Science desk
< March 21	<< Feb \| March \| Apr >>	March 23 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 22

Tides

I was reading my textbook which tried to evaluate the height of the tides. To do this, they borrowed what was apparently a method devised by Newton, and "pretend[ed] that two wells full of water run from the surface of the Earth to the center, where they join. One is along the earth-sun axis, and the other is perpendicular." I'm having trouble seeing how this can in any way help in determining the equilibrium height of tides. After all, the ocean runs along the surface of the earth, not through it. 173.179.59.66 (talk) 00:50, 22 March 2010 (UTC)[reply]

Yes, I agree that this sounds like an unreliable model. While tides do affect solids as well as fluids, they deform far less than the fluids do. StuRat (talk) 02:40, 22 March 2010 (UTC)[reply]

Stu is misunderstanding the thought-experiment. There's no assumption that the solids will deform.

Suppose you have two lakes that are at somewhat different levels, and you dig a canal connecting them so that water can flow freely between them. If the canal and the lakes are deep enough that nothing empties out, then the water will flow from the higher-surface lake to the lower until the level is the same in both lakes, right? Okay, now instead of the canal, dig an underground tunnel connecting the two lakes. The water will still flow in the same way, leaving the lakes at the same level, right?

Now imagine that you have two lakes, one at the South Pole (which thanks to global warming is now above the freezing point) and one at the Equator. Join them with a tunnel through the Earth and the water will still try to flow to the same level, right? That's true even if the tunnel down goes all the way straight down to the center of the Earth and makes a right-angle turn to rise back to the surface. (Pretending that you could actually build such a tunnel and that water would remain liquid in it.)

Well, that's the layout that the thought-experiment is about. Two deep wells meeting at the center of the Earth. And the point is that when we say the water reaches the "same level", the idea of what is the same level is affected by the tidal force. Because the water at the equator is pulled toward the Sun if the well is pointing toward the Sun, then it will rise higher to reach what "feels" to it like the same level.

And similarly with the ocean -- it tries to reach the same level everywhere, but its idea of what is the "same" level is affected but the tidal force. And what's more, it's affected in exactly the same way as in the thought-experiment configuration with the wells or tunnels. So the simple configuration described in the experiment tells you how high you could expect the ocean tide to rise, if things like friction, and the shape of the seafloor in coastal areas, did not have a big effect on it.

--Anonymous, 05:51 UTC, edited 18:55, March 22, 2010.

Okay, that makes a bit more sense. But who's to say that the effect of having the oceans connected along the surface will be the same as have the two lakes (at the equator and at the South Pole) connected underground? Is there a more direct way of calculating the tidal height? 173.179.59.66 (talk) 20:34, 22 March 2010 (UTC)[reply]

If they are connected than water can flow between them, so gravity will make sure the weight AKA the height of the water will be exactly the same in both. You can experiment with it - take a long transparent tube, fill it with water, and raise/lower the ends and you will see the water level will always be exactly at the same height on both sides. (It's a great tool for surveying, and it's odd we have no article on it.) Ariel. (talk) 07:50, 23 March 2010 (UTC)[reply]

How did the Atlas ICBM navigate?

How did the SM-65 Atlas ICBM navigate, i.e. how did it know where it was and where to go? Was it a) remote-controlled, or b) steering autonomously, or did it c) simply follow a fixed pre-programmed flight schedule without any adjustments? In case of a) or b), how did it know its location in the absence of GPS? Did it have a camera built in, maybe to observe the stars? Or did the ground stations follow it by radar and use option a)? AxelBoldt (talk) 03:16, 22 March 2010 (UTC)[reply]

Our article describes the guidance of the Atlas D variant as radio-based (an inertial system with ground-supplied course correction information), and the E and F variants as having true inertial guidance systems. This page (see 'System Operation') has more detailed information about the specific navigation equipment. TenOfAllTrades(talk) 03:34, 22 March 2010 (UTC)[reply]

Just a note that celestial guidance was not around until the 1970s (used first in the UGM-73 Poseidon). Missile guidance has more general information on technologies developed to get the missile from launch to target. Note that the Atlas D had a CEP of 1.8 nautical miles and Atlas E-F had 1.0. So their accuracy was not exactly pin-point (though still impressive for that generation of missile), but when you have a 4 Mt warhead on the tip of it, it doesn't have to be. --Mr.98 (talk) 14:57, 22 March 2010 (UTC)[reply]

I thought the Snark missile used celestial guidance too... 24.23.197.43 (talk) 23:16, 23 March 2010 (UTC)[reply]

Cloud base measurement

Here's a question for all the weathermen in here: Suppose that you have to measure the cloud base at your location in support of an important aerial mission. Suppose also that you don't have an operating ceilometer (what the hell, no article?!) at your disposal, so you have to find another way to do this. Further suppose that it's nightttime, so you can't use any method that relies on visual observations by ambient light alone. Given these conditions, how would you do this? Thanks in advance! 24.23.197.43 (talk) 03:37, 22 March 2010 (UTC)[reply]

I before E fixed your link :). Well our Cloud base article says The height of the cloud base can be estimated from surface measurements of air temperature and humidity. . But stuffed if I know how. At least it sounds possible. Vespine (talk) 03:56, 22 March 2010 (UTC)[reply]

Unless this is what a celiometer is, you could shine a laser at the cloud base and time how long the light takes to get back. --The High Fin Sperm Whale 03:57, 22 March 2010 (UTC)[reply]

Unfortunately that's exactly how a ceilometer works, so this method is not available in my case. Also, what is a ceiling projector and how is it used? I've got a vague feeling that it might be one of the solutions to this problem. 24.23.197.43 (talk) 05:06, 22 March 2010 (UTC)[reply]

Actually just google Cloud Base Calculator, i've just found a bunch. Vespine (talk) 04:23, 22 March 2010 (UTC)[reply]

Oh, that's how it's done -- you shine a ceiling projector at the cloud base and measure the bright spot's angle of elevation with an alidade or theodolite, and then take the tangent of that angle to get the cloud base height. Well, thanks anyway :-) 24.23.197.43 (talk) 05:13, 22 March 2010 (UTC)[reply]

Sort of like the way the Dambusters measured the altitude of the bombers above the lake water during Operation Chastise, only pointing the other way. --Anonymous, 05:54 UTC, March 22, 2010.

Exactly. 24.23.197.43 (talk) 06:03, 22 March 2010 (UTC)[reply]

Resolved

-Pete5x5 (talk) 15:54, 22 March 2010 (UTC)[reply]

Resolved? Nobody mentioned base reflectivity - which is a standard data product output from a NEXRAD or a WSR RADAR! We have Composite reflectivity, which explains more detail. Note that there may be a cone of silence (i.e., the airfield may be so close to the RADAR that the RADAR can't image the cloud base at such short range), so in practice, a combination of observations are used. Nimur (talk) 16:23, 22 March 2010 (UTC)[reply]

Thanks for the info, but in my case it's a remote outpost, so there's no WSR radar either. I think I'll stick with the ceiling projector for now. :-) Clear skies to all of you 146.74.230.104 (talk) 20:08, 23 March 2010 (UTC)[reply]

Number of conjugated double bonds and photon energy absorbed

The wiki page says, "With every double bond added, the system absorbs photons of longer wavelength (and lower energy), and the compound ranges from yellow to red in color. Compounds that are blue or green typically do not rely on conjugated double bonds alone."

I thought it should be the opposite, because the more double bonds we have in a conjugated system, the more overlapping p-orbitals there are. So the system can absorb or bare with a greater amount of energy (The electrons have much more space to travel around when they are excited, so the system can take in more energy). It doesn't make an intuitive sense as to why a conjugated system with more double bonds absorbs photon of lower energy. —Preceding unsigned comment added by 70.68.120.162 (talk) 04:34, 22 March 2010 (UTC)[reply]

Because it takes less energy to get those conjugated electrons racing around back and forth? 24.23.197.43 (talk) 05:16, 22 March 2010 (UTC)[reply]

More space for the electrons means higher uncertainty about their positions which allows for lower uncertainty about their ~~momentums~~ momenta which permits lower excited states which allows absorptions of lower energy photons. Dauto (talk) 05:40, 22 March 2010 (UTC)[reply]

Right. Personally, I prefer to visualize this a bit differently, though. Imagine you have a potential well with some discrete energy levels. If you connect two wells like that, each level will split into sub-levels. The more wells you connect, the more components the original levels split into, and the broader the absorption line (band) becomes. Benzene is transparent (absorbs UV only), tetracene is orange (absorbs UV and green/blue), and graphite & amorphous carbon absorb UV and any wavelength in the visible range. The two explanations (Dauto's and this one) are not contradictory, of course. Hope this helps. --Dr Dima (talk) 07:34, 22 March 2010 (UTC)[reply]

I'm really confused. "More space for the electrons means higher uncertainty about their positions" Okay. "...which allows for lower uncertainty about their momentums" I don't understand this. "...which permits lower excited states which allows absorptions of lower energy photons." Oh my goodness. It doesn't make any sense at all.

"Imagine you have a potential well with some discrete energy levels. If you connect two wells like that, each level will split into sub-levels." Why? "The more wells you connect, the more components the original levels split into, and the broader the absorption line (band) becomes." I just don't see it intuitively. "tetracene is orange (absorbs UV and green/blue)" Why is tetracene orange if it absorbs green/blue? Shouldn't its color be some mixture of colors that are not green/blue, since objects' color is determined by the colors they don't absorb? For example, if something does not absorb only green, then its color will be green".142.58.129.94 (talk) 17:25, 22 March 2010 (UTC)[reply]

From the point where your confusion sets in, it seems that you have too little knowledge of quantum mechanics to understand the full explanation (which is no shame, but a bit more than can be remedied in a refdesk exchange). So here is a non-quantum half-story: think of the sequence of conjugated double bonds as a conductor that works like an antenna to pick up electomagnetic radiation. Now, the longer an antenna is, the longer is the wavelength it is tuned to. So more bonds means a longer antenna, which means that the antenna can pick up longer waves. (In neither case does the antenna reach an entire half-wave of visible light, but that is where quantum stuff comes in).

And yes, orange is "a mixture of colors that are not green/blue/UV". –Henning Makholm (talk) 17:45, 22 March 2010 (UTC)[reply]

It still doesn't make intuitive sense, but I'll just take what you said for granted. I'm concerned that I have "too little knowledge of quantum mechanics". I've taken college intro physics, chem, calculus, bio and organic chem. Are intro physics/chem supposed to equip you with enough knowledge of quantum mechanics? If not, what can I do now to get such knowledge? Taking an upper level course is not an option since I don't want to risk my GPA. —Preceding unsigned comment added by 142.58.129.94 (talk) 18:42, 22 March 2010 (UTC)[reply]

Conjugation/resonance should have been covered in orgo (electrons delocalizing along the whole system rather than being in specific alkene-locations). The specifics of UV/vis is often not. But the relationship of wavelength to energy should have been covered in physics (longer wavelength is lower energy). So now you have electrons resonating in a longer space, which is analogous to a wave, and therefore lower energy. DMacks (talk) 18:50, 22 March 2010 (UTC)[reply]

Sure you've heard of Heisenberg's uncertainty principle that states that the higher the uncertainty in the position is the lower the uncertaity in is momentum can be, haven't you? Dauto (talk) 18:52, 22 March 2010 (UTC)[reply]

If you really want to understand thaose things you really need to take those advanced courses. There is no royal road to science. Dauto (talk) 18:52, 22 March 2010 (UTC)[reply]

Radiocarbon dating

Near my home in California a fair number of old pictograms are to be found. One set of them is rather controversial at the moment, with one faction claiming the drawings are of considerable antiquity (one to two thousand years), while another faction claims they were drawn by local kids in the 1930s. Radiocarbon dating has been discussed; how would that work with stuff painted on the boulders? What would they test for decay? --jpgordon^{::==( o )} 05:46, 22 March 2010 (UTC)[reply]

They would test the material used to make the pictograms. With powdered plant pigments or charcoal paintings, radiocarbon dating may work (depending on human/environmental contamination). If they are chalk scrawlings, then radiocarbon would be next to useless. caknuck _° needs to be running more often 07:27, 22 March 2010 (UTC)[reply]

Looks like red ochre to me. --jpgordon^{::==( o )} 17:17, 22 March 2010 (UTC)[reply]

Carbon dating won't work with red ochre. If there are other metals present (like cobalt), they might be able to figure something out using the radioactive isotopes of those; other than that (since radioactive iron isotopes aren't present in sufficient quantity), I don't know how they can do this... 146.74.230.104 (talk) 20:14, 23 March 2010 (UTC)[reply]

The requirement is not just a decaying isotope, but also some mechanism by which the composition differs as of the moment it is used as an art medium vs the source. For example, I'm not sure that the cobalt in the paint would be different from cobalt found in whatever ore the artist presumably got it from. Radiocarbon dating works because the starting isotopic mixture is known at the time of death and only decays from there whereas until death the carbon ratio is constantly being re-set. See radiometric dating for lots of examples. DMacks (talk) 20:39, 23 March 2010 (UTC)[reply]

Sabbath feeling

Is there a name for it when you somehow got the feeling that you shouldn't do any (specific kind of?) work? Quite unreligiously I get this feeling whenever I have to do my tax declaration, even when I'm sure I'll get money back. I can see it sometimes in other people, too. In some fantasy/fiction stories there are alien races with a limited supply of "vital energy" and every time they make a decision, it decreases.

I'm sure there is a name for that, perhaps an article. 95.115.136.225 (talk) 09:33, 22 March 2010 (UTC)[reply]

Procrastination? (This really belongs on the language desk - it's hardly a scientific matter - maybe I'll move it...tomorrow). SteveBaker (talk) 10:28, 22 March 2010 (UTC)[reply]

Religious guilt? 67.243.7.245 (talk) 13:14, 22 March 2010 (UTC)[reply]

I used to get that, but it stopped when I started dressing more apropriately for the cold weather, that lack of energy and feeling it gradually draining away. Or is it more like the way I feel when I've been working almost non stop all day for a week or two. I call that work tireness, though I am sure that isn't the official name, if there is one.148.197.114.158 (talk) 15:25, 22 March 2010 (UTC)[reply]

Well, it could even be simple conditioning. Perhaps you've conditioned yourself mentally for work aversion (Gandalf61) on sabbath days. 67.243.7.245 (talk) 23:10, 23 March 2010 (UTC)[reply]

Work aversion ? Gandalf61 (talk) 15:53, 22 March 2010 (UTC)[reply]

Theory of relativity

Would somebody kindly explain Theory of Relativity (and related concepts) in simple layman's terms? I have been struggling to grasp the concepts for some time now. Everything I can find is written so complexly that I understand next to nothing. Also, something about String theory would be useful, I have had mentioned that in conversation few times, and it's been sufficiently awkward. 203.206.49.48 (talk) 15:59, 22 March 2010 (UTC)[reply]

Have you looked at Theory of relativity? It is hard to explain this physical theory in simpler terms than the article uses. What specific concept(s) are you struggling with? Nimur (talk) 16:26, 22 March 2010 (UTC)[reply]

You should take one step at a time. If you're struggling with relativity, than you should concentrate on that and leave string theory for later. Relativity is vast subject. You probabily should start by reading time dilation, length contraction, and relativity of simultaneity. Let us know if you find those articles hard to read. Dauto (talk) 16:29, 22 March 2010 (UTC)[reply]

If you're interested in the whole gamut of relativity and string theory, I'd recommend getting one of the many popular books about it. Brian Greene's The Elegant Universe (1999) is well-known for its accessibility and readability in explaining relativity, quantum theory, and string theory. It's not totally up to date anymore, but at the level of your interest it is probably one of the better launching points. I find Wikipedia is a little spotty on things like this—its topical coverage swings between the exceedingly general and the exceedingly technical. --Mr.98 (talk) 17:05, 22 March 2010 (UTC)[reply]

Out of the articles cited above you probabily should start by making sure you understan the section time dilation#Simple inference of time dilation due to relative velocity. Dauto (talk) 18:33, 22 March 2010 (UTC)[reply]

Of course, we also have the of the article.~A H 1^(TCU) 23:18, 22 March 2010 (UTC)[reply]

Simple English Wikipedia has an article on it: simple:Theory of relativity. — DanielLC 23:34, 22 March 2010 (UTC)[reply]

DanielLC, you stole my input!!! ~A H 1^(TCU) 02:04, 24 March 2010 (UTC)[reply]

As has been brought up many times on the Science Desk before, Simple English means that the vocabulary is reduced and the sentences are short. However, this does not always improve comprehensibility for complicated topics or unusual conceptual ideas - especially the non-intuitive modern physics ideas. Nimur (talk) 01:20, 23 March 2010 (UTC)[reply]

There's an Introduction to special relativity article and an Introduction to general relativity article, but I haven't read them so I don't know if they are helpful. Rckrone (talk) 03:56, 23 March 2010 (UTC)[reply]

Perhaps it would help to read Galilleo's theory of relativity first, then move on to the more advanced versions. Perhaps it would also help to learn everything that went before as well. This is one of the problems with science now, it takes so long to learn everything that has already been done, before anyone can start work on adding to it. 148.197.114.158 (talk) 08:38, 24 March 2010 (UTC)[reply]

Angular frequency in x-ray crystallography

I've already read over the wiki page for angular frequency, but how is angular frequency different from normal frequency and what does it mean in terms of photons when used in x-ray crystallography? How is sin(2pie*x) different from sin(x)? —Preceding unsigned comment added by 142.58.129.94 (talk) 16:25, 22 March 2010 (UTC)[reply]

It is merely a matter of units - are the angles being measured in terms of radians or cycles? The conversion is simply a multiplication by a constant scale factor; the difference can be summarized by saying that angular frequency yields fewer messy terms in the algebra; while frequency in cycles (or hertz) is more intuitive and works better for engineers or scientists who are reading off results from a machine calibrated that way. Nimur (talk) 16:29, 22 March 2010 (UTC)[reply]

Emission (Fluorescence) in Polar and hydrophobic environments

Emission energy decreases (i.e. emission wavelength increases) in polar environment ("red-shift"), and it's the opposite in hydrophobic environment ("blue-shift"). Is that because, in polar environment, the molecules stabilize the excited molecule and help dissipate some of its excited energy through non-radiative means, thus decreasing the amount of energy emitted? —Preceding unsigned comment added by 142.58.129.94 (talk) 16:26, 22 March 2010 (UTC)[reply]

I haven’t had time to read these two through but they may have some answers.Vibrational transition Molecular electronic transition. Just curious: Why are you asking?--Aspro (talk) 18:50, 22 March 2010 (UTC)[reply]

Note: This is presumably a continuation of Wikipedia:Reference desk/Archives/Science/2010 March 11#Fluorescence in polar and nonpolar environments. DMacks (talk) 18:52, 22 March 2010 (UTC)[reply]

Thanks for that Dmacks. I am out of my depth on this. However, page 21 of this pdf [1] talks of “Energy loss and quenching.”The Wikipedia articles to look at, are I suppose: Quenching (fluorescence) and a radiationless mechanism known as Förster resonance energy transfer & Vibrational transition & Molecular electronic transition. However, there must be a simpler way of explaining it but I suppose the answer looks if its going to be ‘yes’ and the pdf may make the bumbling prof’s mumblings more intelligible. . . or on the other-hand it might not.--Aspro (talk) 19:59, 22 March 2010 (UTC)[reply]

First thing to note, what you are describing are rules of thumb, not rules of physics. The presence of a solvent may or may not alter the fluorescence, and in some cases may even act in a way that is opposite to what you describe. That said, I'll try to explain why the rules of thumb make sense. The electronic ground state of most large molecules exhibit a high degree of symmetry and a low degree of polarization. By extension, most electronic excited states are more polarized than the ground state. Since fluorescence emission is associated with the transition from a meta-stable excited state to the ground state (usually) the question is really, how is the energy of that transition affected by interactions between the medium and changing degrees of polarization. In a polar environment, the solvent rotates in response to polarization in its neighborhood, hence attracting those local charge separations. The result is that it is generally easier (takes less energy) to reach a polarized excited state of an initially non-polar molecule than it would be to reach the same state in a non-polar solvent. If you reduce the energy required to reach the first excited state, then you reduce the transition energy. Hence the light associated with that transition will be red-shifted. Dragons flight (talk) 05:42, 23 March 2010 (UTC)[reply]

Is E2 (17beta-estradiol) a dioxin?

Is 17 beta-estradiol a form of dioxin? —Preceding unsigned comment added by ValkyrieKnight (talk • contribs) 20:29, 22 March 2010 (UTC)[reply]

Estradiol does not appear to have the dioxin (chemical) structure. DMacks (talk) 20:36, 22 March 2010 (UTC)[reply]

No, it's not a dioxin, but it is a steroid (specifically a sterol). – Clockwork Soul 21:01, 23 March 2010 (UTC)[reply]

Orbiting junk

I was discussing with a mate about the future civilisation of super-intelligent insects, and the World Aero-Space Program (WASP) gets underway and they get to Mars. What are the chances they'd find the junk (i.e. the Spirit and Opportunity rovers, and so on) we'd left there? Also, I'm sure I heard that our own artificial satellites are forever having to move around to avoid floating space debris. Wouldn't that space junk be just as moch of a problem for an arthopod's satellite TV and permanent space station? Orbits have a tendency to decay...so...would it all be gone by then? Vimescarrot (talk) 21:08, 22 March 2010 (UTC)[reply]

Oh...And I'm sure this question has been asked before, but I couldn't find it in the acrhives...Will keep looking, though. Vimescarrot (talk) 21:08, 22 March 2010 (UTC)[reply]

And the time-scale we had envisaged for the evolutionary processes needed to make this possible was 20 million years from the breakdown and eventual extinction of the human race. --KägeTorä - (影虎) (TALK) 21:13, 22 March 2010 (UTC)[reply]

They would only find the rovers, etc., if they were doing Martian archaeology. They would be buried in dust by then. The various things we've landed on the Moon (which has no atmosphere, so the dust only moves around when there are meteor strikes) would probably still be there, though. Objects in Low Earth Orbit will decay due to atmospheric drag within a few years, certainly within the millions of years you're talking about. Objects in Geosynchronous orbit (those are the two orbits we have put significant amounts of things in) may well still be around in millions of years (over that time scale, all kinds of things will affect them, but I can't think of anything that would guarantee them to be gone). --Tango (talk) 22:05, 22 March 2010 (UTC)[reply]

Kessler syndrome may reduce the big bits to bits small enough for the solar winds to blow much of the orbiting junk away.The kinetic energy at orbit speeds can exceed what is required to vaporise the material.--Aspro (talk) 22:18, 22 March 2010 (UTC)[reply]

Spirit is likely to be found: it's stuck on a fairly exposed ridge, so 20 million years from now, it'll have some interestingly-shaped sand formations built up on and around it, but it'll still be an obviously-artificial object. Opportunity is currently wandering around on a dust plain, looking at interesting craters. Most likely, it'll wind up stuck in a large crater somewhere, where it will be buried over time. --Carnildo (talk) 01:06, 23 March 2010 (UTC)[reply]

Looking at images like these: [2][3][4] One can easily find exposed rocks that are relatively small. Unless those formations happen to be geologically recent (which seems unlikely, but not impossible), then one would have to conclude that it takes a very long time for dust buildup to bury objects on Mars (at least for some sites). In locations where rocks stay exposed on Mars for millions of years, then I would say it is likely that the rovers would similarly stay exposed. Though the details probably do depend on where the rovers end their lives. Dragons flight (talk) 07:15, 23 March 2010 (UTC)[reply]

Even if they do not get buried, they can be sandblasted into nothing over millions of years of martian dust storms. Googlemeister (talk) 13:28, 23 March 2010 (UTC)[reply]

Wikipedia:Reference desk/Archives/Science/2010 March 22

Contents