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You didn't do anything wrong, just incomplete. As @Sebastian alluded to, the work will be $$W=E_{3R}-E_{2R}=T_{3R}+V_{3R}-T_{2R}-V_{2R}$$ where the subscript denotes orbital position. You already found the $V_{3R}-V_{2R}$ term, now just use the centripetal motion formula $$\frac{mv^2}{r}=\frac{\gamma Mm}{r^2}$$ to find the difference in kinetic energy ...


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However it makes sense that gravity can't travel faster than light because of the force-carrying photons Whilst it makes sense that gravity can't travel faster than light, we don't actually know this for sure. What we do however know is that the force of gravity is not conveyed by photons. Even electromagnetic force is not conveyed by photons - hydrogen ...


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The gravitational field of a spherical body of mass $M$ at a distance $r$ from its center is $$\frac{GM}{r^2}$$. This is a Newtonian Mechanics answer, but it is an extremely good approximation when dealing with objects of small mass like the Earth, the Moon. etc. For higher mass and energy scales we use the Theory of General Relativity, which describes the ...


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This is an experimental fact. A measurement. Physics is about making mathematical models, called theories, that describe the measurements and predict new phenomena with success. The physics model describing (and predicting ) measurements/gravity is called Newton's gravitational model. It is very successful. Keep in mind if you go on to study physics that ...


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Planets do not move in fixed orbits, they move in very finely balanced stable orbits. What you see in the sky today is the result of 4 billion years of thumping, bumping, gravitational games and a lot of stuff falling into the Sun. It's all a very delicate balance of velocity, vector and gravity and it's not all that difficult to upset. While Jupiter won't ...


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I'll try to rephrase the question. If a planet only experiences one external force directed toward the Sun, why does it orbit instead of crashing into the Sun? This question falls under the category of "uniform circular motion," so if you want more detail, you can look up that chapter in a physics text. The main idea is the following: The force of gravity ...


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Potential energy has absolutely nothing to do with stress-energy or pressure. The following reference is a good source about the origin of the pressure term in the stress-energy tensor: "Momentum due to pressure: A simple model" by Kannan Jagannathan in American Journal of Physics 77, 432 (2009);  http://dx.doi.org/10.1119/1.3081105 Potential energy ...


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That discussion is about a spherical shell and does not apply to the Earth nor to the sun. In the latter situations assuming uniform mass distribution we know that time is dilated by $t(\infty)/t(r)=\sqrt{g_{00}}=\sqrt{1 - \frac{2GM}{rc^2}}$, where $r$ is the distance from the center. For a spherical SHELL however we have to go back and forth between the GR ...


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As you point out, there is no way to make artificial gravity work everywhere in a spherical shell by use of the centrifugal force. The obvious fix is to use a cylinder instead, and use the "endcaps" (top and bottom of the cylinder) purely for energy generation. Or, drop the endcaps entirely and make do with an open cylinder. If it's also very short it will ...


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For clarity let's work with a Lorentzian signature. Our $g$ is a metric for a 2 dimensional Lorentzian manifold $M$. It is well-known that any two dimensional pseudo-Riemannian manifold is conformally flat, that is $$g = e^{2\omega}\eta$$ Where $\eta$ is the flat 2D Minkovski metric. Your Lightcone gauge example You didn't define your $x^\pm$s but I ...


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Not all galaxies are spiral, there are elliptic and irregular shaped ones also. Search for them on Google and images abound of non spiral shaped ones. A more interesting question is: why are galaxies, in general, laid out in a flat plane, like a dinner plate? According to studies dating back over 40 years ago, spiral galaxies are gravitationally unstable ...


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Alpha decay, which is why it generally comes from natural gas fields where the alphas are trapped with the methane (in the US there are several in Kansas, Oklahoma, and Texas with reasonable fractions). There is a long history of helium in the US, where the Bureau of Land Management has been responsible for the helium reserve since the days of giant ...


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Turns out that due to orthogonality relations of Hermite-Gauss poly's, Hermite-Gauss modes are orthonormal, so $$ \int \int u_{n,m} \left(u_{n',m'}\right){}^*dxdy=\delta _{m,m'} \delta _{n,n'} $$ Then a's can be found by multiplying both sides by conjugate of u and integrating ...


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Energy is required to push something over a distance. In the case of the elevator, when it's not moving a brake can be engaged and the power removed and the elevator will sit just fine. That's because it doesn't take any energy to keep something still. But wait, if all I want to do is keep the helicopter still, then it doesn't require any energy? Sort of. ...


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The pressure at any point is the total weight of everything above an area of a square metre. So for example the atmospheric pressure at Earth's surface, 101325 Pa, means the total weight of atmosphere above a square metre at the surface is 101325 N. So when you ask for the pressure at the centre of the Earth, the way to calculate this is to work out the ...


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We were having a conversation with a peer about stupid ways of interpreting theories. Once you understand a few things about gravity and relativity, you understand just how stupid it is. Gravity on a plane in a bidimensional space can be interpreted as the acceleration of spacetime towards an object. Like John Rennie said, spacetime is not a thing. ...


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Spacetime is not a thing so it can't accelerate. Spacetime is a manifold equipped with a metric. However, in order to describe events in spacetime we construct coordinate systems, and coordinate systems can be accelerating. For example the Gullstrand-Painlevé coordinates describe the geometry around a black hole and they accelerate towards the black hole. ...


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If I take a plane to the equator, and travel east until I come back to where I started, have I travelled in a curved path or a straight line?


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If you mean within an earth radius, then the moon would break apart, pretty darn fast too. More here: http://en.wikipedia.org/wiki/Roche_limit and here: http://www.astro.washington.edu/users/smith/Astro150/Tutorials/Roche/ The earth/moon Roche limit is about 1.49 earth radii (from the surface, not the center), so a bit under 10,000 KM. If you have the ...


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The earth's orbital radius is about 150 million km, or 1 AU. The moon is already closer than that. If you mean "an orbit smaller than the earth's diameter" it would put the moon inside the Roche limit, and the results would be quite exciting to watch. Too bad no one would survive.


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You can't have a "ball of ice with the mass of sun", because the ice in the middle of the ball wouldn't be strong enough to support the weight of the ice on top of it. Instead, the ice would collapse under its own gravity. This would cause the pressure and temperature inside the ball to increase until the water molecules that make up the ice would break up ...


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Consider what happens to the total energy of the gas as it shrinks. By the virial theorem $K=-P/2, w$e can write the total energy $E$ of the gas cloud as $E=K+P=-P/2+P=P/2$. Additionally, according to equation (2) from the webpage, $P\sim -N^2/V^{1/3}$, so $E\sim -N^2/V^{1/3}$. Thus, as volume the volume decreases, the total energy becomes more and more ...


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The Baez article is strongly misleading in that it applies simplified concepts and arguments appropriate for gases with short-range interactions to a system with many particles interacting purely gravitationally. Systems where short-range forces dominate (Lenard-Jones, van der Waals forces... ) such as rarified gases can be described in such simplified ...


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Short term, the ice would be vapourised as it fell. It would mix with the sun and form a bizarrely metal-rich star of twice the mass. Such a star would have a much more opaque envelope. This leads to (once an equilibrium is reached) the final star being much less luminous and cooler than a 2 solar mass star of more normal composition. It would probably be ...


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In our universe, there are only four known forces which are given names such as "Strong", "Weak", "Electromagnetic", and "Gravitation". And, there seems to be an almost continuous report in the news of evidence or rumors of evidence for a 5th force but so far insufficient to be considered seriously. Of the four known forces, three of them are stronger than ...


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There is a force that is preventing us form falling through the Earth or through the floor of your house. It is called normal force. It is not possible in this way to make an object of zero mass, because mass isn't dependent on the force: my mass is exactly the same on Mars or Jupiter.


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This would be a highly energenic event, a gravitational collapse in combination with the initial inward velocity of 1000km/s (which is greater than the escape velocity at the surface of the sun). There would be some type of nova event initially because the hydrogen already present in the sun would be compressed by the infalling new material, greatly ...


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Here's another angle to this: it depends on your latitude. If you are standing somewhere near the equator, you're also experiencing circular motion around the Earth's axis of rotation with an angular frequency of 2*Pi/86400 and a velocity around 463 m/s. In the rotating frame of reference you'll suddenly find a centrifugal force $$|\vec a| = \omega^2 R = ...


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If ice is "all around the sun" I fail to see how it can be moving at a velocity of 1000 m/s inwards. The mass of the sun is $2\cdot 10^{30}\mathrm{\;kg}$ and the radius $7\cdot 10^{8}\mathrm{\;m}$. The thickness of a shell of ice with that inner radius and mass would be (assuming the usual density of ice of about 0.9x that of liquid water) approximately ...


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The person would be accelerated upwards briefly, then continue to float upwards at constant speed until his head hits the ceiling. When standing on a floor, the ground is pushing up on you with a force equal to your weight. Your shoes and your feet aren't perfectly rigid. They are compressed by this force and will act a bit like a spring if this force ...


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First and most important effect of having two sun sized bodies in our solar system very close to each other is that this new system will throw the planetary motion off its course, and there will be chaos(Noticeable chaos right at the moment when lets us assume the ice appeared 1000 km away from sun out of nowhere). Gravitational pull will be twice as much as ...


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The first thing you'll notice is that the Sun stops shining. It still produces heat and light, but everything is stopped by the thick layer of cold ice. The Sun however is not completely cold. The core is still active, even more than before. You doubled the mass, so the Sun has a higher pressure and thus can fuse easily hydrogen atoms together. The net ...


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When Miguel Alcubierre proposed a "warp drive" solution resolving the prohibition in Einstein’s special relativity on traveling faster than light the problem turns out to be that anybody in the path of the incoming space ship gets fried. That’s the conclusion of a group of University of Sydney physics students who have re-examined the maths of the ...


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I will bring an example from classical electrodynamics. In EM(electromagnetism) you have to consider that the fields(electrical and magnetic) also have energy and momentum. A classical example is to apply the third law of Newton(each action has an equal counter-action) to two moving charges. Then you'll conclude that the third law does not hold-thus the ...


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If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear. For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the ...


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I) For a single$^1$ spherically symmetric thin-shell collapse with (rest) mass $m$, the geometry inside the shell is flat Minkowski space, cf. e.g. this Phys.SE post; and outside the shell the geometry is Schwarzschild geometry with Schwarzschild radius $R_s=2E$ in natural units, where $$E~=~ m\sqrt{1+\left(\frac{dR}{d\tau}\right)^2} - \frac{m^2}{2R} ...


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The answer from a Newtonian perspective: TL;DR: Objects do feel gravitation, but only if they're very big, or if the gravitational field is very strong. Suppose you are in a spacesuit and are orbiting the Earth. Your feet are pointed toward the Earth, your head into space. Because gravitation is a 1/r2 force, the force on your feet is slightly stronger ...


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From a Newtonian perspective, the difference between being accelerated by gravity in freefall (which includes orbits) and being accelerated in a car has to do with the fact that you only "feel" accelerations when the external force is only being applied to one part of your body, rather than accelerating every particle equally as with gravity. For example, if ...


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https://www.youtube.com/watch?v=uENITui5_jU How about this experiment, there is some conection between sound and gravity?


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Let's start with a basic example. For the case $m=0$ the operator (1) can be rewritten as $$ \mathcal{O}^{\mu \nu, \alpha \beta} = \Box \left[ \frac{1}{2}(\eta^{\mu \alpha} \eta^{\nu \beta} + \eta^{\mu \beta} \eta^{\nu \alpha}) - \frac{1}{2}\eta^{\mu \nu}\eta^{\alpha \beta} \right], $$ after gauge freedom is fixed. In momentum space we have \begin{align} ...


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The electrostatic forces acting on Earth are completely negligable because the net charge on Earth is relatively close the zero.


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I'll talk about 3D space. If you have Newtonian Gravity, then only rest mass acts as a source, and it is an instantaneous action at a distance type force. You won't get a stronger force based on them moving faster. But they will still radiate electromagnetically, so there will be an electric force pushing them apart and a gravitational force pushing them ...


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No, or at least not under normal circumstances. Calculating the gravitational force between light beams turns out to be rather complicated, so let's use an analogous but simpler system. Instead of a spherically symmetric pulse of light consider an explosion that throws out a spherically symmetric cloud of light particles. At any time $t$ the cloud of ...


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No, they are not the same. If we use a quantum field theory to describe gravity then we get a theory with the graviton as a gauge boson propagating in a flat spacetime. We expect this theory to be an effective theory that is useful only where spacetime is not highly curved. So gravitons are (probably) not a fundamental description of how gravity works, and ...


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What is the shape of a self-gravitating rotating body? I didn't read through all answers but skimmed them instead. So far as I've read, are missing a piece. As @Alan Rominger pointed out the gravitational acceleration at any point on a self-gravitating sphere varies with latitude, like it does on the Earth. But the shape the body is complex and does not ...


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Yes, the force indeed is proportional to the mass. (Since $F_g=\frac{Gm_1m_2}{r^2}$) ( $m_1, m_2$ are masses of the bodies.) But this doesn't really matter, because say, for $m_1$, $$m_1\vec a=\frac{Gm_1m_2}{r^2}$$ If you notice, the acceleration of the body is independent of its mass, like you mentioned. Why do you think this will be any different for a ...


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The formula you've referenced, $\vec a_c = -\omega^2\vec r$, must define $\vec r$ as a function of the angular position, which depends on time (uniform motion and all). Something like $\vec r(\omega t) = r(\hat x \sin \omega t + \hat y \cos \omega t)$. $\omega$ essentially represents the angular speed, which means for constant $\omega$ you have an orbital ...


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Imagine you are on the northern hemisphere on Earth (assuming it's a perfect sphere). Now go north with constant speed: you can just go straight north you don't need to steer. Now go east with constant speed: this is something different now, in order to remain on the same latitude circle you must steer northwards constantly. If you don't see why, try to ...


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Original facetious answer involving a ghost cloth Since the material doesn't experience any force other than gravity the cloth will fall through the post, through the planet, come out the other side and then fall again. It will oscillate about the planet's centre, exhibiting perfectly undamped simple harmonic motion, assuming the planet has constant ...


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Upshot: The following two facts can be used to argue that the scalar correlator simplifies in the special case described above. When the scalar action is Weyl invariant, then the scalar equation of motion is covariant and we can use a Weyl transformation to simplify the equation. There is a Weyl transformation that maps AdS to the upper half plane of flat ...



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