and (0,1/2) denoted by dot and no-dot indices. Vector (spin 1) representation in spinor indices is (1/2,0) X (0,1/2) = (1/2,1/2) which has 4 dofs. If the theory is massless, the theory is gauge invariant, there is one gauge dof, which can be ...


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Yes. General Relativity is perfectly capable of showing how a large oblate spheroid of gas moving in a particular direction contracts to form an oblate spheroid shaped star moving in that same direction that shines and emits light. And it is perfectly capable of making models that predict all the observations that any observer would make. Frankly, its a ...


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Newtonian gravity doesn't predict anything traveling at speed $c$ as neither the constant $c$ nor any constants you can use to make the constant $c$ even appear in Newtonian Gravity. Maxwell has $\epsilon_0$ and $\mu_0$ from which you can make $c$. And General Relativity and Special Relativity have $c$ directly as a constant of the theory. But Newtonian ...



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