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New sig! Thanks!

Is there any user guide for these hidden features?

The causes of what you feel are exactly the same in all three scenarios. With one exception, you do not feel gravitation. (That one exception: You would feel extreme tidal forces; these would cause you to undergo spaghettification. You'd feel yourself being pulled apart.) Local experiments (e.g., what you feel) cannot sense gravity. You feel forces other than gravity. You feel the floor pushing up on the parts of your body in contact with the floor. That's the normal force. You feel this when you push against a wall or pick up a book. This normal force induces stresses and strains on your body as that force propagates through your body. You feel those stresses and strains. Your inner ear contains the biological equivalent of an accelerometer. It works just like an accelerometer does. Some parts are (somewhat) free to follow an inertial path, other parts are more or less rigidly attached to your skeletal structure, and yet other parts sense the stresses that result when the accelerometer case (or your skeleton) is forced to follow a non-inertial path. So what's an inertial path? This is one of the key places where Newtonian physics and general relativity differ. In Newtonian mechanics, Newton's falling apple does not follow an inertial path. It does in general relativity. Einstein described his realization of this key distinction as his happiest thought.

Patent it then, or try to. Good luck with that. You'll need it.

That 33 feet is in water, not air. I said "It's a big problem with diving because all that's needed to be at twice atmospheric pressure is to dive down to 33 feet." Diving, as in diving underwater.

You're looking at pressure the wrong way. Imagine building a house of cards with a roof comprising one card. Now put a weight on top of that roof. One pound? Good luck. 128.6 pounds? Not a chance. The roof is not bearing a load of 128.6 pounds (2.5 inches × 3.5 inches × 14.7 psi). Pressure doesn't work that way. It acts on every surface. It acts on the front and back of each of those cards in that house of cards. The net force on each card is near zero. What pressure does is to make us lighter. Imagine suiting up in a vacuum suit, weighing your self in vacuum, and then weighing yourself again at STP. The latter measurement will be a tiny bit less than the first. The same thing happens with balloons filled with helium. A helium filled balloon will fall in vacuum. They rise in air because of buoyancy. You can feel pressure if your body is not in equilibrium with the surrounding air. This happens if you drive down a mountain too fast; you have to pop your ears. It's also a big problem with diving underwater because all that's needed to be at twice atmospheric pressure is to dive down to 33 feet. Come up too fast and you'll feel a good deal of pain, and possibly sustain some damage. Come up slowly and your internal pressure will have time to keep pace with the changes in pressure.

How do you stand something up right in a gravity free environment? What does that even mean?

No, I'm talking about the 'Delete' button, which no longer exists. Suppose I post a reply but I don't like what I see, so I wan't to delete it. There is no way to do that.

Lists don't work right. See http://http://www.scienceforums.net/topic/71250-when-do-we-feel-gravity/page-2#entry717870 Multiquote doesn't work right. Nothing to show cause it doesn't work right. Only one post shows up. Where's the delete button? When I posted something that looked completely wrong, I would sometimes delete the post and start over. Now I can't do that. I never have liked the "edited by" stuff. Don't show that the post has been edited if the edit was made less than a few minutes after the post was submitted.

It's exactly the same thing. Suppose you wake up in a strange room. You sit up and try to figure out where you are. The last thing you remember is going out to a bar with some friends on December 21 to celebrate the end of the world (or maybe not the end of the world). How can you tell, just based on feeling your butt on the floor whether It was the end of the world. The evil aliens who brought about the end of the world kidnapped you, brought you up to their rotating space station, and are about to do unthinkable things to you. The space station's rotation gives the illusion of normal Earth gravity. It was the end of the world. The benevolent aliens who foresaw the end of the world rescued you and brought you up to their rocket, which is currently accelerating at 1g toward planet X. It wasn't the end of the world. You had too much too drink last night and your friends wouldn't let you drive home. One of them took you to their place, where you promptly crashed on the floor. Aside: This new interface bites. Completely.

Gravity gradient is what drive the tides. And yes, it is physically possible to detect gravity gradient. It's what a gravity gradiometer does. Note that while accelerometers cannot measure gravity (no local experiment can measure gravitation; J.C. Macswell is correct), a pair of rigidly connected accelerometers can measure the gradient in gravity between the pair. The European Space Agency's GOCE satellite uses three pairs of accelerometers to measure the gravity gradient tensor.

You can't just think in physics. You have to do the math. So, do the math. Assume a body is orbiting circularly and the Sun loses a tiny bit of mass. I'll use the Sun's standard gravitational parameter [imath]\mu\equiv GM_{\text{sun}}[/imath] instead of mass; it makes the math easier and it's what is observable. Initially, the relation between velocity, the Sun's mass, and orbital radius is given by [math]v^2 = \frac \mu a = \frac \mu r[/math] Now suppose the Sun gains or loses a bit of mass: [imath]\mu \to \mu + \Delta \mu[/imath]. Note that the increment here is positive; losing mass means [imath]\Delta \mu < 0[/imath]. Per the vis-visa equation, [math]v^2 = \frac \mu r = (\mu + \Delta \mu) \left(\frac 2 r - \frac 1 {a'}\right)[/math] where [imath]a'[/imath] is the semi-major axis length after the mass loss. Denoting this as [imath]a'=a+\Delta a[/imath] with [imath]a=r[/imath] yields [math]\frac \mu a = (\mu + \Delta \mu) \left(\frac 2 a - \frac 1 {a+\Delta a}\right)[/math] Grind this through and you'll get [math]\frac{\Delta a}{a} = -\,\frac{\mu}{\Delta \mu}[/math] In other words, a loss of mass ([imath]\Delta \mu < 0[/imath]) causes the orbit to expand. There are other effects besides mass loss from solar wind on a planet's orbit. There's also mass loss from radiation. This currently dominates over mass loss from the solar wind. When the Sun was young it was the other way around. However, mass loss is mass loss. It doesn't matter how the Sun is losing mass. Mass loss causes orbits to expand. Effects other than addition to mass loss include Radiation pressure (radial component) The radial component of radiation pressure exerts an anti-sunward force on objects. Gravitation is of course inward; both are inverse square law forces. Define [imath]\beta[/imath] as the ratio of the outward force from radiation pressure to the inward gravitational force. [imath]\beta>1[/imath] for extremely small particles. Radiation pressure ejects very small particles from the solar system. [imath]\beta[/imath] deceases as particle size increases assuming constant density. Eventually increasing particle size will make the inward gravitational force dominate over the outward radiation pressure force. At this stage, the radial component of radiation pressure becomes a no-op. The anti-sunward component of solar radiation pressure merely makes the Sun appear less massive. Radiation pressure (tangential component) Thanks to the finite speed of light, sunlight hits an orbiting body with a non-zero tangential component directed against the body's velocity vector. This creates a drag on the body called Poynting-Robertson drag. This causes small particles (but not so small as to be ejected by the radial component of radiation pressure) to spiral inward toward the Sun. Because radiation pressure increases with the square of radius while mass increases with the cube, this Poynting-Robertson drag becomes smaller and smaller as body size increases. It is incredibly small for planet-sized bodies. Solar wind drag The solar wind, like radiation pressure, has both radial and tangential components. The effects of each component are very similar to those of radiation pressure. Yarkovsky effect The Sun warms the sunward facing side of an orbiting body. This results in an anisotropic distribution of the thermal radiation from the orbiting body. If the body rotates, the outgoing radiation is more intense on the dusk side of the planet than the dawn side. This results in an acceleration in the dawn direction. The Yarkovsky effect mades a body with a prograde rotation spiral outward from the Sun; a body with a retrograde rotation with spiral inward. Gravitational radiation Per general relativity, an accelerating object will radiate energy in the form of gravity waves. This makes very massive objects orbiting very close to one another spiral inward. (1993 Nobel Prize in Physics). In our solar system, this is essentially a non-effect. The effect on the Earth, for example, is to cause it to spiral inward at about 10-15 meters per day. For planetary sized objects, all of the above are essentially non-effects, completely dominated by the spiraling out due to solar mass loss.

The article you cited later discussed the presence of water in sunspots, not the solar wind. Sunspots are relatively cool regions on the surface of the Sun, cool enough that molecules can form. The solar wind is borne from the Sun's outer atmosphere and has temperatures of nearly a million kelvin, or more. There is no water in the solar wind; it's too hot. Another problem with this hypothesis is abundance (or lack thereof). Oxygen is but a trace element in the Sun itself; about 0.078% of the atoms in the Sun are oxygen atoms. Oxygen is an even smaller component of the solar wind. That said, there is a related hypothesis that does posit the solar wind as the source of water on the Moon, and possibly other bodies as well. Oxygen or water is not needed. All that's needed is protons, which along with electrons are the dominant component of the solar wind. Per this hypothesis, the protons in the solar wind interacted with the oxygen already present in a body's surface or atmosphere to form water. Yang Liu, et al., Direct measurement of hydroxyl in the lunar regolith and the origin of lunar surface, Nature Geoscience 5, 779–782 (2012) http://www.nature.com/ngeo/journal/v5/n11/full/ngeo1601.html

Every bit of this post is nonsense.

The colon operator creates a set. In for i = 1:n-1 the colon operator creates the set of integers between 1 and n-1, inclusive, and then iterates over the elements of that set. The colon operator in u=a(i,: ) the colon operator is equivalent to u=a(i,1:end). This takes a slice of the matrix a.

That's because you are a mathematician. Mathematicians have yet a third meaning of the word "theory". Game theory, graph theory, knot theory, number theory, string theory: Here the word "theory" does not mean something that has been tested against reality. It instead means "mathematical body of knowledge". Mathematics is not beholden to reality, so mathematicians don't particularly care whether some mathematical theorem is realistic. That the Banach-Tarski theorem cannot be used to replicate the Earth is irrelevant. It's a cool theorem with solid logic behind it, and that's good enough. Science is beholden to reality. In the context of science, string theory is not a scientific theory. Not yet, anyhow.

It's sloooow. Very, very slow.

http://www.gnuplot.info/faq/faq.html#SECTION00056000000000000000

If you cross the event horizon, yes, there is no escape. Otherwise, no. I strongly suggest you stop reading bad sci-fi and bad pop sci, and start learning some real science instead. Wrong. Suppose your trajectory takes you close to, but not inside, the event horizon. Sans relativistic effects, you'll just climb right back out of the black hole's gravity well with no energy required. The only energy required would be that needed to overcome those relativistic effects. Black holes are not mythical monsters that suck everything towards them. That's a pointless point. How exactly are you going to get to that "stationary position relative to the sun"? The Earth is orbiting the Sun at about 30 km/second. You can't get rid of that just by wishing it away. It takes energy, a lot of energy. It takes a lot more energy than is available with any existing technology. Wrong again. Going toward the Sun is extremely expensive. Let's go back to my example of sending nuclear waste into the Sun. Starting from a circular orbit about the Sun, a delta V of 25.7 km/sec would be needed to attain an elliptical orbit that just grazes the surface of the Sun if that initial circular orbit was at 1 AU (Earth's orbital radius). It gets more expensive closer in. From Mercury's orbit, the required delta V is 37.5 km/sec. Compare that to a delta V of 9.6 km/sec to put the vehicle on an elliptical orbit with a 1AU aphelion and then a 7.6 km/sec delta V to circularize to 1AU. That's a total of of 17.1 km/sec, which less than half the 37.5 km/sec needed to go toward the Sun. Wrong again. MESSENGER used gravity assists and conventional chemical rockets. BepiColombo will use ion propulsion as well as gravity assists and conventional rockets, but it will still use conventional rockets for that final orbit insertion at Mercury. Ion thrusters don't have near enough oomph to accomplish this task. Short of some huge breakthrough in propulsion technology, we will never send people to Mercury. Never. Even if humanity does develop that requisite breakthrough technology, this will not change the huge cost in getting to Mercury. That cost is always present. It is far cheaper to send machinery and people to practically any other place in the solar system.

This is completely wrong. Learn some orbital mechanics. I suggest you start with learning about Hohmann transfers, http://en.wikipedia.org/wiki/Hohmann_transfer_orbit. Trajectories involving gravity assists are not "timely", particularly when the gravity assists involve flying by Earth, then twice by Venus, and then Mercury three times. It took MESSENGER over 6½ years from launch to orbit insertion at Mercury. This, too, is wrong. Gravitation is a conservative force. It takes just as much energy to get there as it does to get back.

Learn some orbital mechanics. Every once in a while people come up with the brilliant idea of sending nuclear waste into the Sun. This is just silly. It's silly because it would take less energy to send that nuclear waste on an escape trajectory out of solar system. People just don't understand how expensive it is energy-wise to go down into a gravity well. This also applies to getting to Mercury. It is extremely expensive to get there, extremely expensive to stop there. If it weren't for Venus, we could not do it. The reason it took so long for MESSENGER was that it had to make a very large number of gravity slingshot maneuvers.

Emphasis mine: Do you realize how hard it is to get to Mercury? There's a reason NASA (and nobody else) has sent all of two missions to Mercury, while practically every space faring nation has sent multiple missions to Mars. Getting even close to Mercury an extremely expensive endeavor. Placing a vehicle in orbit about Mercury is monstrously expensive. Landing on Mercury would be even more than monstrously expensive.

That begs the question, how many different physical things are there? I'll start with the International System of Units (SI), which has seven fundamental units. One of those, the mole, is obviously dimensionless; it's just the number of items that constitutes a mole of such items. Another, the candela, is in reality a derived unit that pertains to how the human eye works. That leaves five. Is this the right number? Before answering this question, I'll first look back to Newton's second law prior to the SI. Newton's second law does not say [imath]F=ma[/imath] (force is mass times acceleration). It says that force is proportional to the product of mass and acceleration: [imath]F=kma[/imath], where k is some constant of proportionality. Is this proportionality constant k a dimensionful or dimensionless quantity? The early view in the development of the metric system was that this proportionality constant is dimensionful. From this perspective, constructing the units of force, mass, length, and time so that this proportionality constant has a numeric value of one was merely a convenient mathematical trick that simplified the math and removed a source for error. The modern view is that [imath]F=ma[/imath] is not just a trick that hides this proportionality constant. This proportionality constant is dimensionless, and the proper value is one. [imath]F=ma[/imath] is the "right" way to define force. From this point of view, the need to express Newton's second law as [imath]F=kma[/imath] with [imath]k\ne1[/imath] means that the underlying definitions of force, mass, length, and time are inconsistent. Now let's look at velocity: Is velocity dimensionful or dimensionless? If time and distance are fundamentally different quantities, the interpretation of the Lorentz transformation as a hyperbolic rotation of space-time coordinates has to be viewed as a trick that just happens to get the math right. Viewing the Lorentz transformation as a rotation is not a trick if one views time and distance as fundamentally the same thing. Some physicists prefer to work in some set of normalized units, systems where the speed of light has a numeric value of one. If velocity is a dimensionful quantity, choosing units of length and time such that the speed of light has a numeric value of one is just a mathematical trick that makes some calculations easier. If velocity is a dimensionless quantity, this choice not a trick. There is one universally agreed-upon velocity by all observers of some phenomenon, and that universally agreed-upon velocity is the speed of light. This is the natural value that makes a system of units consistent. From this perspective, time and distance only appear to be different because the SI ultimately is an inconsistent set of units, just as is the English system with its inconsistent representations of force, mass, length, and time. So instead of five fundamental units, perhaps there are only four? Another one of those fundamental units, temperature, bites the dust quite readily. Temperature is essentially energy per unit mass. That leaves but three: time/distance, mass, and charge. Is this the right answer? Rather than go any deeper, I'll let three physicists make their cases. Michael J. Duff et al, Trialogue on the number of fundamental constants, JHEP03(2002)023 doi:10.1088/1126-6708/2002/03/023 Preprint at http://arxiv.org/abs/physics/0110060 In this paper, Lev Okun argues that the correct number is three, Gabriele Veneziano argues that the correct number is two, and Michael Duff argues that the correct number is zero. More and more physicists are taking Duff's point of view, that there are no dimensionful quantities.

Bubba's. East Texans are obviously subhuman.

Axiom of choice, well-ordering theorem, and Zorn's lemma are equivalent. Of course they are. That statement about the axiom of choice, the well-ordering principle, and Zorn's lemma is a joke by Jerry Bona, and a pretty good one. Exactly.