# Tag Info

1

A little reflection will show that the law of the equality of the inertial and gravitational mass is equivalent to the assertion that the acceleration imparted to a body by a gravitational field is independent of the nature of the body. For Newton's equation of motion in a gravitational field, written out in full, it is: (Inertial mass) * ...

2

By Newton's Universal Law of Gravitation, the force between two object due to gravity $F = \dfrac{GMm}{r^2}$ where $M$ and $m$ are the masses of the two bodies attracting each other. Let's say $M$ is the mass of the earth and $m$ of the object we're dropping. Using $F = ma = \dfrac{GMm}{r^2}$ We can rearrange to find $a = \dfrac{GM}{r^2}$, independent ...

1

Partly this answer is just gathering together the comments above, though there are a couple of points that haven't been mentioned. Firstly, as mentioned in the comments electromagnetic waves do gravitate and the links in the comments cover this well. In the early universe (for the first 47,000 years after the Big Bang) EM radiation was the dominant ...

5

The problem is that when you write $r$ and $t$ in the equation for the time dilation you are using the Schwarzschild radial and time coordinates, which are part of a system of coordinates that works well far away from the black hole but fails at and within the event horizon. To make things a bit clearer we'll rewrite your equation as: $$t_0 = ... 5 The value r = \frac{2GM}{c^{2}} defines a special surface in the Schwarzschild spacetime called the event horizon. Observers inside this radius cannot be stationary with respect to points very distant from the black hole, and they cannot communicate with any observer outside this radius, so the notion of time dilation doesn't make sense for them. 3 The moon orbits around the sun, but so does the earth. They orbit together with the moon's orbit perturbed by the nearby earth. If fact, despite their different masses they experience the same acceleration, so it shouldn't be surprising that they are bound to the same orbit since they are bound to each other (i.e. at basically the same distance from the ... -1 The answer below supports an incorrect idea. See comments. The strength of the gravitational effect on an object (e.g., a spaceship or a moon) depends both on the mass of the celestial object and the distance from that celestial object. I suspect the difficulty you're experiencing has to do with not accounting for this second effect. In essence, the Earth ... 2 I mean why should the gravity of a less massive object dominate the gravity of a more massive one? Within the Hill sphere of the Earth, objects can orbit the Earth, because in the non-rotating frame of reference centered in the Earth (moving with acceleration around the Sun, so the frame is non-inertial), the Sun's gravity force is for the most part ... 4 This is not my field, but the question is interesting so I'll give you my best answer. The sphere of influence of an isolated astronomical body (in this case the binary stars, treated as a unit) is not well-defined; therefore, some of the forthcoming argument requires more information about the context in which you are studying the binary system (like the ... 2 Is there an equivalent formulation of classical electrodynamics in terms of action at a distance that is completely equivalent to the formulation in terms of fields (Maxwell's equations)? Yes, but only if special boundary conditions on the fields are assumed. For example, if the fields are purely retarded (wiki: Retarded and Advanced Potentials), one ... 2 If I understand your question correctly you assume that the spaceship is driven by some kind of engine giving it the necessary speed to revolve around the earth. As the astronaut does not have such an engine you believe he should fall back on the earth. If this interpretation is not correct, maybe you could make your point a little more clear. There are two ... 4 Remember that for the astronaut's spaceship to be in a steady orbit, it must be moving around the earth at the appropriate velocity v, where \frac{GMm}{R^2} = \frac{mv^2}{R} (i.e. the gravitational pull of earth is matched by the force needed to accelerate the astronaut in a circular orbit), where M is the mass of the earth, m of the ship, and R is ... 2 There seems a lot of conjecture in any event. Venus could have been a meteor, with an innate spin, that swung by the Sun and have been captured into our Solar systems anticlockwise orbital arrangement. Retaining her original spin momentum, clockwise relative to the others. 4 Black holes this small will have very high Hawking temperature:$$ T_H = \frac{\hbar c^3}{8 \pi G M k_B} \approx 10^{20}\,\text{K},  So, before this black hole can fall down even the diameter of an atom it will evaporate through Hawking radiation. As a result, the 1 tonne of black hole mass would be converted into the energy of very high energy particles ...

2

As dmckee said in his comment, the black hole would fall towards the center of the Earth. To specifically answer this portion of your question: How dense would rock have to be to form a barrier? There is absolutely no density of rock or anything else that would stop or even slow it down. Even if you created this microscopic black hole on the surface ...

3

This is a more complicated question that you probably realise. This first point to make is that the speed of light is always locally $c$, that is, if you measure the speed of light at your location you will always get the result $c$. The problem comes when you measure the speed of light at some location distant from you. To measure the speed of light ...

0

The following articles may be helpful. This is actually an active scientific topic. http://www.symmetrymagazine.org/breaking/2009/02/19/most-extreme-gamma-ray-blast-also-probes-quantum-gravity http://www.sciencemag.org/content/323/5922/1688.abstract http://www.sciencemag.org/content/early/2013/11/20/science.1242353.abstract

0

There is no experimental evidence on whether light travels slower in a gravity field. Some quantum gravity theories require light to be slower in an intensive gravitational field while others not so. So, it is to be determined by experiments or astronomical observations. Light travels in glass as fast as in vacuum. Because microscopically, glass is nothing ...

5

Spin 2 just means that the gravitational field is given by a metric field and general covariance, which is the nonlinear expression of a massless spin 2 representation of the Poincare group. The latter appears when linearizing around the Minkowski metric and dropping all interactions. See the classical paper by S. Weinberg, Phys.Rev. 138 (1965), B988-B1002 ...

1

On your first question: absolutely, energy gravitates (or induces curvature in spacetime) the same way that mass gravitates. If you read general relativity, you will learn that it is in fact the Stress-Energy Tensor that is the source of gravitational interaction (or equivalently spacetime curvature). Energy can be localized very easily; a parallel-plate ...

3

Unlike electrical or magnetic field, which acts differently on particles of different charge, for the gravitational field there is equivalence principle, which means that electrons and nuclei would experience the same acceleration due to gravity. There is, of course, the overall shift of energy level if the atom is observed from the place with the different ...

7

No. Newtonian gravity is a conservative vector field (or conservative force) which mean that energy that you extract from the field has to be put in first. This is technically stated as the work done around any closed looped is equal to zero. For example, you raise your pet cat up 1 meter (you do work against gravity) you let go and gravity does the same ...

0

If you remove a bit of moon and leave the rest with the same velocity, it will continue to follow the same orbit. Since you are carrying away some of the mass, as you carry it away it will exert a gravitational force on the moon, which could change the velocity. You are correct that cutting the mass in half will cut the gravitational force in half, but so ...

0

I believe your confusion comes from a misunderstanding of the designation of a force as "centripetal". Any calculation of centripetal force is telling you how much force is needed to make a circular motion take place. This doesn't create the force. There is no guarantee that a force of the calculated size and direction actually exists! You need to go ...

1

Is teleparallelism an alternative to the introduction of a metric? Teleparallel gravity still comes with a metric - just take the tetrad field as orthonormal basis and there it is. The main difference between GR and teleparallelism is that the former uses curvature, the latter torsion to model gravity. According to Kleinert, there's actually a type of ...

1

There is a sense in which metric theories of spacetime are "general". I simplify to four dimensions, but the argument generalizes to higher dimensions. Consider a particle whose path is parameterized by four coordinates $x^{a} = (t(s),x(s),y(s),z(s))$. We wish to describe the motion of the particle, given that at s=0, each of these functions has a known ...

2

The answer to your question: Would it be correct to assume that the particle has a stronger gravitational field [...]? is no, it would not be correct. Here is why. Comparing gravitational field in special relativity to its Newtonian limit means trying to take an ill-defined limit. If one wants to include relativistic corrections such as relativistic ...

-2

At the event horizon, the person getting sucked in will see light twice as fast. Then as he falls in he will eventually see light 3 times as fast, then 4 times as fast, then 5 times as fast, until light seems to be infinitely fast and time starts to grow infinite to him, and probably he will seem to take a split-second to fall in, then he will be destroyed ...

6

Gamma rays are affected just like light rays, so they will be subject to a gravitational red shift and they will be bent by gravitational fields just as visible light is. It's important to be clear that in a gamma ray burst the gamma rays are not generated by the black hole. The process of forming the black hole heats the interior of the star to incredible ...

0

Probably Mt. Chimborazo has the MOST gravity, as being in the "fattest" part of the planet and as one of the highest mountains you will have LOTS of mass generating more gravity (remember the more mass, the more gravity).

1

Hint: You can approximate the water in front of the wiper as being contained in a box being subjected to the given accellerations. The steady state angle of the surface will be perpendicular to the vector formed by the the gravity accelleration vector minus the lateral accelleration vector (i.e., water surface is normal to the net force). E.g., ...

3

From your point of view it would be instant. Sun is there, then poof! The sun is gone and so is its influence. We only know this event happened ~8.5 minutes before we saw it because we're so clever :-). However there's no way to detect it early and warn ourselves because no information could reach us any faster than the sun's extinguished light and ...

2

Gravitational waves travels at the speed of light, thus you would feel it at the same exact moment you saw the sun disappear. By general relativity, spacetime acts like a trampoline being bent by a central mass. When the mass is removed the trampoline does not go back to the unbent state instantaneously.

0

A scale has two options to work. a) It either acts as a pendulum with gravity counteracting the weight imbalance, or b) there is a return spring that tries to level the scale if it is not horizontal. Either way the effect is the same. The angle is proportional to the imbalance. In your case the details do not matter, and so you get to pick the design (the ...

0

English mathematician Sir Isaac Newton published Principia, which hypothesizes the inverse-square law of universal gravitation. He deduced that the forces which keep the planets in their orbs must be reciprocally as the squares of their distances from the centers about which they revolve. If he was reasoning this way about forces (F), he was also doing so ...

2

Wiki says : "Einstein and Fokker observed that the Lagrangian for Nordström's equation of motion for test particles, $L = \phi^2 \, \eta_{ab} \, {u}^a \, {u}^b$ , is the geodesic Lagrangian for a curved Lorentzian manifold with metric tensor $g_{ab} = \phi^2 \, \eta_{ab}$ ." [Remark : here $u^a = \frac{dx^a}{ds}$, so no "dot" here on the $u^a$, I think ...

0

Yes, for example, an object with a constant applied force and an oppositely directed speed dependent force has a terminal speed. The sum of forces is: $F_{NET} = F_A - kv$ Clearly, when the speed is large enough that $kv = F_A$, the net force is zero and the object stops accelerating - the object has reached terminal speed. See this hyperphysics article: ...

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