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No. You can't feel gravity. What you feel as weight is the force of the ground pushing up on you. Take this force away and you will feel a bit queazy. There is very little variation in gravitational acceleration during a roller coaster ride. It is the extreme variations in the normal force during a roller coaster ride that gives you that sick to the stomach feeling. From a Newtonian perspective, you don't feel gravitation because it affects everything equally (acceleration due to gravity of a test mass is independent of mass). From a relativistic perspective, you don't feel gravitation because it is a fictitious force. There is something you do feel from those orbits. (You missed the biggest, which is the Earth and Moon orbiting one another.) What you missed are tidal forces. The Earth as a whole accelerates toward the Moon, or the Sun, or the rest of the galaxy. So do you, but by a slightly different amount than does the Earth as whole. The difference between the acceleration of the Earth as a whole toward these other objects and your acceleration toward these other objects results in something that you can feel. These are tidal forces and they are small.

As stated, the problem is unsolvable. You apparently have the wrong wording. The delivery cost should be $0.05 per gallon per mile.

You're right. I should have justified what I wrote. So, on a sphere we have [math]x^2+y^2+z^2=a^2[/math]. Thus [math] \begin{aligned} y^2 &= a^2-(z^2+x^2)\qquad(1) \\ z^2 &= a^2-(x^2+y^2)\qquad(2) \\ x^2 &= a^2-(y^2+z^2)\qquad(3) \end{aligned} [/math] Taking partial derivatives with of equation (1) respect to x, equation (2) with respect to y, and equation (3) with respect to z yields [math] \begin{aligned} \frac{\partial y}{\partial x} = -\frac x y \\ \frac{\partial z}{\partial y} = -\frac y z \\ \frac{\partial x}{\partial z} = -\frac z x \\ \end{aligned} [/math] Thus [math] \frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z} = \left(-\frac x y\right) \left(-\frac y z\right) \left(-\frac z x\right) = -1 [/math]

I can't help you much with that source of the data used by Einstein and others. As far as I can tell, it is that 120+ page booklet published back in 1882 by the US government printing office that is the source of the observed precession and the calculation of the portion of the precession attributable to Newtonian mechanics. Einstein's paper references this booklet, as do several other papers I happened to find. None of those papers re-performed the calculations of the parts of the precession attributable to Newtonian mechanics. That Simon Newcomb could have made such a blatant mistake is ludicrous. Then again, I can't get at that paper. It's certainly not on the 'net. It's too old. What you can find on the 'net are descriptions of how JPL, the Paris Observatory, and the Russian Institute of Applied Astronomy calculate their ephemerides. All three are now fully relativistic and do their calculations in a solar system barycenter frame. If Tsolkas is right, those groups would have found that they can't make their calculations match observations (some of which are very, very old; thousands of years old). You're a scientist: Would these groups have conspired to keep these findings hush-hush or would they have been in a race to reveal their findings to the world?

This guy has been tilting after windmills and cracking pots for a quarter of a century? Goodness. You'll have to go way, way back. S.Newcomb, “Discussion and results of observations on transits of Mercury from 1677 to 1881”, Astr. Pap. am. Ephem. naut. Alm., 1, 367-487 (U.S. Govt. Printing Office, Washington, D.C., 1882). I suspect that Newcomb did not "explicitly mention consideration of sun's motion relative to the barycentre." There is no need to do so. A heliocentric frame works just fine, as does a geocentric frame. (I suspect he did the latter given that you'll often see general precession cited as a part of the overall precession of Mercury.) Use an origin other than the solar system barycenter and you'll get fictitious forces. There's nothing wrong with that so long as you account for these fictitious forces. The name for these fictitious forces in the astronomy community (and in aerospace as well) is "third body effects". Why do you need to go so far back into history? Simple. This is a solved problem. Scientists don't get paid to revisit solved problems. What they get paid to extend the state of the art. Here you will find plenty of references regarding modeling the solar system from the perspective of a barycentric frame and in a fully relativistic framework. All three of the leading planetary ephemerides, the DE series from JPL, the EPM series from the Institute of Applied Astronomy, and the INPOP series from the Paris Observatory, do just that. Let's suppose that Tsolkas is right. That means that - Astronomers and physicists from the 1800s on up to Einstein's time (and later). All of them. - Modern astronomers who supposedly model the solar system to very high precision using a fully relativistic formulation are lying. Or we could suppose that Tsolkas is wrong.

@Obelix: Tsolkas is a crackpot. Of course astronomers account for the gravitational acceleration of the Sun toward the planets. BTW, it's "precession". @Gozzer101: You are talking about Doppler spectroscopy. @mathematic: There is no single "usual method". There are instead a number of different techniques used by different investigators. @questionposter: No, it doesn't. The Kepler mission uses the transmit method. Confirmation by Doppler spectroscopy is done elsewhere.

I presume you're talking about those eight steps used to derive conservation of energy. I would call that "freshman physics calculus." There's a couple of steps in there that would make a mathematician cringe a bit. That said, those quicksteps are easily rectified. On the other hand, why not just start with conservation of energy? As far as the original question is concerned, you can often get away with interpreting Leibniz notation as a ratio. It works fine -- some of the time. Other times, not so fine. Consider a point on a sphere [math]x^2+y^2+z^2=a^2[/math]. What is [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math]? Treating these partials as ratios would suggest the answer is 1. It isn't. It's -1.

This is wrong. The electrostatic force, like gravitation, is a 1/r2 force. However, the electrostatic force cannot explain gravity. This is wrong as an explanation for gravitation, for at least two reasons. 1. If it was correct objects would repel one another rather than be attracted to one another. 2. This is just wrong, period. Matter on the macroscopic scale is electrically neutral, or nearly so. Were it not we would see electrostatic discharges on a massive scale all over the Earth. Consider, for example, the act of lifting a boulder from the surface of the Earth. The kind of charge needed to explain the observed gravitational attraction between the boulder and the Earth would result in an electrical discharge between the boulder and the Earth. I'll do that for you. 1. Explain how planets can have moons. For the electrostatic force between two objects be attractive the charges on those two objects must be of opposite signs (otherwise the electrostatic force is repulsive, not attractive). Suppose your conjecture is correct. Since planets orbit the Sun, all the planets must have an electrostatic charge that is of the opposite sign of that of the Sun. The same goes for the moons that orbit some planet. That means the moons have the same sign charge as that of the Sun. The Sun would shoot the Earth's moon out of the solar system. The same would happen for many other moons. 2. Explain how people are still pulled Earthward when they are flying in an airplane. (The same object regarding the moons applies here.) 3. Explain why we don't see electrical discharges every time some object is lifted from the surface of the Earth.

No, there is conservation of angular momentum. There is no such thing as conservation of angular velocity. The angular momentum of a solid body is given by [math]\vec L = \mathbf I \vec{\omega}[/math]. That [math]\mathbf I[/math] is a tensor. Because angular momentum remains constant, a change in any of the components of that tensor means there has to be a corresponding change in the angular velocity.

10 cm is about 3 milliarcseconds, so yes, a tad short of 17 degrees. Greenfaerie, there are 3.6 million milliarcseconds in a degree. The claim is that the Earth's principle axes shifted by about that much (actually, 14 cm). Since angular momentum is a conserved quantity,with all other things being equal, this shift in the Earth's inertia tensor means that there has to be a corresponding counter-shift in the Earth's angular velocity. The problem is that all other things are not equal. Similar but slightly smaller shifts were predicted for the Feb 27, 2010 Chilean earthquake and the Sumatran quake of Christmas 2004. There were no detectable shift that rose above measurement and process noise from these previous quakes. Some experimentalists explain this lack of an observable shift in a way that is kind to the theorists. There are other confounding conditions, etc. etc. Other experimentalists aren't so kind. Effects as large as these should be observable (measurement error is about 0.1 mas for polar motion) -- but only if the effects are real.

Across the board? What do you mean by that? Those certainly are not the only notations used, even for three space. You will see [math]\hat x, \hat y, \hat z[/math] to denote the unit vectors along the x, y, and z axes, [math]\hat r, \hat {\theta}, \hat{\phi}[/math] for the spherical unit vectors, and more generically [math]\hat u_1, \hat u_2, \hat u_3[/math] to denote an arbitrary set of unit vectors. That's just a starter. And of course i, j, and k only pertain to three space.

Bingo!

And Yogoslavia. We should be careful not to count our democratic chickens before they are hatched. We don't know yet if those eggs contain chickens or baby rodans.

There are a couple of issues with this thread: 1. You are mixing up a lot of different concepts, lemur. The end result is that you are confused. 2. The answer to the quoted question depends on what you mean by "weight" and what you mean by "falling". Regarding orbits: One challenge is getting past the seemingly paradoxical fact that to transfer a vehicle from a circular orbit to a higher circular orbit, the vehicle's velocity needs to be increased twice and yet at the end of the process the vehicle's velocity is smaller than it was originally. The resolution of this apparent paradox is that you are extending your Earth-bound common sense to regimes where it doesn't apply. Orbital mechanics is a bit counterintuitive at times. Some suggested reading material: The wikipedia article on Hohmann transfers, http://en.wikipedia.org/wiki/Hohmann_transfer_orbit (google "Hohmann transfer" for more), and the wikipedia article on the vis viva equation, http://en.wikipedia.org/wiki/Vis-viva_equation. Regarding weight: This term has multiple meanings. One meaning is the magnitude of the gravitational force vector exerted on some body. This is the definition of weight that you will see in most introductory physics texts and in almost all aeronautics engineering texts. In this sense, the "weight" of an astronaut on the International Space Station is about 90% of the astronaut's weight on the ground. Another definition is the value read by an ideal spring scale, so some call this "scale weight". In terms of Newtonian mechanics, this is the magnitude of the net non-gravitational force acting on some body. In terms of general relativity, this is the magnitude of sum of all real (non-fictitous) forces acting on some body. Some introductory physics texts and most general relativity texts use this definition of "weight". In this sense, the "weight" of an astronaut on the International Space Station is zero.

It doesn't explain annihilation, either. From whence comes the energy in annihilation? In the standard model the energy arises precisely because both a particle and its antiparticle have positive mass.

Just because you do not fully understand antiparticles does not mean that we (read: scientists) do not.

You and I might not be living in parallel universes, but we sure are focusing on different aspects of Horgan's article. It appears you took focused on the title and inferred something from that. I focused on these two sentences: Telling the lay community that something that is inherently unobservable is true (and Greene does write that way) is, IMO, a disservice to science. One interpretation of the many worlds interpretation of quantum physics is that those many worlds are real parallel universes. MWI is but one of many metaphysical interpretations of quantum mechanics. The many worlds = multiple universes is a metaphysical interpretation atop a metaphysical interpretation. My opinion: Do your QM calculations in any way you want, but don't pretend that the interpretation you choose to view the counterintuitive QM world is "real". It is your calculations that are real. If you want, you can call this the meta-shutup and calculate interpretation of quantum mechanics.

Why the paranoid hostility? And for that matter, why haven't you (or anyone else) defined what "fairness" means? Without a definition of fairness all we have is people talking past one another and people dragging goalposts all over the place.

ANSI receives a good chunk of its funding from the US government.

We must be living in parallel universes here. I don't see how Tyson's argument has anything to do with Horgan's complaint against Greene's book. Horgan's complaint is that Greene is no longer being a scientist when he talks about non-communicating universes. He is being something that IMO is the antithesis of being a "scientist" -- the word "religious proselytizer" comes to my mind.

I have the same problem as does Horgan with respect to some aspects of theoretical physics. I'm not a fan of the many-worlds interpretation. Quantum mechanics is weird enough as-is. Just shut up and calculate. Leave the stupid metaphysical interpretations out of it, please.

You are creating a false dilemma here. Perfectly? Almost certainly not. Perfection is an illusory goal. More to the point, what observed problem in physics are you trying to solve with your nonsense of primal mass, and how does your nonsense notion explain that phenomenon? Non sequitur. Just because you don't understand something doesn't mean no one understands it. There are multiple ways of looking at annihilation events. I'll look at an electron-positron annihilation; other annihilation events are similar in form. Here are three points of view: Mass is a form of energy, specifically bound energy. From this point of view, what is happening in annihilation events is just conversion of one form of energy to another. The energy of the photons created from the annihilation event is given by the relativistic c. Note that this formula falls apart if the mass of either particle negative. Energy is a form of mass. From this point of view, what is happening in annihilation events is just conversion of one form of mass to another. The intrinsic mass of the photons created by the annihilation event is exactly equal to the intrinsic mass of the electron-positron pair. The intrinsic mass of a system of particles follows directly from the mass-energy equivalence formula. Mass and energy, while different, are nonetheless related. Specifically, they are related by the mass-energy equivalence formula. Note how the mass-energy equivalence formula plays a key role in each of these interpretations.

Why are you opening yet another thread on this same tired nonsense?

Your post looks rather out of place, steevey. Were you replying to a post in some other thread?

That is a part of the problem. However, (1) citation needed! and (2) are you comparing apples to oranges? The states and the municipalities in the US also collect taxes, some of them rather significant amounts of taxes. A fair comparison would be the total tax burden in the US versus the total tax burden in those other countries. Otherwise you are indeed comparing apples to oranges.