Wikipedia:Reference desk/Archives/Science/2008 November 5

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November 5

Do Angel fish come from Africa or South America?

I knew the commonvariety angel fish is a member of the cichlid family, but thought it to be of the African cichlid variety. In watching an informational TV show, I heard theirs were from the Amazon, not the Nile or, more likely, one of the 3 lakes of Africa where nearly all the available, more common variety Africans in stores can be acquired today.

Question: Do Angel fish come from Africa or South America?Vaylem (talk)

Hi Vaylem, the freshwater angelfish article is here (not to be confused with the Marine angelfish group. The intro makes it "All Pterophyllum species originate from the Amazon River basin in tropical South America." Read on, I might have missed something, Julia Rossi (talk) 08:31, 5 November 2008 (UTC)

Arrow shields

I seem to recall seeing in some depiction of medieval warfare a portable wooden wall used by sieging armies to protect against archers (especially when they were in a protected position and could not simply be attacked). Assuming that I didn't dream it, what are these implements called and where might I find more information about them? --Tardis (talk) 02:11, 5 November 2008 (UTC)

No, you didn't dream them. They come in many forms, with various names, including tortoise, cat, weasel and sow (the last giving rise to the famous quote "the sow has farrowed" when the Scots destroyed the English sow at Berwick in 1319). Some stand alone, others form part of an engine (such as a ram). Siege engine offers a poor start. A web search offers a lot of disinformation as well as good stuff, but Bradbury's Companion is a good start. Minor info can be found in various books such as this. Konstantion Nossov's Ancient and Medieval siege Weapons ISBN 186227343X is good, if you can get hold of a copy. Gwinva (talk) 03:09, 5 November 2008 (UTC)

Mantlet has some nice diagrams, but not much else. Clarityfiend (talk) 03:47, 5 November 2008 (UTC)

There was a mention of such a shield used in more recent times in the movie "Treasure of the Sierra Madre". A gang of Mexican bandidos started to build one so they could attack three American prospectors who were holding them off with gunfire. I forget what the shield was called, but it was made of logs. The bandidos had to vamoose before they completed the shield because the Federales (Mexican fedaral police) arrived. The actual appearance of the shield was not shown. It's a good B/W movie, well worth seeing, though I thought the ending was weak.

["Oven logic"] and microwave power levels

Inspired by the above-linked page on the TV Tropes Wiki: can one assume that a microwave-cooked item intended to be cooked at N minutes at 1/M power would cook equally well in N/M minutes at full power? If not, what factors in the microwave's working or the food itself make the difference? 69.107.248.192 (talk) 03:17, 5 November 2008 (UTC)

I expect there will be a clear failure of time-intensity reciprocity when the duration is too long. (This is true of vision and of photography). Edison (talk) 06:05, 5 November 2008 (UTC)

If the only method of heating was via the excitation of polar molecules (water and carbohydrates) by the microwave, then you may have a case. However, lots of heat is transfered via other methods, principly conduction within the food you are cooking. For some foods, cooking at longer times under lower power settings is indeed more efficient than cooking at shorter times at full power, since the process of conduction can be rather slow. Also, there is a difference in how the power is applied. Many microwaves don't simply emit a constant energy at lower power, what they do is oscilate between on and off states given the power setting. Thus, at 50% power, the microwave may run at full power for 10 seconds, then off for 10 seconds, and so on, while at 75% power, they may run on for 15 seconds and off for 5. Some do have variable power magnetrons, and can directly apply lower power at a constant rate, so the type of microwave oven will greatly effect how it heats the food at different power settings. --Jayron32.talk.contribs 17:25, 5 November 2008 (UTC)

The problem is with the evenness of cooking. The reflection of the microwaves off of the metal lining of the oven causes interference of the waves - and that means that the cooking is rather uneven. If you cook for a short period on full power then the energy cooks some parts of the food too much and other parts too little. Running for a longer period at lower power has two benefits: Firstly the turntable rotates more times - resulting in more opportunities for every part to be moved through some of the hotter regions - secondly (more importantly, I suspect) is to allow heat from the hotter areas to be conducted through to the colder areas. This is especially important when defrosting food because ice is a good insulator and the heat doesn't flow easily from the hot, watery, melted bits into the solid, frozen bits. Hence, defrost at low power - and (in most ovens) with frequent stops and starts to allow time for the food's temperature to equalise before zapping it some more. You can easily demonstrate this 'patchiness' by either turning off the turntable - or flipping the turntable upside-down so it doesn't rotate anymore - then get some of those chocolate chips that you'd put in cakes or cookies and make a neat little row of them across he middle of the oven. Zap them (at full power) for maybe 20 seconds and then quickly look inside to see what what happened. You'll find that at intervals of about 6 centimeters, there are chips that didn't melt - and in-between there are chips that are just a puddle of chocolate. This is cool because at 12.5 GHz (the frequency of the microwaves) - the wavelength of the radio waves is 12cm. The turntable helps that a bit - but the food close to the center of the turntable doesn't move around much (imagine those 6cm hotspots in a stationary pattern) so it's not a perfect solution. Where possible, I always put things I'm cooking as far from the center of the turntable as I can - so it cooks more evenly. But cooking on lower power for more time works well too. SteveBaker (talk) 17:58, 5 November 2008 (UTC)

Agree with the conduction argument, which is why defrost settings tend to be low power: the microwaves only heat the outer few mm of the object, and at high power, the outside would be blackened by the time the inside had thawed. As well as this conduction, some of the chemical processes involved with cooking (boiling off, caramelisation, and others) depend on time as well as energy delivered. If you apply N watts of power to a beaker of water for a minute, you will not boil off as much of it as if you applied 60N watts for a second. —Preceding unsigned comment added by DewiMorgan (talk • contribs) 20:20, 8 November 2008 (UTC)

Violation of conservation of mass in positron emission???

In positron emission, a proton decays into a neutron, electron, and neutrino. Since the mass of a proton is less than that of a neutron, does that mean that energy is converted into mass in the reaction? Beta decay#β+ decay says that some of the binding energy goes into converting a proton into a neutron, but that's not really what I'm looking for. So, does the extra mass result from the conversion of energy into mass? --70.242.149.190 (talk) 03:44, 5 November 2008 (UTC)

an isolated proton is essentially stable. It does not "decay" into neutron, positron, and neutrino, unless you provide energy for the process. Even a proton plus an electron will not combine into a neutron plus neutrino without additional energy being provided. Indeed, if $\mathrm {p} ^{+}+\mathrm {e} ^{-}\rightarrow \mathrm {n} +{\nu }_{e}\,$ was occurring spontaneously, most of hydrogen in the Universe would have long decayed by electron capture; which, luckily, is not the case. In a beta-plus decay (positron emission) and in electron capture processes the required energy is supplied by the other constituents of the decaying atom. --Dr Dima (talk) 08:48, 5 November 2008 (UTC)

Short answer: yes. Axl ¤ [Talk] 21:48, 5 November 2008 (UTC)

The strength of a monkey

How strong is a gorilla or a chimpanzee? That's a very generic question, I know, but could you still give me some idea? What is the highest recorded level, the lowest recorded level of strength? Who holds those records? How strong are they compared to strong, very well physically developed humans? Thanks. —Preceding unsigned comment added by 24.7.36.125 (talk) 07:20, 5 November 2008 (UTC)

The article Chimpanzee mentions: "the average chimpanzee has over 5 times the upper-body strength of a human male". In Gorilla "Adult males range in height from 165-175 cm (5 ft 5 in – 5 ft 9 in), and in weight from 140–204.5 kg (310–450 lb). " Taller than females and much heavier. So fairly intimidating implications for humans, though google gets comments that human strength moves such as weightlifting maybe are not matched by a gorilla for whatever reasons. Yeah, right. Julia Rossi (talk) 07:39, 5 November 2008 (UTC)

Just to be pedantic chimps and gorillas are apes not monkeys. Interestingly it looks like chimps have much faster reactions as well as being much stronger. Basically it looks like speed and strength have taken second place to intelligence for us. Dmcq (talk) 13:48, 5 November 2008 (UTC)

Yeah, chimps can and will kick your ass. Common Chimpanzees can, especially in captivity, become agressive and mean. There was a zoo/park about a mile from where I grew up, Benson's Wild Animal Farm. They had a chimp in a cage who was known, in fits of rage, to tear an inflated basketball in half with his bear hands. I wouldn't want to tangle with that. Bonobos are supposedly more docile and less aggressive, supposedly more suited to human interaction, that is, if you can keep them from humping everything in sight. But AFAIK, their strength is equally as formidable, so I don't think I'd want to mess with a Bonobo in a bad mood either. --Jayron32.talk.contribs 17:16, 5 November 2008 (UTC)

the JFK shot

I was watching a movie with a story that revolved around the notion that a rifle could be fired to hit a minuscule target a mile away,do rifles that accurate and powerful exist ? —Preceding unsigned comment added by Blak thot (talk • contribs) 08:33, 5 November 2008 (UTC)

How minuscule a target? According to sniper, people have been killed by rifle fire at ranges of well over a mile. Algebraist 08:54, 5 November 2008 (UTC)

The difficulty over that distance is that the wind takes a massive toll on accuracy. If conditions are at all erratic (wind shifting at all) then you'll miss by a huge margin. --98.217.8.46 (talk) 13:39, 5 November 2008 (UTC)

When I was in the Marines, I had to qualify every year on the rifle range, which included accurately hitting a black circle about the size of a human head from 500m (about 1/3 mile) using an old crappy M-16 with no scope of any kind. Weather conditions were not considered - I had to qualify if it was nice our or if there was a hurricane blowing in. Given a scope and a good rifle, hitting the same target from a mile would be trivial. -- kainaw™ 13:43, 5 November 2008 (UTC)

The question of target motion should also be considered. Even a supersonic bullet will take two or three seconds to cross a mile distance. Can the shooter accurately predict where the target will be? — Lomn 14:14, 5 November 2008 (UTC)

If we're talking about JFK, since he was travelling in a car I would presume he was travelling at a fairly constant velocity so likely a good sniper could account for his motion. I'm not saying I believe he was shot from a mile away though Nil Einne (talk) 09:47, 6 November 2008 (UTC)

From Sniper rifle#Accuracy, the FBI spec for sniper rifles is an accuracy of at least 0.5 minutes of arc (MOA). At 1 mile (1600 meters), that's a target circle of about the same size as a human head. More expensive weapons can have an accuracy of 0.25 MOA or better. Subject to the conditions above regarding wind and target motion, hitting a human head from a mile away is readily achievable with commercially-available rifles.

For the purposes of a movie, the odds can also be improved by luck, the opportunity to take multiple shots, and artistic license. (For reference, incidentally, JFK was shot from the book depository window 265 feet (81 meters) away. With any moderately well-maintained rifle, in clear weather, and a slow-moving target, the shot wouldn't be difficult.) TenOfAllTrades(talk) 14:42, 5 November 2008 (UTC)

If you are ever in the area - you can visit the museum and look out of the window next to the one that the shot was supposedly taken from (the actual window is behind a glass partition so you can't get to it). It certainly looks do-able - but having to escape through that building seems awfully tricky. But now go and actually stand on the grassy knoll - behind the fence...it's extremely compelling. The shot looks a million times easier - the escape route is really convenient - everything about it seems WAY better. Although all of the evidence does seem to point to the book repository as the place the shots came from - I've gotta say that if I were planning it - I'd want to be standing on the grassy knoll. SteveBaker (talk) 17:41, 5 November 2008 (UTC)

I read this New York Times article on Model M82A1 .50 BMG (12.7 x 99mm) sniper rifle,[1] which was a really great read. Throughout the article they talk about hitting targets from about 2 miles away. At that range, not only do you have to take into account wind speed, but the coriolis effect. The rifle itself, bullets "fired from single-shot or semiautomatic rifles, it exits the muzzle at about 3,000 feet per second, has an effective range of 7,500 yards -- or more than four miles -- and at interim distances can do a stupendous amount of damage." -- MacAddct1984 ^{(talk • contribs)} 19:12, 5 November 2008 (UTC)

Coriolis effect, really? That sounds implausible to me. Do you have a source for that? Friday (talk) 19:27, 5 November 2008 (UTC)

It sounds implausible to me, too. The effect would probably only be a few millimetres at most over that kind of time and distance. --Tango (talk) 19:44, 5 November 2008 (UTC)

My back-of-the envelope calculation suggests the coriolis force could create an error on the order of 5 ft. That's an error you'd want to take into account. Algebraist 19:51, 5 November 2008 (UTC)

Best I could find with my quick 5 minute search is this. I do concede that it may be more hype than fact, especially when things like wind speed have much greater effect. -- MacAddct1984 ^{(talk • contribs)} 20:16, 5 November 2008 (UTC)

I get 4 feet, so certainly of an order you would need to account for (that's the maximum, obviously it depends on your latitude and what direction you are firing it). You learn something every day. --Tango (talk) 22:53, 5 November 2008 (UTC)

At the equator, there'd be zero effect from coriolis: both shooter and target would be spinning at the same speed. Largest difference in speeds would be if the shooter were standing at one of the poles, in which case the target one mile away would be moving around the shooter in a disc two miles wide, at 365 degrees a day, or 2pi miles/day (from 2*pi*R) A mile is 5280 ft. Above, it was stated sniper bullets have 3000ft/sec muzzle exit velocity and 4 mile range. So, call it average 1000ft/sec over the first mile (no evidence, but it seems a conservative estimate), so would travel a mile (5280ft) in 5.28 sec. Divide 1 day by 5.28 sec to get the proportion of the revolution: 1/16363th. So the effect on the bullet of a polar assassin will be pi miles / 16363 = 0.000191993684 miles, or almost exactly 12 inches or 1 foot. I think that would increase with the square of the distance, as something 2 miles away would be moving 4pi miles/day (2*pi*2), and the bullet would have travelled for twice as long (assuming it hadn't even slowed). Twice as long to hit something moving two times faster means it gets to move 4 times the distance, or 4ft at 2 miles at the pole, with an average bullet velocity of 1000ft/sec. I admit I'm surprised the effect could be so large. DewiMorgan (talk) 20:51, 8 November 2008 (UTC)

Temperature of mixture of methanol and water

Hello all,

I'd like to calculate the temperature of a mixture of equal volumes of -80C methanol and RT water. I get -17C when I use:

nm x Cm x |Tim-Tf|= nw x Cw x |Tiw-Tf|.

But I'm wondering whether I need to adjust for energy of dissolution (I don't know if that's a real term) or something like that? How would you solve this problem? Thanks! 141.14.217.217 (talk) 08:47, 5 November 2008 (UTC)

See enthalpy change of solution (unfortunately there is no value for methanol-water in the article). Icek (talk) 12:05, 5 November 2008 (UTC)

You may want to head to the library and pick up either a CRC Handbook of Chemistry and Physics or a Merck Index. One of these handy reference books is likely to have a table of standard enthalpies of solution. --Jayron32.talk.contribs 17:39, 5 November 2008 (UTC)

Thanks for the tips, I'll give that a shot, any further suggestions always welcome, of course. Btw, 17C is a typo - it's meant to be 12C. 141.14.217.217 (talk) 11:42, 6 November 2008 (UTC)

Stress, Breaking force and Cross sectional area

Hi could someone please give me a hand with this question - its driving me mad!

Question: Calculate the force needed to break a nylon wire with a diameter of 0.8mm. The breaking stress of nylon is 5x10⁷Pa.

Ok so this is what I've done so far:

Recognised that to calculate the force needed to break the wire is calculated by multiplying the Breaking stress by the Cross sectional area.

Then I calculated the cross sectional area by doing: Pi x 0.4 x 0.4 = 0.502 mm². To convert this into metres squared I simply divided 0.502 by 1,000 to give 5.02X10^-4m². Whereas my textbook says the answer is 5.02x10^-7m². Whats going on there?

Finally To calculate the force required I multiplied the force needed to break the wire by the cross sectional area: 5x10⁷ x 5.02x10^-4 = 25,100 Newtons - whereas my textbook gives the answer 25.1 N.

Please help! —Preceding unsigned comment added by 139.222.240.178 (talk) 14:50, 5 November 2008 (UTC)

The problem is with your area calculation and conversion. First, change the 0.4mm to m before multiplying - notice that you have to divide each radius by 1000 to convert to m. If you find area first, then convert, you need to take into account that the unit is square mm (mm*mm) and convert accordingly.PhySusie (talk) 15:48, 5 November 2008 (UTC)

Thanks a lot :)

Ancient bacteria in ice

As you may know, some glaciers are melting as a consequence of global warming. What I notice is that, these glaciers are formed million years ago. When the water froze at ancient times, bubbles inside may trap some ancient bateria in it. It should not be amazed that some bateria are still surviving, because they are strong enough to adapt the low-temperature environment. The problem is that, after all these years of seperation, if these bateria are released during the melting of glaciers, can modern organisms be immune to them? Will they become some sort of super germ?--Lowerlowerhk (talk) 14:53, 5 November 2008 (UTC)--Lowerlowerhk (talk) 14:53, 5 November 2008 (UTC)

Bacteria that have evolved to survive in temperatures well below 0 C are unlikely to do very well in the human body at 37 C. There are millions, perhaps billions, of species of bacteria to be found on the earth (estimates from Wikipedia article), so a few more highly-specialised species are unlikely to add much peril. Currently scientists are investigating the bacteria in Lake Vostok under the Antarctic icecap which may have been trapped there for 500,000 years, but they are more concerned with preventing contamination from outside than in stopping the bacteria escaping and killing us. --Maltelauridsbrigge (talk) 15:37, 5 November 2008 (UTC)

I could be wrong, but given that the bacteria has been isolated for such a long time and thus have not been exposed to the rest of the world, I would guess that we're more dangerous to them than they are to us. Think Native Americans (who were isolated from the rest of the world) and European explorers. 216.239.234.196 (talk) 16:00, 5 November 2008 (UTC)

I love this question! I guess it depends whether they've been living in the ice or somehow been hibernating. If the former - then they'll have evolved so spectacularly to living in that extreme niche that they'll likely die as soon as the ice melts. If the latter (more likely to be true of a virus than a bacterium) - then they may still be in severe trouble because all of the things they evolved to do are no longer optimal. If they evolved in a period when there was much more oxygen in the air - or when there were no flowering plants or grasses - then they might be unable to rely on a high density of whatever plants were around back then. It's a really interesting question though - and it's definitely something worth thinking about. Someone out there must be studying these things. SteveBaker (talk) 17:33, 5 November 2008 (UTC)

Bacteria in ice don't evolve. Being in ice is a prison that doesn't allow the bacteria to grow or divide at all. A fraction of the bacteria (perhaps ~1%) that fall onto glaciers are able to adapt to the cold by expressing cold-tolerant proteins and mechanisms that already exist in their genomes. Under the right conditions, some bacteria can remain viable even after 100s of thousands of years imprisoned in ice. In general these bacteria pose no risk to anyone and are only slightly older versions of existing strains.

Bacteria beneath ice, i.e. in sub-glacial lakes and soils, do have opportunities to grow and evolve. The oldest of these environments might have been isolated for tens of millions of years, providing plenty of time for profound evolution. However those bugs would have adapted to a very cold, low nutrient environment, and are unlikely to be dangerous to warm-blooded creatures like ourselves. Also, scientists are coming to realize that there are a lot of sub-glacial streams and rivers that may allow bacteria from under the ice reach the ocean. So it is possible that some of these bacteria are being released all of the time, with no visible consequences. Dragons flight (talk) 18:12, 5 November 2008 (UTC)

Regarding the OP's concern about lack of immunity, it's worth noting that vertebrate immune systems have the potential to recognize virtually any microbe, whether we (as a species) have encountered it recently or not. It's likely that these bacteria will carry molecular patterns that our innate immune system can recognize, the proteins will almost certainly contain epitopes that our T cells can recognize, and other molecules may be recognized by our B cells and perhaps even our NKT cells. --Scray (talk) 00:34, 6 November 2008 (UTC)

Dead pigeon

I have a dead pigeon in my back yard. If I bury it, how deep do I need to go? WAYB (talk) 15:56, 5 November 2008 (UTC)

9 to 12 inches, then firmed down should be plenty to prevent anything from digging it up. 86.4.187.55 (talk) 16:17, 5 November 2008 (UTC)

Don't touch it though - that whole "bird flu" thing hasn't gone away. Use something with a decent length of handle to drop it into the hole. SteveBaker (talk) 17:28, 5 November 2008 (UTC)

Why would a dead animal have avian influenza? Viruses require lihosts. I'd be more concerned about other diseases involving bacteria. But Steve's point stands, put a bag over your hands or something so you're not making direct contact. —Cyclonenim (talk · contribs · email) 23:53, 5 November 2008 (UTC)

In many jurisdictions, local animal control officers will come and dispose of the remains of dead wild animals for free. This is because dead animals can pose a significant human health risk. Pigeons are especially likely to carry diseases that are transmissible to humans, so I would treat it with care. Dragons flight (talk) 17:58, 5 November 2008 (UTC)

Authorities in some areas collect dead birds to check if they died from West Nile Disease. It wiped out all the birds in my area a few years ago. Edison (talk) 03:50, 6 November 2008 (UTC)

Did the local bird population recover? Plasticup ^T/C 15:31, 6 November 2008 (UTC)

Effect of Ultraviolet (UVA) on Crystal Growth Versus Visible Light

Since it is known that the shorter the wavelength the greater the "penetration" into an object (organic or inorganic. For example, xrays or gamma rays. In the case of high or frequent xrays and gamma rays this causes damage or mutations in oarganic substances.

Based on this premise, am I correct that UVA would effect the growth on crystals (from a liquid solution)in a way different from visible life. The effect could be accelerated, slower growth and/or effecting the crystals' structure. Am I on the right track?

Thank you JM

Not sure that it will. The damage that high-energy radiation does to, say, living tissue is also a function of the complexity of the tissue; there's a lot of really large, really finicky compounds that don't take well to being bombarded with high energy. Crystal growth is actually a relatively simple process; I don't see where the energy of light is likely to effect it in any meaningful way. Crystal growth is largely a function of a few factors; the rate of cooling of the solvent, rate of evaporation of solvent, availibility of nucleation sites, etc. will all effect crystal growth. I just don't see where light enters into the picture in any meaningful way.--Jayron32.talk.contribs 17:45, 5 November 2008 (UTC)

I'm inclined to agree with Jayron32. About the only way that UV exposure is likely to affect crystal growth is to hinder it, by breaking down the molecule that you were hoping to crystallize. (Not a serious problem with simple, stable compounds like sodium chloride, but more trouble for fragile biomolecules like proteins]].) In principle, I suppose that insoluble breakdown products could also act as nucleation sites for crystal formation, but that's more likely to give you a sludgy amorphous mess of uselessly tiny crystallites. TenOfAllTrades(talk) 23:15, 5 November 2008 (UTC)

I read of an experiment where citrine and aquamarine crystals were subject to 8% black light (test group). They had dendritic crystals form in that group where no dendrite crystals were found in the control group without UV light. The person believedthat UV radiation had an energizing effect on the sensitivity of the crystals.

Would crystals of copper sulfate possibly have this effect?

JM

Space trash

Reading this story about a dumped refrigeration unit falling to Earth, I noticed that the item took 469 days to come down. That sounds like a very long time for a large piece of junk to be a potential hazard in orbit. I would have thought astronaut Clayton Anderson would have stood on top of it and pushed it towards the Earth as hard as possible using his feet. As a rough guesstimate, I would think he could get it to a velocity of ~1 m/s towards the Earth's surface - getting the fridge approx 300 km closer to Earth (and well into the atmosphere) in just a few days. So, have I got my rough guestimate of the astronaut's strength terribly wrong, or did he just push it aside gently (perhaps due to some NASA dumping policy)? Astronaut (talk) 18:12, 5 November 2008 (UTC)

The object weighed 635 kg. Let's say with his space suit, etc. he might weigh around 150 kg? If he uses a force to accelerate a mass one direction at 1 m/s, how fast is he going to accelerate himself (with 1/4 the mass) in the other direction? He would then have to use some sort of energy to keep from flying off into outer space. Even if he leaned against the space station (a much more massive object), there's still the equal-and-opposite force problem, which would have to be counteracted to avoid changing the orbit. Better to just strap a small thruster onto the tank and launch it to avoid altering the motion of "everything else" too. DMacks (talk) 19:36, 5 November 2008 (UTC)

It was thrown (literally) by the station's robotic arm at about 30 cm/s. Dragons flight (talk) 20:29, 5 November 2008 (UTC)

There is so much space debris about, one more piece for a little over a year doesn't make a lot of difference. Also, applying forces to objects in orbit doesn't work quite as you would expect, I think you would be better off pushing it backwards, rather than down. --Tango (talk) 19:40, 5 November 2008 (UTC)

Well, for one, a push yielding 1 m/s down isn't actually "down". The trash gets closer to Earth on one side of its orbit, but not on the other. What that describes is more or less a partial transfer orbit. Per our orbital mechanics article:

If a brief rocket firing is made at only one point in the satellite's orbit, it will return to that same point on each subsequent orbit, though the rest of its path will change. Thus to move from one circular orbit to another, at least two brief firings are needed.

As a result, you can't simply say that the object moves towards Earth at 1 m/s. Ultimately, atmospheric drag is the primary force responsible for re-entry. The junk doesn't have the booster rockets that the ISS does, so it crashes. — Lomn 19:41, 5 November 2008 (UTC)

If you push is hard enough, it would still work since it would skim the atmosphere (or, rather, a thicker part of the atmosphere - the ISS is still well within the exosphere) once an orbit at perigee and that would lower the apogee until enough of the orbit is in the atmosphere for it to deorbit. --Tango (talk) 19:47, 5 November 2008 (UTC)

Right. But the point is, it's not just a matter of "push it downward at 1 m/s and it will continue descending toward the Earth at 1 m/s indefinitely", as the previous poster seemed to think. The push just puts it into a different orbit. If the original orbit is circular, the best technique is to push backwards along the orbit, which will make it an ellipse with the apogee where you did the push and a lower perigee on the other side. Make the perigee low enough and the orbit will decay reasonably fast. I don't have time to attempt the math now, but it would take a fair bit more than a 1 m/s push to produce a substantial lowering of the perigee. --Anonymous, 23:14 UTC, November 5, 2008.

As space debris goes, this one isn't much of a problem. It's big, it's easy to track, and it isn't surrounded by a cloud of small particles. Avoiding it is simply a matter of not intersecting its orbit when the refrigeration unit is there. --Carnildo (talk) 22:25, 5 November 2008 (UTC)

To get an object in low earth orbit to reenter the atmosphere (in the next orbit) they fire the retro-rockets not to directly push it downward, but retrograde to the direction of motion. A decrease in velocity of about 1% is all that is needed to achieve timely reentry. That would be a velocity change of about 77 meters per second. Edison (talk) 03:48, 6 November 2008 (UTC)

With regard to throwing the refrigerator with the robot arm, I cannot understand the situation with regard to the equal-and-opposite effect. In the case of burning rocket fuel, hot and expanding gases exert a force on the rocket engine. In the same way, an exploding bomb will smash a nearby window. That action of expanding gas does not occur with the robot arm.

Neptune and Pluto disc color

Since Neptune is 1.5 times farther away from sun than Uranus, it would be 1000 times dimmer than Earth, and Pluto is 2000 times dimmer than Earth how can Neptune still get vivid blue color, and Pluto get apricot color. Bartlett gave me this gives the ture color of planets. I thought Neptune suppose to be dark indigo and almost look nothing 300 meters from space, just feel a swirling mass is coming to me almost black until I come like severals of miles down. Pluto's surface suppose to be almost BLACK, oribting by Pluto's surface I thuoght I will just see a ground over a black-out homes over a deep midnight with no light. --FRWY 23:31, 5 November 2008 (UTC)

You are confusing hue and brightness. Obviously Neptune is dim. So dim in fact it can't be seen from Earth unless you use tools like a telescope to increase the amount of light reaching your eye. A 6-inch telescope, for example, captures roughly all the light entering its 6-inch opening. That is much greater than what would normally enter through the 1/8 inch iris of your eye. Since you have increased the amount of light being captured, Neptune appears brighter. Add to that, many images of astronomical objects are accumulated exposures over many seconds to many minutes (or even hours). That accumulation of light further increases the brightness.

On the other hand, Neptune and Pluto aren't really all that intrinsically dim as humans perceive things. They are illuminated by 1/1000 and 1/2000 of the sunlight reaching the Earth respectively. That's roughly the difference between sunlight and 100 W bulb at 10 feet. A single 100 W lighting up a 10 ft space is certainly dim, but it isn't dark. If a human were nearby, they shouldn't have any trouble seeing Pluto and Neptune. Dragons flight (talk) 00:29, 6 November 2008 (UTC)

• Why is it most source say true color of neptune and Pluto is vivid blue and brown?I menat when you orbit it on spaceship.--FRWY 00:40, 6 November 2008 (UTC)
• Did you mena I will see a blue neptune and brown pluto when I am flying orbiting around it. What does intrinsically mean?--FRWY 00:43, 6 November 2008 (UTC)

Find something that is vivid blue when seen in sunlight. Now take it into a room with only a single 100 W lamp 10 feet away from you for lighting. That's roughly what Neptune will look like to someone orbiting. You will still see it fine. Dragons flight (talk) 02:31, 6 November 2008 (UTC)

I think that calculation is not quite right. At neptune's distance, the sun is approaching a point-source of light. It will still, obviously be the brightest star in the sky, but it will look more like a bright point than a big yellow disc. I wouldn't imagine that you could see much of neptune at all in ambient sunlight; it'd likely look to you like a dim, gray sphere; if anything at all. You'd probably notice it more by the way it blocks out background stars more than as a colored object... --Jayron32.talk.contribs 03:05, 6 November 2008 (UTC)

The sun at Earth delivers ~1300 W/m^2. At Neptune it gives ~1.4 W/m^2, which is again like a lightbulb at moderate distance. Of course to do it right, you really need to express it in candela since apparent brightness is not directly proportional to power, but the sun and incandescent bulbs in that regard are relatively similar so it is a fair approximation. Neptune won't be bright, but it would be visible. For comparison, moonlight provides ~0.001 W/m^2 of illumination. A moonlit night is fairly dark (by most people's standards), but you can still see objects. By contrast, the Sun will still provide substantially more illumination at Pluto than our moon provides us at night. Dragons flight (talk) 03:28, 6 November 2008 (UTC)

From Neptune, the Sun ought to have a angular diameter of 1 arc minute, which is just resolvable by someone with 20/20 vision (in fact, that's the definition of 20/20 vision). So, I guess it's pretty much the smallest thing you can describe as a point source. --Tango (talk) 18:12, 6 November 2008 (UTC)

We've addressed this issue twice before for this same OP. We don't need to do it all again. SteveBaker (talk) 18:51, 6 November 2008 (UTC)