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I don't have enough physics knowledge, so please correct me as you see fit. Short answer: both. Long answer: the "actual" reason can be said to be the warping of spacetime as Hawking describes. But what is this warping for? Isn't it because of the gravity? The gravity affects particles as it warps the spacetime, and these two are just different ways of ...

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I believe it is the warping of spacetime. Because gravity described by general relativity is't like a normal force you would think of it is the bending of spacetime that is gravity. And light moves constant in a vacuum. PLEASE PLEASE PLEASE correct me if I am wrong because I love information.

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Ignoring air resistance, all objects near the Earth's surface fall at the same rate of acceleration, approximately 9.8 meters per second squared, whether of the same mass or not. So, when released at the same time, the object closer to the ground would reach the ground first.

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Newtonian Mechanics give us some equations that would described mathematically that experience of yours. So, when you drop a ball of mass $m$, the ball will only be acted by the gravity force (we are making the assumption that the air drag is negligible). So the ball will experience an acceleration $g$ downwards. By Newton’s Second Law we come with the ...

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While your question is rather straight forward, if you know the math which is involved (see NiveaNutellas answer), you probably found out a rather interesting thing: If you drop two identical objects, the first object from 0.2m and the second from 1m, it will be rather simple to tell, which one reaches the floor first. However, if you drop the first object ...

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The time $t$ it takes an object to fall to the ground from the height $h$ on earth is $$t=\sqrt{\frac{2h}{g}}$$ With $g\approx constant\approx9.81\frac{m}{s^2}$ As you can see, as you increase the height, the time also increases. For a height of $1m$ it takes about $0.45s$. For a height of $2m$ it takes about $0.63s$ To make this more obvious, you could ...

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I think you are mixing up several related things. A more basic question is why do we need a quantum theory of gravity in the first place! This is where the idea of the graviton comes from. Historically there have been many attempts to "unify" gravity with the other forces. And it is this desire that leads down the path of putting gravity in the standard ...

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OK, I figured it out myself. $$\begin{eqnarray}\delta_{gct}(\xi) B_\mu^{\>\>A} &\;=\;& \mathcal L_\xi B_\mu^{\>\>A} \\ &=& \xi^\lambda\partial_\lambda B_\mu^{\>\>A} + B_\lambda^{\>\>A}\partial_\mu \xi^\lambda \\ &=& \xi^\lambda\partial_{[\lambda} B_{\mu]}^{\>\>A} + \partial_\mu(\xi^\lambda B_\lambda^{\&... 1 If "regions of lower energy density" are, e.g., rigid spherical shells filled with vacuum; and if "regions of higher energy density" are, e.g., filled with CO2 gas; then yes, gravity repels regions of low energy density. Of course that's what buoyancy is and it's why hot air balloons and helium balloons float. You didn't specify what you really mean by "... 0 Force doesn't matter (or if it does, it matters indirectly). Energy does, but not force. For example, suppose you're trying to get a spacecraft into orbit, and you calculate the velocity needed is 10 km/s. How do you accelerate an object to 10 km/s? You could apply a lot of force (by F = ma, this is equivalent to high acceleration) for a brief time, ... 0 For an object of mass m to maintain a circular orbit you need an inward force F_\mathrm{inward}=\frac{mv^2}{r}. This force is now gravity F_\mathrm{inward}=F_\mathrm{gravity}=G\frac{mM}{r^2}. Sovling for v, you obtain$$ v=\sqrt{\frac{MG}{r}}$$. It is worth noting that this result is independent on the mass m of the orbiting object. Now, time for ... 0 A circle is defined in the following way: The ratio of the perimeter C=2\pi r and the diameter d=2r is constant. This constant ratio is \pi. Note that this definition holds only when Euclidean space i.e. flat space-time is considered. When we consider the curved space-time of general relativity, this definition doesn't hold as the space-time is no ... 0 It's the Doppler equations Einstein had an argument (paraphrased from memory), that if you drop a flashlight from a height, and have it switch on just before it hits, the light emitted immediately before the impact is blueshifted by the flashlight's downward velocity. However, if the flashlight is instead switched on briefly at the top of its fall, while ... 1 Gravitomagnetism is a general term used for the field effects associated with the relative motion of masses. Under traditional C19th Newtonian theory, students were taught that the principle of relativity only applied to bodies with uniform rectilinear motion. If we accelerated or rotated, we could tell that we were "really" accelerating or rotating (and by ... 0 Old question, but I wanted to contribute. I think you are correct. You are actually echoing a point of view that was historically held by Einstein himself. In a letter written in 1950 to Max von Laue he stated: It is true that in the case of the R^i_{klm} [the Riemann curvature tensor components] vanish, so that one could say: “There is no ... 0 Each photon will be at rest relative to the other, since they both travel in the same direction at the same speed. A photon at rest has zero frequency, hence zero energy. So, at least to first order, there should be no gravitational interaction between the two photons within their mutual rest frame. I'm not sure what an outside observer would see. It ... 0 From my understanding:. Photons have a rest mass of zero, but the electromagnetic waves do carry energy which causes gravitational pull. Since light isn't affected by gravity because its rest mass is zero, but the space light is moving on gets curved. This is what causes light beams to curve around massive objects. The two light beams wont be pulled into ... 0 They do not interact because their rest masses are zero. You can not attach to the photon neither an inertial nor a gravitational "mass" as$$ m \neq\frac{\hbar \omega}{c^2}$$See for example on this topic https://arxiv.org/abs/physics/9907017. This part is pure kinetic part and is related to the momentum of a photon in the energy–momentum relation (with ... 0 The initial gravitational frequency is just determined from the initial orbital parameters (masses and separations) that are given in the table that comes up when you follow the link you provided. These give an initial orbital frequency (given roughly by Kepler's third law for binaries which are reasonably well separated to begin with), and for gravitational ... 5 If your question is asking whether it's possible for an asteroid to be naturally captured and put into orbit around the earth, the answer is "yes". But it requires a very specific set of conditions. The incoming asteroid would need to be deflected by the Moon into a direction and speed that corresponds to an orbital trajectory. That means that its speed, ... -3 The moon itself was an asteroid that became natural satellite of Earth. As for 'spin' in the case of the moon, the periods of spinning around the moon's axis and spinning around the earth are the same as one month. That is why we only see one side of the moon. The dark side of the moon is usually concealed from the public (unless you are Pink Floyd), just ... 3 In general relativity the term “gravitational field” is a little ambiguous. There are at least three quantities that could qualify. The one that is most central to GR is called the metric. The one that best describes the curvature aspect of GR is called the Riemann curvature tensor. The one that is closest to the Newtonian gravitational field is the ... 0 Yes, it does. For instance, if an observer measures a G-field as g^\prime in a reference frame at rest with respect to the G-mass (planet), yet far enough from it, the observer who moves at v perpendicular to the field lines measures it as g=g^\prime/\gamma^2, where \gamma=1/\sqrt{1-v^2/c^2}. If the motion is not perpendicular, i.e., for any ... 1 1) You answer it yourself, the total reaction force is zero at every point, regardless of position, so a non-rigid object will not deform 2) If the sphere is rigid, the shell will not move inwards, but if it is made of individual particles, it will start (or try) moving inwards, which will create a pressure on inward shells, but this pressure will not be ... 0 But, how do gravitational waves propagate unless there were some form of fluidity to the space between bodies? There is a very fluidic like behavior of wht we refer to as the "fabric" of spacetime. Even with local large mass gravitational distortion, once the object moves from a given reference point the fabric returns to its previous contour like a boat ... 1 The ball sitting on top of a membrane image is wrong in many respects. A better image would be to show just a membrane (i.e. a curved two-dimensional surface) and colour different parts of the surface differently to show what is there, such as matter or vacuum, for example. Such an image can serve to illustrate the spatial situation for a static spacetime. ... 1 To answer this question, one has to choose how to quantify "the warping of space-time". In general, it takes 20 numbers (independent components of the Riemann curvature tensor) at every point to specify how spacetime is warped! A reasonable approach is to look at the Ricci scalar curvature R, which boils down this complexity into a single number. In a ... 0 The people on the larger planet would marvel at how quickly those on the smaller planet got things done. Those on the small planet would marvel at how long those on the larger planet live. 1 You are making some (not unreasonable) assumptions about the early universe that are not supported by any evidence. For example, we’re not sure that gravity was once unified with other forces and separated from them around one Planck time, although that seems plausible or likely to many physicists. But yes, in General Relativity gravity acts on any kind of ... 1 You really should search Google first. 1) Check out Wikipedia. They are the ratios between the various spherical deformations and the deforming potentials (since astronomical bodies do not have perfectly spherically symmetric gravitational potentials, so you'll get higher order terms in a spherical expansion of the potential). 2) I am pretty sure they have ... 2 As you can see from the above example the potential function can be broken into components just like force functions.In the example demonstrated above the actual axis is the diagonal line and the force function has been broken into two components along two axes inclined at 45^° to the diagonal axis,consequently the potential function along the diagonal ... 1 The most likely place to find a difference would be in the quantum realm. We've studied the equivalence of gravitational mass and inertial mass on the large scale for quite some time. We've looked at stars, black holes, and all sorts of large scale things. We find them to be the same at these scales. While we can't really talk about liklihood of ... 3 In general relativity, we don't measure the strength of the gravitational field as an acceleration g, the way we would do in Newtonian gravity. The equivalence principle tells us that g can be anything we like, based on the frame of reference we choose. Instead, we measure gravitational fields using measures of curvature. These measures of curvature ... 0 The maximal magnetic (and electric) field is likely set by the Schwinger limit where QED effects become appreciable and pair production starts. The corresponding limit for gravity is when quantum gravity starts mattering. This presumably starts to happen when we approach the Planck scale, for example when the field causes accelerations on the order of the ... 2 The no-hair theorems tell us that the properties of a black hole are limited to mass, charge, and angular momentum. A black hole has no memory of what went into it, other than these three quantities. 1 What are the theoretical/experimental problems with using negative gravitational energy and positive EM energy of the vacuum to balance one another? Energy is : In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation ... 0 Buoyancy is lost in the case you describe. It often happens with drill pipe in oil and gas wells with the drill string getting harder to pull out. 3 No, it would not, if there is no gap. On the contrary, the pressure of the fluid would keep it firmly stuck to the surface. It is the principle behind the whole notion of "suction cup" ! The atmospheric pressure is exactly what makes a suction cup sticking to the surface it is sticking to. 1 To make this question meaningful, you would have to say what this velocity is to be measured relative to. Furthermore, it has to be measured relative to something local, not something distant. See How do frames of reference work in general relativity, and are they described by coordinate systems? Relative to some other material object that is also at the ... 0 You are not saying it, but I assume you are talking about an object with rest mass. An object with rest mass can never reach the speed of light as per SR. The velocity of the dropped particle is hiding in the \gamma. After some algebra we can find:$$ v = c \sqrt{1 - \left[ \frac{1}{\frac{GM}{c^2} \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\...

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I think you are too attached to the usual "vector boson" theories. From the Effective Field Theory point of view, Einstein's gravity is just the theory of a massless spin 2 particle (the graviton), which $\textbf{must}$ be embedded into a symmetric rank-2 tensor. In such theory, all particles that carry energy interact with the graviton. The potential in ...

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You are correct, and this is the equivalence principle. As per GR, the inertial and gravitational mass are equivalent. In the theory of general relativity, the equivalence principle is the equivalence of gravitational and inertial mass, and Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body ...

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Gravity is a vector force in disguise. The metric $g_{\mu\nu}$ is the "multiplication" of two tetrad vectors $e_\mu$ and $e_\nu$. The vectorness is embedded surreptitiously inside metric. What Dirac serendipitously stumbled upon in 1928 were the "square roots" of the flat metric $\eta_{\mu\nu}$, where vierbein/tetrad vectors $e_\mu$ are simply Gamma matrix ...

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Let the potential energy if the system be zero at infinity then gravitational potential energy of the system is given as$$E=\frac{-GMm}{r}$$ You have missed the negative sign in your answer which accounts for the error.When the system is released from rest when it is at a separation of $2r$ then kinetic energy of the system increases which comes at the cost ...

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You're missing a negative sign. The gravitational potential energy $U$ of two masses $m$ and $M$ separated by a distance $r$ is given by $$U=-G\frac{mM}{r}$$ As the objects get closer together, the gravitational potential energy becomes more negative, which is another way to say that it decreases. As the potential energy decreases, the kinetic energy ...

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You have a sign error. The potential energy for gravitation is lower the closer two objects are to each other, not higher. The potential energy of two objects separated by $r$ is $$U = U_0 -\frac{GM_1 M_2}{r}$$ where $U_0$ is the potential energy at infinite separation. $U_0$ is conventionally taken to be zero, but it doesn't really matter what value it ...

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I don't see it any more paradoxical then hibernating yourself until the calculation is complete. If you have an event A in spacetime with coordinates (t,r) and another event B with coordinates (t',r'), then the amount of time passed for an observer going from A to B would depend, in general, on his trajectory. You don't even need gravity for this, special ...

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Force is a concept coming from the classical level, is always a vector, and is given by the vector dp/dt. The spin of the particular gauge boson does not touch the concept of force even at the quantum level. In any interaction the exchange has a dp/dt, and it is a force whether a Higgs particle comes out, or any other particle or complex of particles from ...

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I don't have a lot of knowledge about this, but this is what I understand -- Gravity affects spacetime by stretching and distorting the amount of space and the length of time, and speed affects your perspective of time as you slowly get closer to the speed of light, or in this sense, the speed of causality, the "information" around you seems to be moving ...

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