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4

This is really a difficult problem, but possibly not for the reason you imagine. The following naif criterion seems at first highly appropriate: take any pair of stars, subtract the center-of-mass motion, compute the kinetic ($T$) and gravitational ($W$) energies, and check whether $T + W < 0$. If so, the pair is bound, otherwise it is not. How can ...


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No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the ...


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The "theory" describing the black hole interior (in the classical approximation) is the same theory that implies the existence of the black holes, namely the general theory of relativity. As the OP correctly said, the singularity at the event horizon is a coordinate singularity – one that is an artifact of a bad choice of coordinates. When a coordinate ...


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The "buoyant force" does not arise from a principle, but is what remains of the gravity force when you subtract from it an average hydrostatic pressure. Let's assume that you have some reference density $\rho_0$. Then if $\rho=\rho_0$ everywhere and the fluid is at rest, $p = \rho_0 g z$. Now let's call this baseline pressure $p_0(z)$, without loss of ...


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Yes. You're exactly right, deviations from no-hair do occur for example after BH mergers --- and hints of the "quasi-normal" mode ("ringdown") were observed in the LIGO detection. The no-hair theorem is constructed for a static, stationary BH (i.e. fully settled). In general, deviations from no-hair (magnetic fields, asymmetry, etc) will be radiated away ...


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Negative energy or mass is not forbidden in Relativity, but gravity is not a force but geometry, so if you have a negative mass it would repell positive mass as well as negative mass, just like positive mass would attract negative and positive mass all together. If you place a positive and a negative mass near each other the positive mass would attract the ...


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If you see the expression of gravitational force: it has a coefficient, two mass terms and one distance term. Now, G (gravitational constant) ~ 6.6x10-11 SI Units m1 & m2 (suppose mass of an electron) ~ 9.1x10-31 kg r (distance between two electrons) ~ 10-12 m Gives us force in range of 10-47 - 10-46 N which is very very less. And there are other ...


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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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No, it cannot be so, for many reasons. One thing is, the amplitude of electron wave function in atom drops exponentially, which would make this effects vanishingly small, and would make any gravitation forces drop exponentially with distance.


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Yes, the acceleration of a freely falling particle anywhere else would be slightly altered. If the person jumping would be right next to the freely falling object, and one could neglect the maximum speed of propagation $c$, the falling object would be slightly pulled towards the jumping person. And thus change its falling acceleration. If this happens at a ...


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Any object with mass that accelerates (is it linear or angular acceleration) produces gravitational waves, though in most occasions those will be much too small to be detected. As @CuriousOne pointed out, same happens with electromagnetic waves and accelerating charges. The gravitational waves that can be detected usually come from very massive objects (such ...



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