# Tag Info

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i have no answer so i put you a question that how we fly with one jump

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1) If we talk about the Newton's concept then there is gravitational force as centripetal force and its reaction as centrifugal force. And centrifugal force will save it to fall at the center. 2) But if we talk about Einstein's concept of geodesics then there is no gravitational force or centrifugal force. Then falling bodies have to follow geodesics ...

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Objects not being affected by forces move on straight lines in euclidean space. If space is curved, there will not be straight lines, and particles follow trajectories that are the next best thing to straight, eg. paths whose length is the shortest. Knowing curvature, these can be calculated, for example by using variational principles (you can calculate ...

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Now, why do the objects falling towards the Earth move along the geodesic paths with no acceleration? A body in a free-fall moves with acceleration g, so, why is it written like that? To understand the passage, we must make two crucial observations. (1) To person at rest on the Earth's surface, a free-falling object is accelerating towards the ...

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The strong analogy is between gravity and fictitious forces. Of course centripetal force is the special case of rotating reference frames. In Newtonian Physics you have $F=ma$, that is valid in an inertial frame. Suppose that we want to describe a particle that is moving with constant velocity, so $a=0$. If you want to study the same system in a ...

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Suppose you and I start on the equator, a kilometre apart, and we both head exactly due North in a straight line, so we head off in exactly parallel directions: Now we know that in Euclidean geometry parallel lines remain the same distance apart. But if you and I measure the distance, $d$, between us we find that $d$ starts off at 1km but decreases as we ...

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Insofar as the gravitational force points towards the 'centre of gravity' of an object (like the earth or the sun), it can be considered a 'centre seeking' or centripetal force.

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This is definitely not the case with gravity. It has been shown after many years of observation and evidence gathering, and through the study of galaxies, solar systems, moon systems etc, that the gravitational force obeys the inverse square law at arbitrarily large distances. This is to the best of our knowledge how gravity operates.

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Only the center of mass of the planet moves on a geodesic line, objects on its surface do not and they do experience tidal forces. The same is true for the ISS. True microgravity experiments can only be performed at the center of mass of the station, which, of course, keeps shifting all the time as people are moving around, air circulates trough the modules, ...

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Of course it does. A mass moving trough a constant distribution of other masses will accelerate them and create a kind of gravitational wake. As a result it will lose velocity relative to this background. Read up on the Virial theorem and its importance for the dynamics of galaxies: http://en.wikipedia.org/wiki/Virial_theorem

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Also in simple terms. In a universe that is infinite with a uniform distribution of dark matter you are always in the center of that distribution. Gravity in a spherical shell, if you work it out, always goes to zero inside a spherical shell. This is why gravity in the earth doesn't go to infinity. Approaching the earth gravity increases with distance by ...

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The short answer is that a uniform matter distribution with Newtonian mechanics is an ill defined mathematical problem. The reason is equivalent to why $\int_{-\infty}^\infty |x|^{-1/2} dx$ is not defined: you only get a value with additional structure (e.g. a zero-point, or a rule for regularizing non-absolutely convergent series). While a uniform mass ...

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Calculating the path that a light ray takes in a gravitational field is a complicated business, but for the special case of a light ray coming from infinity and escaping to infinity there is a convenient approximate formula for the angle, $\theta$, the light ray is deflected: $$\theta \approx \frac{4GM}{r_0 c^2}$$ Where $M$ is the mass of the ...

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The stress-energy tensor, is up to multiplicative factors, can be defined by $\frac{\delta S}{\delta g^{\mu\nu}}$, where $S$ is the action and $g_{\mu\nu}$ is the metric. When people talk about the graviton, they talk about quantizing the metric around it's classical solution, so we consider field values $g_{\mu\nu} = g^{(c)}_{\mu\nu} + h_{\mu\nu}$, where ...

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Your logical error: you are confusing the direction of time and the direction of entropy. Elementary physical processes may be time-reversible. Complex processes are (nearly!) never reversible, due to the second law of thermodynamics. So your game should be about elementary processes. Also (simplified) models may be time reversible in your game, for ...

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You can describe Coriolis effect If one walks along meridian, strange force pulls him aside. If one walks along parallel, gravity appears stronger or weaker. If one jumps, he lands into different place. All these effects the grater the faster planet rotation. If it rotates to make Earth-like weights, effects will be noticeable.

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It is not known yet. Gravitons are from quantum mechanics model, while stress-energy tensor is from General relativity (GR) model. Two models are not connected until quantum gravity created. Also, gravitons were never observed, so they are pretty hypothetical. Simultaneously, it is known, that metric tensor is "generated" by stress-energy tensor. ...

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One issue is since it's a sphere, the spinning is only like gravity along the plane of spin. At the poles, there will be no such effect, and part-way to the poles, the force will be less, and at an angle to the surface. That's why a hollow "ringworld" is slightly more plausible than a hollow sphereworld. This would be a welcome question with ...

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In a physics sense, the acceleration would still be downward (time is squared in acceleration, so reversing time doesn't make the square negative), but it's the second law of thermodynamics that's really being reversed. Instead of getting nudged off of a table, falling to the ground, and bouncing succesively smaller bounces, a ball would be shot from the ...

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No ,a photon has no mass but it (still)can have momentum according to the equation E^2=(pc)^2+(mc^2)^2 and yes,light can be effected by gravity because of its energy(due to momentum ) simply you can say its not nessesary to have mass for a photon to be effected by gravity.

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Imagine a flowerpot sitting on a ledge. A breeze blows the pot off of the ledge, and it falls to the ground. When it hits the ground it shatters into a bunch of pieces, it kicks up dust, it makes a sound, it vibrates the ground, and the shards come to rest. The time-reversal of this is that some pieces of flowerpot are sitting on the ground. Suddenly a ...

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One of the problems you will encounter is causality. Imagine you have a ball resting on the ground. Without already knowing how it behaved in the past you cannot uniquely define the next frame of your game. You cannot tell if the ball should: move upwards vertically. move upwards in any direction. roll on the ground towards any direction. do nothing. ...

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There is no mistake. The laws of physics themselves are reversible in time, but the solutions not necessarily so. Thus, the "behavior" of the universe itself does not show symmetry under time reversal, primarily due to the second law of thermodynamics. The second law is about the behavior of the solutions, is not a fundamental law in itself. In your specific ...

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The gravitional constant is so small because before there exists a force, there are some substances which compute together to form forces, so the sum total of the masses of there atoms, which is very small, gives the value of the force constant. You can prove this theoretically.

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Your claim that nothing will happen because they initially have no velocity is where things start to go wrong. In fact everything in relativity has the same speed (the speed of light). A particle that looks to be "at rest" in some reference frame simply has all of its velocity pointing in the "time direction". This is an intuitive reason why you would see ...

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If it had a foundation in the laws of physics, I don't think we would call it magic... The word "magic" seems to be used for an impossible un-explainable process (if it was explainable, it might not be interesting enough to be used in a movie and to be called magic). You will have to close your eyes for many physical laws to accept it as such. Magic can, ...

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In Reverse Time, objects would experience a inversion of velocities and accelerations. Instead of a falling ball moving toward the ground and gaining velocity, it would be moving away from the floor and losing velocity. When it returned to the spot it was dropped, its velocity should equal 0. To achieve this, change the sign on it's velocity along the ...

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Proporties that are independent of time will not change when time is reversed, forwards or backwards, or changed in any other way. Anything dependent of time will run backwards. In general we may say that any change of any proporty (causing a difference $\delta f(t)$ over a time difference $\delta t$ between "before" $t_{1}$ and "now" $t_{2}$, that is ...

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To make such a game and have it make sense, you would have to use a much simpler physical system than anything you would reasonably encounter on Earth's atmosphere. For heat-dissipating systems, the laws of thermodynamics give time a direction: the direction of increasing entropy. Thus an object that falls to earth dissipates its energy, mostly as heat. You ...

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The direction of the gravitational force would not change under time reversal. Your object would feel a force downward, just as it does usually. It might be easier to imagine you had a movie of an object under the influence of gravity. Drop the ball from rest some distance above the floor. You'll see it move downward and speed up. You'd interpret this as a ...

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In reverse time, it would seem gravity is still attractive. But the quick impulse force that stopped the motion of something falling would instead send it upward at some inital velocity, which gets slowed by gravity. Otherwise, the object would accelerate from rest away from earth. My hunch would be that all conservative force fields remain the same ...

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How accurate do you need to be? The problem with these calculations is that massless, frictionless pulleys are usually out of stock at Acme Mail Order. High school physics will give you the tension in the rope, a different set of high school equations will tell you how much extra tension is needed for a certain acceleration. A rough metric is add 10% for ...

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Whilst the question is not a resource request, I would recommend Edward Witten's paper on the topic published in 1988, titled, 2+1 Dimensional Gravity as a Soluble System. In the paper, Witten shows: $2+1$ dimensional gravity with or without $\Lambda$ is soluble classically and at the quantum level $2+1$ dimensional gravity is related to a Yang-Mills ...

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Classically they are clearly topological. The metric does not appear, and you don't need a metric for integration on manifolds to make sense. Now in dimension 3 you can cast the Einstein-Hilbert action into a Chern-Simons theory as you say. The connection takes it values in the Lie algebra of the Poincare group. In higher dimensions you need to use higher ...

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The gravitational Chern-Simons action is topological, yes. The gauge connection encodes the field of gravity and since it is being integrated over, the result does not depend on a metric. (In the expressions you write maybe the vielbein contribution is missing? Or maybe you mean to have absorbed it in the notation.) Notice that it's just the usual ...

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It cannot disrupt the gravity of the Earth. There are other infinitesimal possibilities of doomsday scenarios though these stories have been debunked many times. The report ruled out any doomsday scenario at the LHC, noting that the physical conditions and collision events which exist in the LHC, RHIC and other experiments occur naturally and routinely ...

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Forget gravity for a little while, and put yourself in the pilot seat of an aerobatic airplane. The plane has so-called "elevators" at the tail or "empenage", and those are connected to the stick that you hold. What the elevators do is determine your angle of attack - the angle at which your wing meets the air stream. If you pull back, the elevators move ...

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The problem with such a maneuver is that you're kicking in some strong g-forces there. Once you you hit an angle of 60$^\circ$ (the common banking angle for a commercial airliner), you are hitting 2G's. Past that, it goes exponentially: (source--load factor is the hypotenuse to gravitational & centrifugal forces. Note that the typical person can only ...

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The Anthropic answer to this question is that if gravity were a lot stronger, then the evolution of the universe would have proceeded in a different way, it would have collapsed just after the Big Bang. One can speculate that all possiblities really exists, but we can obviously only find ourselves in those universes with laws of physics that are compatible ...

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The static electrical force between two particles of electric charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb’s law of electricity, which says that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force is repulsive for like ...

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"According to Newton's law the negative mass should be repelled" -- Nope, in both Newtonian physics and in general relativity, negative mass would be attracted gravitationally to positive mass, although negative mass would exert a repulsive gravitational effect on positive mass (but if the negative mass is small compared to the mass of the black hole this ...

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From my knowledge, gravity is infinite and extends throughout all of space. It diminishes as distance increases but is still present everywhere. So given enough time, no matter where something is in the universe, it would accelerate due to the gravitational force of something in the universe. You must mean "the effect of gravity reaches to infinity" . ...

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Terrific question. You had it right in your first sentence: “the same amount of energy must have been released during the Earth's history,” but then it gets a little mixed up when you look at various energies, some of which aren’t related to the question at hand (for example, the current internal energy contributes positive mass-energy to the Earth, rather ...

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The rest of the energy went into space. Without that energy loss the planet would not even have condensed and the gas/dust cloud would have stayed a cloud. Having said that, the details of these condensation processes in planetary clouds seem to be non-trivial and, from what I have read, are not fully understood, as of yet.

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In order to understand how gravity affects objects at different points within or on a sphere, see the following link: http://en.wikipedia.org/wiki/Shell_theorem This basically states that inside a sphere, the gravitational force of shells further away from the centre than you cancel each other out. Considering your initial question about pressure at the ...

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This is outside of my area of expertise, but I figure I'll take a stab at it. You elaborate in a comment: We assume capillary action of flow of fluid, as any type of pressure may alter composition of the paper. The paper is similar to paper towels, except the sheets are folded. The height $h$ of a column in a capillary with radius $r$ is given by $$... 4 You are correct that as you get very close to the center of the earth, the value of g can become arbitrarily low. If you could somehow create a space there, you could potentially float in it because you would not be pulled in any particular direction with respect to the earth. But while gravity is not strong there, it is strong in other places (like your ... 4 The Einstein equation says:$$ {\bf G} = {\bf T}  where $\bf G$ is the Einstein tensor that describes the curvature, i.e. the gravity, while $\bf T$ is the stress-energy tensor. So the origin of gravity is the stress-energy tensor. This is typically dominated by mass, but includes less obvious contributions like pressure and momentum. Actually solving ...

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The case on the left is correct. Gravitational gradient goes in the direction of the source of gravitational force. Think about the extreme case: if the object was leaning almost all the way over at 89.9$^o$, the center of gravity would necessarily be almost directly below the center of mass.

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If the density is constant, it is guarantedd that M and G will fall into the same vertical line. However, if the density is not constant, it is not difficult to find a counterexample that will have G and M on diffrerent vertical lines.

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