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October 30

Venusian atmosphere

What is the atmospheric density profile, for the last lower five kilometres of the Venusian atmosphere? I want to investigate the properties of the supercritical CO₂ ocean, such as the rate of change in the terminal velocity, or the hydro/aerodynamics. If a tungsten sphere enters the atmosphere at sufficient speed, can it undergo two succesive sonic booms, once upon entering the atmosphere, once upon entering the ocean? Plasmic Physics (talk) 00:00, 30 October 2012 (UTC)[reply]

The article says that the density at the surface is 65 kg/sq.cm; at higher altitudes, it should decrease as a logarhythmic function of altitude, as is the case with Earth's atmosphere. 24.23.196.85 (talk) 02:24, 30 October 2012 (UTC)[reply]

Or, more plausibly (on two counts), 65 kg/m³. —Tamfang (talk) 03:32, 30 October 2012 (UTC)[reply]

Sorry, messed up the units. 24.23.196.85 (talk) 04:25, 30 October 2012 (UTC)[reply]

That won't apply in this case. Plasmic Physics (talk) 02:38, 30 October 2012 (UTC)[reply]

What I mean is that while Earth's atmosphere is in a homogenous state and can be modelled by a singular logarithmic function, Venus' atmosphere is in a heterogenous state, and must be modelled by two functions. Plasmic Physics (talk) 06:31, 30 October 2012 (UTC)[reply]

Even a temperature/pressure profile would be nice. Plasmic Physics (talk) 08:17, 1 November 2012 (UTC)[reply]

What is the farthest known planet from earth including dwarf planets?

What is the farthest known planet from earth including dwarf planets? Neptunekh94 (talk) 01:40, 30 October 2012 (UTC)[reply]

Last time I checked, it was 90377 Sedna -- but this info could be out of date. 24.23.196.85 (talk) 02:01, 30 October 2012 (UTC)[reply]

According to our article, Sedna hasn't been officially recognized as a dwarf planet yet. If the answer is confined to objects so recognized, I believe the answer is Eris. Deor (talk) 02:12, 30 October 2012 (UTC)[reply]

Surely "some" of the known extrasolar planets are more distant... our list of exoplanetary host stars seems very thorough, and lists the Cygnus binary planet system (confirmed on this NASA webpage) at a distance of over five thousand light years. The NASA article states that this system is "among the most distant" planetary system, indicating a healthy scientific ambiguity about which planetary discoveries are currently confirmed. Nimur (talk) 02:15, 30 October 2012 (UTC)[reply]

Rather more speculatively, there are a few candidates for extragalactic planets. Dragons flight (talk) 04:12, 30 October 2012 (UTC)[reply]

Also speculatively, given the cosmological principle, no galaxy in the universe should be unique or distinct from other galaxies in fundamental ways. If planetary formation is found in our galaxy, it should also be found in any of the billions of other galaxies in the universe. So, there are likely uncountable numbers of planets at any given distance from Earth. None has necessarily been specifically identified, but they're clearly there. I don't know the names and current locations of every person in Beijing, China, but I am quite certain they exist. --Jayron32 04:24, 30 October 2012 (UTC)[reply]

I'm sure the OP meant planets in the Solar System. 24.23.196.85 (talk) 04:26, 30 October 2012 (UTC)[reply]

^{[citation needed]}. When you get that mind-reading technology perfected, when are you going to put it on sale? --Jayron32 04:28, 30 October 2012 (UTC)[reply]

In any case, he/she specifically asked about known planets only. 24.23.196.85 (talk) 04:42, 30 October 2012 (UTC)[reply]

Well, we know planets exist in other galaxies, far away from earth, even if we can't name and locate them (see my Beijing analogy above). --Jayron32 04:47, 30 October 2012 (UTC)[reply]

Well, no, we don't know for a fact that there are trillions of planets (even though it is generally believed there almost certainly are). But even if we did know that for certain, that doesn't make any of them "known". Except the very few ones we do actually "know". The range of our knowledge now extends beyond our Solar System, but not very much in celestial terms. It's like the sand on a beach. I am 100% certain there's a virtually uncountable number of grains of sand there, but I cannot tell you the precise measurements of any individual grain, or even if there is a single grain that has precisely some measurements that I arbitrarily specify in advance. I may "know" the beach, but I do not "know" the grains of sand. -- Jack of Oz ^[Talk] 05:26, 30 October 2012 (UTC)[reply]

If I understand the somewhat strange IAU definitions correctly nigher dwarf planets nor extrasolar planets are actualy planets. That would give us Neptune as the most distant planet and Eris as the most distant known dwarf planet at the moment. Gr8xoz (talk) 16:36, 30 October 2012 (UTC)[reply]

Oh my god. I usually respect ref deskers a lot, but most of the above answers are horrible.

1. The farthest known dwarf planet from the Sun, at this moment, is Eris. However, Eris has a highly elliptical orbit with a perihelion of 38.4 AU, which is closer than Pluto is at its aphelion. Sedna, if confirmed as a dwarf planet, is currently closer to the Sun than Eris: 86 AU vs. 97 AU. It won't be this way forever, because Sedna has an aphelion of 937 AU, compared to Eris' 97.5.

2. If you want the farthest dwarf planet from Earth, the answer is Eris, the same as above. The distance between Earth and the Sun is negligible compared to the distance between the Sun and Eris or Sedna.

3. If you include extrasolar planets, according to the Extrasolar Planet Encyclopedia (http://exoplanet.eu/), the farthest known planet is SWEEPS-04 at a distance of 8500 parsecs (28000 light years). Wikipedia's article claims that Kepler-35b is 51000 light years away, when it's actually closer to 5000.

4. If you include highly speculative exoplanets, then see extragalactic planets. Unfortunately they're impossible to follow up on, because their detection depended on a chance alignment of stars that made gravitational microlensing possible.

5. Disregard what Gr8xoz said. Dwarf planets are not planets, but extrasolar planets most certainly are. In fact, one of the primary motivations for defining the word "planet" was that we needed a way to categorize all those extrasolar objects. --140.180.252.244 (talk) 21:15, 30 October 2012 (UTC)[reply]

The oficial definition of a planet is "A planet1 is a celestial body that (a) is in orbit around the Sun,(b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit." See https://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf.

To me that seems to rule out extrasolar planets, can you give any authorative source showing that this is not the case? Gr8xoz (talk) 22:24, 30 October 2012 (UTC)[reply]

That resolution is titled "Definition of a Planet in the Solar System". It starts with "The IAU therefore resolves that planets and other bodies, except satellites, in our Solar System..." It therefore does not apply to any objects outside the Solar System. According to Extrasolar Planet#Definition, which references this IAU position statement:

1) Objects with true masses below the limiting mass for thermonuclear fusion of deuterium that orbit stars or stellar remnants are "planets [...]

3) Free-floating objects in young star clusters with masses below the limiting mass for thermonuclear fusion of deuterium are not "planets", but are "sub-brown dwarfs" (or whatever name is most appropriate).

The IAU never bothered to apply the same definition of "planet" to other solar systems as to our own because it isn't necessary; all known exoplanets are unambiguously planets under either definition. --140.180.252.244 (talk) 22:45, 30 October 2012 (UTC)[reply]

Hey wait a minute everyone, the OP specifically asked about the farthest planet from the Sun including dwarf planets. 24.23.196.85 (talk) 06:04, 31 October 2012 (UTC)[reply]

Thermite reaction

Hello all, I'm very familiar with the Goldschmidt reaction's properties, but there's just one thing I could never understand: How come the aluminum metal reacts with the iron oxide despite the fact that there's an aluminum oxide film covering the aluminum metal and preventing its direct contact with the iron oxide? (I mean, before the mixture gets hot enough to melt the oxide film.) Does this involve some kind of quantum tunneling through the film? 24.23.196.85 (talk) 02:09, 30 October 2012 (UTC)[reply]

I've used thermite welding in one of its' main applications: joining heavy copper cables to stainless steel earth rods. The charge is set off by a flint lighter that produces hot sparks. If you try igniting a mixture of aluminium powder and iron oxide power with sparks, nothing will happen. The thermite welding mixes contain magnesium and other proprietry stuff that the sparks ignite to get the temperature up to white heat - the reaction between the aluminium and the iron oxide then supplie the bulk of the heat. Ratbone 121.215.59.18 (talk) 02:34, 30 October 2012 (UTC)[reply]

What he said. Thermite has a very high activation energy. When I would do it as a classroom demonstration, I always used a chemical starter (ingredients never revealed to students) which provided the initial "burn" to start the reaction. Thermite is highly exothermic, so it is self-sustaining, but the activation energy is so high that a few sparks won't do it. The activation energy is so high that there is little danger of self-starting: premixed thermite powders are shelf-stable. They won't go off spontaneously. --Jayron32 03:27, 30 October 2012 (UTC)[reply]

Heh, but you already let the cat out of the bag. Besides, the kids seem pretty good at figuring it out on their own.[1] (not sure the professors are safe with it though [2] - for ROFL be sure to watch the kids' video first!) Wnt (talk) 19:33, 30 October 2012 (UTC)[reply]

In any case, the article describes the various ignition methods in quite a bit of detail. 24.23.196.85 (talk) 06:01, 31 October 2012 (UTC)[reply]

Thanks everyone! So in fact the initial burn is hot enough to melt the oxide film right off, correct? 24.23.196.85 (talk) 04:40, 30 October 2012 (UTC)[reply]

Sort of. The activation energy is largely spent on liquefying the aluminum, IIRC, so that the reaction can proceed. Reactions don't generally proceed in the solid state; once the aluminum is melted, the oxide "coating" is no longer an issue. It isn't merely that the aluminum has an oxide coating that makes it unreactive, it is that there is no way for the solid aluminum and solid iron oxide to intermix (on the atomic level) to start a reaction. Once there is a liquid phase (which I am fairly certain is the aluminum) the reaction starts, and becomes self sustaining (the initial reaction of the tiny, microscopic amount of aluminum melted by the starter produces more than enough heat to continue to melt the remaining aluminum and allow the reaction to go to completion.) --Jayron32 04:52, 30 October 2012 (UTC)[reply]

Charges on quarks

I know how +1 charge of proton and 0 charge (neutral) of neutron is determined. Their charge is determined by the sum of charges of constituting quarks. But how the charge of a quark is determined. After reading this Quark#Electric_charge, I couldn't get the answer. Sunny Singh (DAV) (talk) 12:10, 30 October 2012 (UTC)[reply]

Charge is just an intrinsic property of quarks, like spin or mass. Goodbye Galaxy (talk) 16:05, 30 October 2012 (UTC)[reply]

Charge is one of the quantum numbers, actually a special class of quantum numbers called "flavours" (which are quantum numbers associated with elementary particles like quarks and electrons). Strictly speaking electric charge is a derivative quantum number, defined by calculation from primary quantum numbers via the Gell-Mann–Nishijima formula. --Jayron32 20:22, 30 October 2012 (UTC)[reply]

To answer the experimental question of how you would measure the properties of a quark, the general answer is usually using a particle accelerator. Using high energy particles (often either electrons or protons), it is possible to probe the structure within a nucleus. The details are fairly technical but you look at the ways that different high energy particles interact and scatter off of a nucleon and provided you use sufficiently high energies you can infer that there are three quarks per nucleon as well details of the properties of those quarks such as mass and charge. Dragons flight (talk) 23:34, 30 October 2012 (UTC)[reply]

Once you know (or just hypothesise) that a proton is two up quarks and one down quark and a neutron is one up quark and two down quarks, then the only values of quark charges that give a total charge of e for a proton and 0 for a neutron are (2/3)e for the up quark and -(1/3)e for the down quark (okay, you also have to assume that charges add up linearly, so there are no interaction effects). Of course, you still need experiments to confirm that these hypothetical quarks actually exist. Gandalf61 (talk) 09:57, 31 October 2012 (UTC)[reply]

Strangelet production engine

Strangelet production seems to require a strange combination (pun intended) of a very high energy environment to pop out strange quarks combined with a much lower energy to have a stable atomic nucleus. So in order to properly destroy the world, wouldn't you need a device like the following?

The first stage is a particle accelerator that is carefully tuned to maximize production of negatively charged strange particles. These are then cooled by bremsstrahlung as they are curved over by a magnetic field towards a positively charged foil of heavy elements, say lead or gold. The strange quarks then merge into the heavy atomic nucleus to form a stable strangelet.

The only remaining piece is to fill out the environmental impact statement for a process that can destroy the world. Hcobb (talk) 14:06, 30 October 2012 (UTC)[reply]

Not sure that this would work, you're only going to add a few s quarks to the gold nucleus and that won't lead to spemething stable. Perhaps better to start with some heavy unstable element injected in the accelerator. If it is accelerated to some huge gamma factor, it will have a lifetime in the lab frame that is not too short. Then you can try to put some s quarks in that nucleus. This can't be done by letting it collide with the nucleus using a beam that travels in the opposite direction, as you would then have a deeply inelastic collision that shatters the nucleus to pieces. You need a small difference in velocity, so the beam will travel in the same direction. Count Iblis (talk) 16:34, 30 October 2012 (UTC)[reply]

The article says you need roughly equal numbers of u/d/s, so doesn't that mean you wouldn't want normal matter in the initiating particle? I'd think (but definitely do not know) that you'd want to collide multiple Lambda baryons and/or Sigma baryons (positive and negative might be handy) in a very brief period of time.

That said, I don't get how we're supposed to have come out of the Big Bang as non-strangelets if this instability really existed.

We could use some remedial education/article development on how the Pauli exclusion principle works in baryons. I know that the proton is the most stable because you can't have three up quarks in the low-energy conformation with opposing spins rather than the delta baryon. But why does it work out that way? From more familiar cases with electrons you expect to be able to pair more particles in a lower energy state when they have opposite spins, not the same spin. Would be much appreciated if someone covers this... Wnt (talk) 18:34, 30 October 2012 (UTC)[reply]

I'll attempt to describe how the exclusion principle works in baryons, but someone had better check my work.

Each valence quark can have a spin in any direction, i.e. any value

\alpha |{\uparrow }\rangle +\beta |{\downarrow }\rangle

where

|\alpha |^{2}+|\beta |^{2}=1

, and any color, i.e. any value

\gamma |R\rangle +\delta |G\rangle +\epsilon |B\rangle

where

|\gamma |^{2}+|\delta |^{2}+|\epsilon |^{2}=1

. The most general wave function is a sum (superposition) of arbitrarily many combinations of spin and color of each quark. Fortunately you can simplify this by expanding all the products using the distributive law and summing like terms, which gets you a finite superposition of states where each quark is either spin-up or spin-down and either red, green, or blue. I'll write

|ddu\rangle |{\uparrow \uparrow \downarrow }\rangle |BRG\rangle

for the state where quark #1 is a spin-up blue d-quark, and so on. I'll ignore overall normalization factors like

1/{\sqrt {2}}

.

For spin-½ baryons, the constraints are:

The spin can point in any direction, but since we're free to choose whatever direction is convenient, I'll pick the +z direction. This restricts us to $|{\uparrow \uparrow \downarrow }\rangle$ , $|{\uparrow \downarrow \uparrow }\rangle$ and $|{\downarrow \uparrow \uparrow }\rangle$ terms.
The total spin is ½. Rather than apply the total spin operator to the wave function I'll cheat and use the fact that I know that $|{\uparrow \downarrow }\rangle -|{\downarrow \uparrow }\rangle$ is a singlet state and therefore multiplying it by $\left|{\uparrow }\right\rangle$ , in any of the three positions, gets you a spin-½ state. This works out to the constraint that the sum of the coefficients in front of corresponding $|{\uparrow \uparrow \downarrow }\rangle$ , $|{\uparrow \downarrow \uparrow }\rangle$ and $|{\downarrow \uparrow \uparrow }\rangle$ terms is zero.
The state is color-neutral. This doesn't mean one each of R, G, and B. It means that if you replace $|R\rangle$ with $\alpha |R\rangle +\beta |G\rangle +\gamma |B\rangle$ where α, β, γ are one column of a unitary matrix with determinant 1, and likewise replace $|G\rangle$ and $|B\rangle$ using the other two columns, the result (after multiplying everything out and gathering like terms) is the same as what you started with. Since $\left({\begin{smallmatrix}0&1&0\\0&0&1\\1&0&0\end{smallmatrix}}\right)$ and $-\left({\begin{smallmatrix}1&0&0\\0&0&1\\0&1&0\end{smallmatrix}}\right)$ are unitary matrices of determinant one, this should restrict the color part of the wave function to $|RGB\rangle +|GBR\rangle +|BRG\rangle -|RBG\rangle -|BGR\rangle -|GRB\rangle$ , which I'll call $\psi _{\mathrm {color} }$ .
The exclusion principle: swapping the position of two quarks of the same flavor must multiply the overall wave function by −1. You get a factor of −1 from the color part of the wave function, so this is equivalent to requiring that the spin part be unchanged. There are three subcases:
1. If all quarks are the same type, you get the constraint that the three spin arrangements must have the same coefficient. It's impossible to satisfy this along with the earlier requirement that the sum of those coefficients is zero.
2. If there are two quarks of one type and one of another, say uud, you can only have $|uud\rangle \left(|{\downarrow \uparrow \uparrow }\rangle +|{\uparrow \downarrow \uparrow }\rangle -2|{\uparrow \uparrow \downarrow }\rangle \right)\psi _{\mathrm {color} }$ .
3. If the quarks have three different flavors, there's no constraint from the exclusion principle. This gives you a two-dimensional space of wave functions ( $\alpha |{\downarrow \uparrow \uparrow }\rangle +\beta |{\uparrow \downarrow \uparrow }\rangle -(\alpha +\beta )|{\uparrow \uparrow \downarrow }\rangle$ times the color and flavor parts) which is divided up between the Λ and Σ⁰ somehow. The Λ is a flavor SU(3) singlet which means it's approximately $\left(|uds\rangle +|dsu\rangle +|sud\rangle -|usd\rangle -|sdu\rangle -|dus\rangle \right)\left(|{\downarrow \uparrow \uparrow }\rangle -|{\uparrow \downarrow \uparrow }\rangle \right)\psi _{\mathrm {color} }$ (but not exactly, I think, since the u-d-s symmetry isn't exact). The Σ⁰ is... something orthogonal to that.

Spin 3/2 is much easier. With the spin in the +z direction, we only have to deal with the

|{\uparrow \uparrow \uparrow }\rangle

spin term. The color part is the same as before, and gives us antisymmetry automatically, so for any combination of quark flavors the wave function is simply the flavors times

|{\uparrow \uparrow \uparrow }\rangle \psi _{\mathrm {color} }

.

Is that right? -- BenRG (talk) 19:19, 2 November 2012 (UTC)[reply]

herpes

Wikipedia cannot provide medical advice or diagnosis, please seek medical help

The following discussion has been closed. Please do not modify it.

I am single and still a virgin but I am having genital herpes.Is it possible to have genital herpes without sexual contact?15:40, 30 October 2012 (UTC)106.218.129.139 (talk)

We cannot answer your question because we cannot ourselves confirm that you have herpes or provide you with any assurances. You may have some other dangerous condition. Please seek medical advice.

μηδείς (talk) 15:45, 30 October 2012 (UTC)[reply]

Germline Mutagenesis of Cesium 137 in Plants and Animals

Hi! I live in Oregon and have been seeing lots of strange mutations in the plants around my house recently. A quick search of the internet shows multiple references to this effect, but nothing is mentioned of this under the wiki "cesium 137" heading? Is this something that has been deemed taboo, or could I add a section on it? Thanks! — Preceding unsigned comment added by 98.232.238.213 (talk) 16:22, 30 October 2012 (UTC)[reply]

You need to give more background information. Why do you make a connection to "cesium 137"specifically? If you have reliable sources that "cesium 137" causes frequent mutations in your area then it should probably be included but i highly doubt that. Note that mutations and deformations can have a lot of different reasons and occurs naturally so any source need to be very specific and reliable. If the radiation levels really are high enough that increased mutation rate can easily bee observed the area would probably have been evacuated. I do not even think there are reliable sources for such effects around Tjernobyl or the Kyshtym disaster. Gr8xoz (talk) 16:58, 30 October 2012 (UTC)[reply]

While I'm finding at least some chatter on various internet forums and discussion groups on this topic, I'm coming up kind of short on sources that would be considered reliable by Wikipedia standards. (See as well Wikipedia's notes for scientific sources and sources on medical topics for additional guidance.)

What sources are you considering using to support your proposed edits to the article? TenOfAllTrades(talk) 17:10, 30 October 2012 (UTC)[reply]

Please take photos of the strange mutations and upload them to Wikimedia Commons. Then people can go through one by one and say what they are. Infections can cause many strange changes in the shape of plant growth, so it could be something common and not genetic at all. Wnt (talk) 18:40, 30 October 2012 (UTC)[reply]

This is a good point—infections and parasites are the most likely explanation for most changes you might see. (Check out the wild effect of parasitic wasps on the shape of these acorns.) Heck, straight climate stress (much of North America experienced severe drought conditions this year, for example) can also produce oddly-shaped plants. Even if there have been germline mutations in your garden plants, however, it's difficult or impossible – and rather improbable, truth be told – to link those changes to miniscule amounts of Cs-137 from Fukushima rather than any other cause. TenOfAllTrades(talk) 19:02, 30 October 2012 (UTC)[reply]

Thinking about it, I suppose I should be a little more blunt: radiation mutagenesis is an astronomically unlikely scenario. First, the Chernobyl exclusion zone looks like a nature park - the animals have higher rates of spontaneous abortion, and a few visible mutants have been spotted (run of the mill stuff like albinism that results from breaking a gene) and the plants may be accumulating mutations, but it's not visibly out of whack. Our article Chernobyl disaster effects says that melanized fungi have taken to growing in the building, but what it says is that this is due to natural selection where the fungi have adapted to absorb energy from the radiation. It's a remarkable claim to me (fungi aren't generally known for their photosynthesis!) but the source checks out and appears in a good journal. [3] Second, a contamination on that scale should be detected, as the Chernobyl event was, by various surveillance systems, though I suspect the odds of Homeland Security dropping the ball are better than average. Now, it is possible that a small scale contamination could occur - things happen like a drum looted from a nuclear plant and used to store water, or a medical X-ray source broken open by metal scavengers. But those things usually announce themselves by the imminent death from radiation sickness of whoever is involved. It can kill plants also (the Red Forest) but that shouldn't look like mutations.

That said, your interest is still commendable. While what you see is probably well known, new plant pathogens do get imported into the country all the time, and action against them almost always is delayed until someone living in the area starts asking questions. Wnt (talk) 19:04, 30 October 2012 (UTC)[reply]

The OP did not mention anything about Fukushima. The radio activity in USA originating from Fukushima is ofcourse several orders of magnitude to weak to be a plausible explanation to the observations. In case of a known Goiânia accident like scenario it could be reasonable to speculate in that direction. Why do you think the OP are refering to Cs-137 from Fukushima, have it been discused in the media recently? Gr8xoz (talk) 21:31, 30 October 2012 (UTC)[reply]

If you do a quick Google search for some combination of plant mutations, Oregon, and/or cesium-137, you'll find that the top hits tend to be Fukushima-related forum posts and fringe/conspiracy outlets. If the OP conducted his research starting from an (understandable, but likely incorrect) explicit initial assumption that he had mutations – rather than just parasites, infections, or drought effects – then the results of his "quick search of the internet" are almost certainly going to lead him down the Fukushima Cs-137 garden path. TenOfAllTrades(talk) 17:37, 31 October 2012 (UTC)[reply]

Could an accident like Fukushima happen in the US due to Sandy?

Comploose (talk) 19:45, 30 October 2012 (UTC)[reply]

Sure, but "something could conceivably happen" isn't a terribly compelling statement. The Fukushima Daiichi nuclear disaster was driven by compounding failures aided by human decisions. In short, the reactors shut down (thus losing the power to cool the reactors themselves), the power grid was severed (thus losing the power to cool the reactors from safely operating remote generators), the backup generators went offline (thus losing the last on-site backup), and officials were slow to irreparably flood the reactors after all of the preceding. Additionally, various prior safety violations have since been noted. One can contrive this sequence of effects at functionally any reactor, as even designs that are big on passive safety still have potential failure modes. A massive natural disaster isn't even necessary, though it makes things more likely. Generally, though, I'd expect a hurricane to be less risky than a massive earthquake / tsunami combo. Most notably, there's advance warning, and several nuclear plants did in fact begin implementing cautionary plans as Sandy approached and the water rose. The end result is that you've got a much better case of seeing a result such as that at Browns Ferry Nuclear Power Plant last year. Same reactor type as Fukushima, emergency conditions due to natural disaster (in this case, an EF5 tornado), and unless you live in the area, you've probably never heard of it. Something indeed "could" happen -- but it's unlikely, and as you've now seen, it didn't. — Lomn 21:18, 30 October 2012 (UTC)[reply]

Apparently some have gotten to the first step - shutdowns. [4] Let's hope they still have a way of powering their cooling systems... Wnt (talk) 21:43, 1 November 2012 (UTC)[reply]

There's always at least two backup diesel generators at nuclear power plants -- when the tea kettle shuts down, they provide power to the cooling system. The problem at Fukushima was that they were flooded by the tsunami, rendering them inoperable -- which is not the case here. 24.23.196.85 (talk) 04:56, 2 November 2012 (UTC)[reply]

Airplanes during storms

Why is it a problem if the wind is over 100 mi/h? Airplanes should be able to deal with much more than that, if it's int he right direction. Comploose (talk) 20:05, 30 October 2012 (UTC)[reply]

(ec) In the United States, it is illegal (14 CFR §121.171) to operate an aircraft for commercial transport of passengers beyond the aircraft's rated operating limitations, which include, among other items, a maximum demonstrated crosswind component (14 CFR § 23.1585). If you aren't operating a commercial aircraft for the transportation of passengers, different regulations apply. Part of flying an airplane is learning which rules apply to each situation. Another part of learning to fly is demonstrating sound judgement, which is specifically evaluated by the FAA. A pilot who operates a commercial aircraft in surface winds exceeding 100 mph is neither complying with relevant laws, nor exercising sound judgement for the safe operation of the aircraft. Nimur (talk) 21:32, 30 October 2012 (UTC)[reply]

It depends on what the wind is doing and on what the airplane is doing. In level flight, at altitude, and with generally horizontal winds, an airplane's ability to fly indeed doesn't care too much about windspeed or direction (its ability to get from A to B may be dependent on whether that's a headwind, though). However, turbulence will generally rise with windspeed, and vertical components (sudden updrafts or downdrafts) are also more hazardous -- and storms, as contrasted with the jet stream, are much more prone to such winds. If the airplane is operating near the ground, particularly in takeoff or landing operations, wind-related hazards are magnified. — Lomn 21:28, 30 October 2012 (UTC)[reply]

You are correct - wind speed at altitude can regularly exceed 100mph ground speed, and that's not a problem. What is a problem is when you have wind speeds of that magnitude near the ground, ans the topography of the ground tends to cause turbulence. It's this turbulence which aircraft try to avoid, as abrupt changes in local wind speed or direction (wind shear) can expose the wings to high angle of attack, and make it easy to stall the lifting surfaces. This is especially true when aircraft are configured for landing and takeoff. Also, all commercial aircraft are only certified to take off and land with a certain crosswind component. This is usually sufficiently high that for day to day operation, operators work easily within these limits by choosing runways that orient the aircraft preferentially with the wind, so that if for example the wind is at 30 knots, but you are landing at 45 degrees to that wind, the crosswind component is only 15 knots. As wind speed gets higher near the ground you need runways that are "more aligned" to the prevailing wind, and that runway might not exist locally. — Preceding unsigned comment added by Happymulletuk (talk • contribs) 21:45, 30 October 2012 (UTC)[reply]

As you noted at the end of your question, "if it's in the right direction". A basic part of the problem is that runways aren't often in the right direction for the wind, or vice versa. Nyttend (talk) 23:14, 30 October 2012 (UTC)[reply]

Also note that steady 100 MPH winds wouldn't be as bad as winds which gust up to 100 MPH. It's the variability that's the worst problem. StuRat (talk) 02:22, 31 October 2012 (UTC)[reply]

You may be interested in this huffington post interview[5] with someone who flies into hurricanes for a living. The two issues seem to be a) as people have suggested, turbulance. The hurricane hunters seem to avoid the worst this by hand flying around it using radar as guidance. b) hail. The hurricane hunters use turboprop planes because ice ingestion by jet engines can cause engine damage. Either of these issues would be enough to make a commercial flight an unacceptable risk, not to mention very uncomfortable for the passengers. I also agree with the point that with 100mph winds at surface level you could not be at all confident of finding a runway with an acceptable crosswind component. Equisetum (talk | contributions) 12:11, 31 October 2012 (UTC)[reply]

I may point out, too, that even the Hurricane Hunters have their limits...see this narrative by Dr. Jeff Masters of Wunderground: [6] Ks0stm ^{(T•C•G•E)} 21:42, 31 October 2012 (UTC)[reply]

Your title is "airplanes during storms". For me, as a pilot, the storm would be a greater concern than a steady 100mph wind; if it were truly steady (no major gusts, no turbulence), and not extending to the ground, it would have more of a factor on how fast one could get to the destination than how safely. Even if it were absolutely steady wind, on the ground I could not physically taxi a light airplane to the runway without the airplane being flipped over, risk taking off by accident, etc. If this hypothetical wind were not almost straight down the runway, the aircraft would get blown off the runway during the takeoff run. Whether the wind is on the ground or at a higher altitude could very well be the decision for whether an aircraft flies or stays parked. As for storms, it is not the 100mph wind that's the biggest problem. It is the fact that you have a lot of very very strong winds moving in different directions or at different rates of speed. While flying in a 100mph steady wind would be smooth flying, flying from a pocket of air that is rising at 60 miles per hour to one that is descending at 60 mph would be quite dangerous; if it gets too bad, it can exceed the aircraft's limits to the point of structural failure - i.e. breaking up. While in normal operation it is generally quite difficult to overstress an aircraft to that point, flying into a powerful storm is a situation where that can happen. Beyond that, there is hail and icing to worry about. Lightning is also an obvious risk, but in reality is not actually as big of a safety concern as one might think. It is the unstable winds and the icing that really cause problems in a storm. Flying into a thunderstorm, or, I would imagine, a hurricane is usually a terrible idea. Also, just because something could be technically possible (taking off into a 100mph headwind straight down the runway, or flying through a storm) does not mean that it should be tried. The point where the risks become unacceptable is reached much, much before the situation gets that bad. Each decision of whether to fly or not must take into account the aircraft's ability and the pilot's ability to fly in the given conditions. Falconus^{p t c} 21:30, 31 October 2012 (UTC)[reply]

To sum up this long paragraph: if the wind is at altitude, it may not be a big problem. If the wind is on the ground, the aircraft cannot fly. In a storm, it is the instability of the air along with icing that are so dangerous, not the 100mph wind by itself. In all cases, which aircraft fly under what conditions will depend on what the aircraft can handle and what the pilots can handle. Falconus^{p t c} 21:43, 31 October 2012 (UTC)[reply]

EEG correlates

What are EEG correlates? The Glenn Wilson (psychologist) article refers to them but doesn't explain what they are; a Google search for the term makes me think that they're somehow related to electroencephalogrammes (is that how you spell it?), but all of the pages I found, like the Wilson article, seemed to assume that the reader already knew what they were. Nyttend (talk) 23:21, 30 October 2012 (UTC)[reply]

Yes, EEG means electroencephalogram. "EEG correlates" means electrical signals from the brain recorded using electrodes glued to scalp. Wilson apparently studied how those signals vary as a function of personality type. Looie496 (talk) 23:46, 30 October 2012 (UTC)[reply]

Can a planet orbit inside a red giant for hundreds of thousands of years ?

That's the claim in this article, labeled "Bad Astronomy": [7]. This is a reference for our extragalactic planet article, where the claim is repeated. This sounds quite absurd, to me. StuRat (talk) 23:59, 30 October 2012 (UTC)[reply]

Why is it absurd? You first need to define what "inside" means, which isn't a trivial issue. Then you'd need to see if, once you pick that boundary, if it is possible for an object to survive inside that boundary. I don't dismiss the possibility as outside of the realm of possibility merely on the face of it. I mean, a star is not merely a giant ball of magic gas that instantly annihilates any matter it touches. After all, the Wikipedia articles Formation and evolution of the Solar System and Future of the Earth notes that many models allow for the Earth to be engulfed by the future Red Giant phase of the Sun, and still survive such an event (for any given definition of "survival"). The Future of the Earth article notes that it would take some 200 years for the Earth to be vaporized from inside the sun, one could envision an object in a stable orbit just skimming inside the outer boundary of a Red Giant surviving many millenium in that state. The Formation and evolution of the Solar System article notes that there are models which allow for the long-term survival of the Earth after being engulfed by the Sun. --Jayron32 00:14, 31 October 2012 (UTC)[reply]

There's a huge diff between 200 years and hundreds of thousands of years. Over that time period, it seems that orbital decay due to drag and/or heating due to drag and radiation ought to either vaporize it or plunge it into the core. StuRat (talk) 00:46, 31 October 2012 (UTC)[reply]

Note this article links to [8], which references BP Piscium. The first article says the upper layers of the star were "nearly a vacuum" - the second makes it sound more substantial, so that the planet forced out a disk, but called the orbiting a matter of "years", not "hundreds of thousands of years". Figure out what's known about BP Piscium (start an article would be nice...) and you'll have your answer. Wnt (talk) 00:21, 31 October 2012 (UTC)[reply]

There's nothing merely about a giant ball of magic gas that instantly annihilates any matter it touches. —Tamfang (talk) 04:58, 31 October 2012 (UTC)[reply]

"Instant" ≠ "hundreds of thousands of years", not by any definition. StuRat (talk) 05:00, 31 October 2012 (UTC) [reply]

Would you accept "relatively instantly"? -- Jack of Oz ^[Talk] 23:40, 31 October 2012 (UTC) [reply]

The boiling point of iron is 3134 K. The surface of our sun is only 5778 K. The surface temp of Betelgeuse is 3140 K. Given the very minor difference in temperature and the extremely low density of the stellar atmosphere, a period of hundreds of thousands of years seems entirely reasonable. I am sure some math wiz can figure out how long it would take to boil away an earth's mass of iron given its heat of vaporization and a given temperature, beginning surface area, and density. That kind of calculus is way beyond me though. Try the math desk, they should find it easy. μηδείς (talk) 19:28, 31 October 2012 (UTC)[reply]

You're neglecting that the planet is orbiting at a speed of tens or hundreds of thousands of miles per hour. Slamming into even diffuse gas at those speeds generates a heck of a lot of heat. Also, the planet wouldn't remain at the surface for long, as this friction would also cause the orbit to decay. StuRat (talk) 20:01, 31 October 2012 (UTC)[reply]

That would have to be taken into account but the "near vacuum" description makes me wonder how strong the effect. I don't have the facts or the math. μηδείς (talk) 22:49, 31 October 2012 (UTC)[reply]

A planet has orbital angular momentum of $L=M\,R_{orbit}\,v={4\pi \over 3}\rho _{planet}\,R_{planet}^{3}\,{\sqrt {GM_{star}R_{orbit}}}$

The drag of the star on the planet imposes a torque approximately $\tau \approx {1 \over 2}A\,\rho _{star\,edge}\,v^{2}\,R_{orbit}$ $={1 \over 2}\pi \,\rho _{star\,edge}\,R_{planet}^{2}\,G\,M_{star}$

It follows that the decay timescale is roughly $T\approx {L \over \tau }={2 \over 3}{\sqrt {R_{orbit} \over GM_{star}}}{\rho _{planet} \over \rho _{star\,edge}}R_{planet}$

For solar mass star and an Earth-like orbit, this comes to $T\approx (2\,{\text{minutes}}){\rho _{planet} \over \rho _{star\,edge}}$

If we expand the sun to 1 AU, it's mean density would be about 1.5×10⁻⁸ times the density of the Earth. That implies a lifetime of about 270 years, not far different from the 200 years mentioned above. However, that's the mean density of the star. The density profile of a star can be approximated as $\rho _{star}(r)\approx 2.5{\overline {\rho _{star}}}(1-(r/R_{star})^{2})$ . In particular, the density equals the mean density around $r\approx 0.8R_{star}$ . If you want the density that the planet is flying through to be 1000 times lower (and hence allow it to last 1000 times longer), that implies $r\approx 0.9998R_{star}$ .

Kinematically, these calculations suggest that it is possible that one could have an orbit located just right so that the planet is barely inside the outer limit of a Red Giant and thus able to persist for hundreds of thousands of years before it decays. However, the orbital positioning that would make such a thing possible is quite narrow. If planetary orbits were randomly distributed circles, there would be only about 1 chance in 5000 of having an orbit in the right position to survive that long.

So far I've neglected the possibility of the planet boiling away. Given the low thermal conductivity of rock, I don't think you'd see much ablation relative to total mass over 100 years. The surface would be a melting / boiling mess, of course, but planets are awfully big and even at 5000 K, propagating the heat would seem to take a long time. Heat the surface to 5000 K, and the back of my envelope suggests that it would still take more than 200 years to remove the first 50 km. Given 1000 times longer you'd probably kill much more of the planet but not necessarily all of it.

All in all, having an object survive hundreds of thousands of years inside a red giant seems plausible, though it would be an unlikely event. Dragons flight (talk) 02:00, 1 November 2012 (UTC)[reply]

Yes, the mean density is nowhere near correct. The density at the outer fringe of an expanded star the mas of the sun will be far, far, far , far less. μηδείς (talk) 02:53, 1 November 2012 (UTC)[reply]

I included a functional form for the density versus radius. At 98% of the way to the surface the density is still 10% of the mean density. Of course the density goes to approximately zero at the very end, but the fall off is pretty abrupt. If you want to invoke special pleading, one can posit almost arbitrarily small drag (which might be necessary to get a planet to last 100s of thousands of years), but for planets that are non-trivially inside the surface, the density is generally a non-trivial fraction of the mean density. Dragons flight (talk) 04:40, 1 November 2012 (UTC)[reply]

Nice calcs. But keep in mind that even if the planet starts out in this perfect orbit, it won't stay there, due to drag. StuRat (talk) 03:00, 1 November 2012 (UTC)[reply]

Feel free to do an integration rather than a constant drag approximation, but I doubt you'll see the dynamical timescale change by more than a order of magnitude. Dragons flight (talk) 04:40, 1 November 2012 (UTC)[reply]

N.B. this question was addressed (same person!) at [9] Wnt (talk) 04:31, 5 November 2012 (UTC)[reply]