Wikipedia:Reference desk/Archives/Science/2015 March 28

= March 28 =

Dropping an Object From Space
What is the heaviest an object may be to survive being dropped from space and land on Earth unscathed? Honeyman2010 (talk) 05:28, 28 March 2015 (UTC)


 * That would depend on its size, its shape, and its composition, as well as on where it landed. In other words, your question is unanswerable as asked, sorry. &#8213; Mandruss  &#9742;  05:35, 28 March 2015 (UTC)

OK, let me ask it this way. If one dropped an empty Coke can from the International Space Station in a trajectory that would bring it back to Earth in the middle of a desert, would it survive without any damage? Honeyman2010 (talk) 05:58, 28 March 2015 (UTC)


 * I think it might, because the density is so low and heat wouldn't build up much because of the thin metal. StuRat (talk) 06:15, 28 March 2015 (UTC)


 * See Atmospheric reentry for our article. Orbit would also be useful.  The answer to the OP's question depends on what they mean by "dropped".  Just throwing the can out of the ISS's airlock wouldn't put it on a re-entry trajectory, it would stay in the same orbit as the ISS, which would only decay after a matter of decades - some sort of retro rocket would be needed.  Putting it on the standard re-entry trajectory that the astronauts use would expose it to temperatures of about 7800 K, well above the melting point of aluminium (933 K).  (Incidentally, having thin walls _increases_ the rate of heat buildup - see heat capacity).  Using a spacecraft-sized retro rocket to bring it in on a subsonic trajectory would restrict its speed to its terminal velocity (according to my very approximate calculations, about 7 m/s), which should allow it to land undamaged on a soft surface.  But there are too many variables to answer the question as stated. Tevildo (talk) 07:34, 28 March 2015 (UTC)


 * The verb "dropped" has no meaning in the context of "from space", especially regarding the International Space Station or any spacecraft in orbit. If an astronaut/cosmonaut on a space walk was to release an object in the same way we would "drop" something on Earth, as oppose to throwing it, then the object would continue in the same orbit as the International Space Station. Cullen328  Let's discuss it  07:46, 28 March 2015 (UTC)


 * An object with a higher surface area to volume ratio should be more effective at radiating heat away, hence limiting heat buildup. StuRat (talk) 07:50, 28 March 2015 (UTC)


 * Think it would be better to put it that the ratio of surface area to mass is great. Have often wondered though whether a paper airplane would make make down to terra firma without getting singed. It would make a good school project as it would be so cheap to do and engage kids interest in science. If it (or a few hundred paper airplanes) were over 10cm in size and aluminum coated, their re-entry could be tracked and possible a few could be recovered -if a suitable reward to the finder was offered. How about that Jimmy Wales? – you could promote this experiment  by offering the finder(s) free editorship to Wikipedia...!--Aspro (talk) 15:55, 28 March 2015 (UTC)
 * To be a little bit pedantic: A reentering capsule would not reenter like a coke can. The center of mass in a capsule is arranged a little bit off the capsule's center line of axis, which provides a little bit of aerodynamic lift. Thus, allowing it to slowdown significantly before hitting the denser atmosphere. Those of us that can remember the return of the Apollo 13 had this drummed into them as to why the reentry angle from Moon to Earth return was so critical.--Aspro (talk) 16:11, 28 March 2015 (UTC)


 * Yes, although they would need to release the pop cans/paper airplanes either during launch or prior to landing, not while in a stable orbit, as noted earlier, to prevent it from becoming more space junk for decades until the orbit decays. StuRat (talk) 16:03, 28 March 2015 (UTC)

Thank you all for your efforts to answer this. I appreciate your knowledge. Honeyman2010 (talk) 08:08, 28 March 2015 (UTC)


 * This actually seems like a better question for What if?, a website by Randall Munroe. At the bottom of the linked page, there's a link to email questions. There's no guarantee your question will get answered, but if it is, it will probably be with more thoroughness than is usually given here. Matt Deres (talk) 12:49, 28 March 2015 (UTC)
 * Funny you should mention that website. I came here because of this xkcd comic about space-exploring squirrels. Somebody on the xkcd forum mentioned that a squirrel's terminal velocity is so low that it can fall from any height and survive. That made me wonder what would happen to the squirrel in the comic if it fell from a point well outside the atmosphere but close enough to Earth to be captured by Earth's gravity and fall more or less straight down. Would it be fried by aerodynamic heating? Let's assume it's an American red squirrel and let's further assume that it can breathe in space (the source in the link makes it clear that the squirrel is conscious halfway to the Sun, so it's clear those squirrels have mastered space suit technology). Sjö (talk) 16:22, 28 March 2015 (UTC)
 * If this hypothetical squirrel just droped perpendicular to earth ( meaning it was not at orbital velocity and thus having the kinetic energy to melt steel). Then at the hight of the ISS this critter would only have to dissipate 2 to 2 ½ kilowatts of energy per kilo of kinetic energy created by the gravity well . So, me  thinks,  it would certainly end up getting cooked on the outside but a still a little bit red raw on the inside. Makes you think though – does anybody have a spare sounding rocket and a pet  squirrel so we can put this to the test?--Aspro (talk) 17:33, 28 March 2015 (UTC)

Falling from space? Cool. The following crudely approximates the effects of falling from space, please do not attempt with actual squirrels.

Let's assume a sphere of mass m and radius r is released at rest relative to the Earth's center of mass from a height h close enough to the Earth that we can assume constant acceleration (g = 9.8 m/s2). Let's further assume the density of the atmosphere decreases exponentially with a scale height of 8 km, that atmospheric temperature can be ignored, and that supersonic effects can also be ignored. The total force (gravity + drag force) on the object is approximately:


 * $$-mg + {1 \over 2}\rho_{air} v^2 0.47 \pi r^2 \approx -mg + {1 \over 6}{\rho_{air} \over \rho_m} v^2 {m \over r}$$

At terminal velocity:
 * $$0 = -mg + {1 \over 6}{\rho_{air} \over \rho_m} v^2 {m \over r}$$
 * $$v = \sqrt{ 6 g r {\rho_m \over \rho_{air}} } = \sqrt{ 6 g r {\rho_m \over \rho_{air,0} e^{-h / 8\, \text{km}}} } = e^{h / 16\, \text{km}} \sqrt{ 6 g r {\rho_m \over \rho_{air,0} } }$$

The power dissipated at terminal velocity would be:


 * $$P = Fv = {1 \over 6}{\rho_{air} \over \rho_m} {m \over r}e^{h / 16\, \text{km}} \left( 6 g r {\rho_m \over \rho_{air,0} } \right)^{3/2}$$
 * $$= \sqrt{6rg^3 {\rho_m \over \rho_{air,0}}}m e^{3 h / 16\, \text{km}}$$

As a rule of thumb, the fraction of the power captured by the object is approximately 1/2 the drag coefficient, which implies at terminal velocity:


 * $$P_{heating} \approx \sqrt{{3\over 8}rg^3 {\rho_m \over \rho_{air,0}}}me^{3 h / 16\, \text{km}}$$

If we assume we are talking about normal objects without heat shields and other technical aids and good thermal conductivity, then we can imagine it has a uniform temperature and dissipates heat via radiation, leading to a quasi-equilibrium temperature of:


 * $$\sigma 4\pi r^2 T^4 \approx \sqrt{{3\over 8}rg^3 {\rho_m \over \rho_{air,0}}}me^{3 h / 16\, \text{km}}$$
 * $$T = \left({ 9m \over 256 \pi \sigma}\right)^{1/4} \left({g^3 \rho_m \over r^3 \rho_{air,0}}\right)^{1/8} e^{3 h / 64\, \text{km}}$$

Finally, let's put in some numbers. Impact velocity reduces to:


 * $$v_{impact} = \sqrt{ 6 g r {\rho_m \over \rho_{air,0} } } = \sqrt{ {9 g m \over 2 \pi r^2 \rho_{air,0}} } \approx 3.7 \text{ m/s } \left( {1 \text{m} \over r }\right) \sqrt{ m \over 1 \text{kg} }$$

At 100 km, the temperature due to drag would be (in crude approximation):


 * $$T \approx 50 \text{K}\left({ m \over 1 \text{kg} }\right)^{1/4} \left({ 1 \text{m} \over r }\right)^{3/8} \left({\rho_m \over \rho_{air,0}}\right)^{1/8} \approx 40 \text{K}\left({ m \over 1 \text{kg} }\right)^{3/8} \left({ 1 \text{m} \over r }\right)^{3/4}$$

Feel free to enter your favorite squirrel / soda can / whatever numbers to see what happens. Personally, I'd suggest unscathed is probably velocities < 10 m/s and temperatures less than 320 K for living things, though obviously higher for inanimate objects or living things with some form of protection. Incidentally, the time required to fall from orbit under these assumptions is approximately:


 * $$ t \approx 4.4 \text{hr} \left( {1 \text{m/s} \over v_{impact}} \right) \left(1 - e^{-h / 16 \text{km}}\right)$$

Dragons flight (talk) 19:35, 28 March 2015 (UTC)


 * P.S. For sufficiently slowly moving objects (e.g. insects) they are actually more likely to freeze in the upper atmosphere than burn up. Dragons flight (talk) 20:41, 28 March 2015 (UTC)


 * That, I think, is a really thorough answer. But it was the heat generated in the supersonic  phase of  deceleration  is what I was  thinking about.  It may not last long but that is that phase where much of the  deceleration and heat is generated. The atmospheric density may increase according to log e but the body (or in this case the squirrel) is still accelerating at G and will be traveling much, much faster than than the speed of sound at STP. Just the  'tail'  of Concord reached over 90°C. Our furry hypothetic friend may have insulating furr but it has one hell of a temp differential to survive.   In this experiment, I would graciously step-aside and let the squirrel jump first. P.S. And as for bacteria - maybe Fred Hoyle was right.--Aspro (talk) 20:50, 28 March 2015 (UTC)


 * The Concorde did about 650 m/s (Mach 2) at 20 km altitude. At the same elevation a falling "squirrel" (r = 10 cm, m = 0.5 kg) would be subsonic at a leisurely Mach 0.25 and dissipate only 1/500 the energy per unit area.  That "squirrel" wouldn't be supersonic below 40 km, and above that point you've lost > 99% of the atmosphere.  Supersonic corrections twiddle with the drag coefficient and the coefficient of heat transfer, but personally I don't think they change the rough picture.  For the current parameter values the squirrel should be more worried about freezing than burning.  Dragons flight (talk) 22:10, 28 March 2015 (UTC)


 * From typical low Earth orbit, an object only has to lose about 1% of its velocity to start reentry. A small retrorocket or even some mechanical device might suffice to fire an object like a Coke can in the retrograde direction and cause it to reenter. The high heat mentioned above of 7800 Kelvin should not be a constant. It should depend on the aerodynamics, shape and area of the reentry package, and on how much mass.. There was the MOOSE project from NASA in the early days of the space age, a system whereby a person could reenter safely, (in theory at least) with a lightweight ablative heat  shield and some foam insulation,  a retrorocket,a space suit, an air tank  and a parachute, all weighing about 200 pounds in addition to the person. When the Columbia shuttle broke up on reentry at about Mach 19 or 20 during reentry, lots of delicate artifacts survived, including a disc drive, a videotape, and a colony of experimental worms, as well as computer equipment in a nylon bag (granted, the shuttle protected the equipment part of the way down until breakup).  It's not clear why a large lightweight metal ball, say, or even an empty Coke can, would rise to the same temperature as a massive ball of the same diameter, or a solid aluminum "Coke can" since the low mass object should decelerate more rapidly and do less work compressing the atmosphere as it descended.  Edison (talk) 00:53, 29 March 2015 (UTC)

There is actually an extremely simple way of estimating this using the impact depth formula. You then just calculate the distance it must travel before the object has pushed away a mass equal to its own mass. This then gives you roughly the distance it needs to travel before it has lost pretty much all is momentum, to surivive re-entry this must happen within the atmosphere. To push away a mass equal to its own mass, the object must travel a distance equal to its diameter times the ratio of its own density and that of air. If the object is made out of Iron and you assume an effective atmosphere of 10^4 meters thickness at the uniform density we have at ground level, then you get a diameter of 1.6 meters for the object. Count Iblis (talk) 15:51, 29 March 2015 (UTC)

gave a really great answer... for an object dropped from rest. But we're speaking of a reentry from orbit. To give the classic case, a bored ISS technician carelessly tosses his Coke can out the window. This is how the problem of space junk gets started :) Now I don't know the truth of claims that Galileo once dropped stones from the Leaning Tower of Pizza and found that the larger ones fall faster, then fudged the results, but one assumes the reverse is true for objects dropped from orbit.  To wit, the ISS loses 2 km/month by aerodynamic drag, being 450,000 kg and having a cross section of very maybe 20 m x 100 m = 2000 m^2 .  I see a "v^2 m/r" in Dragons flight's formula above ... I'm not quite sure what rho is so I'll ignore it (which he kind of does at the end I think).  So a soda can weighs 0.014 kg and has size 4.83 x 2.60 inches = 0.008 m^2, so the mass is 3.1E-8 and area is 4E-6; take square root of the area and the "radius" is just 2E-3 less.  So the drag force on the tin can should be something like 50000 times more than on the space station (???!) and the thing should plummet out of the sky in no time. And here I thought the space station really was just a tin can.

But I still don't know at what point in this transition the tin can is going to get hottest. There's a point where it's moving much, much faster than falling from rest, but much more slowly than the ISS in stable orbit. It may not be in any particular sort of equilibrium at the time? So I'm still undecided on this one. Wnt (talk) 15:57, 29 March 2015 (UTC)


 * I subscribe to Orbital Debris Quarterly, a publication of NASA Johnson Space Center. Every three months, they send a newsletter summarizing all the latest happenings in objects dropping from orbit, including the largest events.  They also provide updated statistical data and insight into the latest and greatest scientific and computational tools for modeling orbital debris.  You can read the archived issues if you're interested in the topic.  Nimur (talk) 12:08, 30 March 2015 (UTC)

Why don't bedbugs carry disease?
Why don't bedbugs carry disease?2601:7:6580:5E3:E8FB:21E2:7FA2:947 (talk) 17:23, 28 March 2015 (UTC)


 * This U. Penn. article indicates that they can. The recent outbreak of Bedbugs in the US as been traced chicken farms, I remember reading that either Delaware or Maryland was ground zero.  In the US Northeast diseases like malaria and Chagas disease are simply not endemic, so the bugs have no pathogens to spread.  (There's also the difference between bug-type mouthparts and mosquito mouthparts.)  No one knows if this may change in the future. μηδείς (talk) 18:47, 28 March 2015 (UTC)


 * Bedbugs tend to feed off the same person or people each night, so there's less opportunity to spread disease than from something that feeds on different people or animals each night. Of course, they do occasionally spread to a new location, and could then spread disease with them. StuRat (talk) 18:50, 28 March 2015 (UTC)


 * Although it is not unequivocally been determined that bedbug don't transmit disease, it is never the less a good question -as other anthropoids (like mosquitoes) do. However, as disease vectors, such as mosquitoes, can easily move onto a new host if it dies and pass on disease from the last host... the bedbug has a very long walk ahead of it if its host dies. So it may be that evolution has selected bedbugs that don't kill their host. Got no references however to back that hypothesis up and on Ref Desk we don't like to speculate. There are other anthropoids and tetrapods (lobster etc.),  isopods (woodlice etc.),  that people can eat raw without any known or recorded risk. So maybe no  zoonotic disease has had found a need to evolve to make them a vector.--Aspro (talk) 19:09, 28 March 2015 (UTC)
 * ITYM arthropods. --Trovatore (talk) 19:18, 28 March 2015 (UTC)


 * Humans would not catch actual diseases that bedbugs themselves suffer from--we are far too different physiologically. We could get diseases for which they might be a disease vector.  In other words, they would consume an infectious agent in the blood of another host, and according to the UoP article I quoted, pass the pathogen along in their feces, with the disease being transmitted when the bedbug host scratched open wounds and rubbed in the feces.  This model has been verified in mice, and is the same method of transmission of Chagas disease by kissing bugs, which are in the same biological order.


 * As a side comment, bedbugs are not that sedentary, they will travel on clothing and infest entire buildings. This is exacerbated by urban living conditions.  They have been found riding the NYC subways.


 * Humans don't eat bedbugs (mice do), and most diseases that are transmitted by eating involve poorly prepared mammals and fowl, not seafood (although, see fugu). Pigs cary flukes, and we can get schistosomiasis by consuming the living eggs of the worm.  We get E. coli infections from the feces of others when they prepare food with unwashed hands (ground beef) or defecate in crop fields.  We get Listeria and Salmonella from undercooked poultry products and contact with reptiles like turtles which are carriers, as well as transmission by unsanitary workers who contaminate meat and dairy products. μηδείς (talk) 21:20, 28 March 2015 (UTC)


 * Avoid eating any chicken prepared by the housewives on Listeria Lane (unless you happen to be desperate). StuRat (talk) 22:23, 28 March 2015 (UTC)


 * Wouldn't that be Wisteria Wane? μηδείς (talk) 01:45, 29 March 2015 (UTC)
 * I think we are getting into Hysteria Lane. KägeTorä - (影虎)  ( もしもし！ ) 11:40, 1 April 2015 (UTC)


 * Hysteria, was she the sister of Hyacinth, Rose, Daisy and Violet who was committed and of whom they do not speak? μηδείς (talk) 15:00, 1 April 2015 (UTC)


 * If so, I'm sure Hyacinth would explain (and have Richard memorize) that she's in a posh upper-middle class health retreat, with room for a pony. StuRat (talk) 17:58, 1 April 2015 (UTC)

Deep sea bases: comet proof?
With all due respect to Aquarius (laboratory), it is not what I'd call comet proof (or even hurricane proof). I entertain some suspicion that militaries might, for various purposes, have adapted their existing deep submersibles to be permanently placed and docked in larger complexes. Questions:


 * Are there any technical barriers to making a large complex several miles under the ocean, provided one is willing to pay proportionally more than the $50 million cost for DSV Alvin? (By military standards I think ten or twenty billion really isn't that much, especially if you can store nukes there you want to keep off the records for arms control)


 * For example, is there a limit to how long something like Alvin could be kept submerged before it gives way to the pressure?


 * Would an undersea complex be protected from a hit by a comet, provided it wasn't near to the impact site? Or would a pressure wave run through all the oceans and crush it like a tin can?


 * If such a base had been constructed, could the military plausibly keep it secret for decades by supplying it with stealthy submarines, etc.? Or would it be detected by some sort of scanning (kind of like  but one hopes with more sophistication ;) Wnt (talk) 23:40, 28 March 2015 (UTC)


 * One wonders if you are writing a sci-fi novel? The two inch thick titanium spherical shell of alvin and other component parts will survive as long as no corrosion affects them. As for pressure waves. Water is an almost anincomparable liquid. The pressure will exhibit itself as a surface wave. If the wave (from say a comet) is a mile high when passing over the base, the pressure will increase by approximately  3,000 p.s.i. In other words, the base would not experience  the same type of  blast wave as happens in air (which is due to compression). Could it be keep secret? If constructed by the US no - as the Chinese would already have copy of the plans before work got started.--Aspro (talk) 00:09, 29 March 2015 (UTC)
 * Although water is approximately "incompressible" for many purposes (as is air, for that matter) it is definitely true that shock waves can travel through water. There is a large literature on the topic.
 * I did not say that water is not compressible at all (which would prohibit shock-waves). Properties_of_water But that shock-waves are of little consequence in water. Also, as Jacques Cousteau ( who also as it happened, co-piloted the first deep-sea Bathyscaphe long before alvin)  recalled in his book The Silent World: A Story of Undersea Discovery and Adventure  that he and his colleges (as military divers) stood in a pool whilst sticks of dynamite where thrown in, closer and closer. They did not experience any shock-waves.--Aspro (talk) 01:08, 29 March 2015 (UTC)
 * Shock waves are actually the main mechanism of damage from sea mines as they are usually made to detonate some distance from the ship. They are also a danger to divers at a larger distance than from shock waves in air.  Divers near the surface (or standing in a pool) benefit from the phase shift when the wave reflects from the surface which will reduce the effects. Sjö (talk) 07:53, 29 March 2015 (UTC)
 * Not necessarily related to compressibility, Chicxulub crater is relevant as a historical precedent. Short Brigade Harvester Boris (talk) 00:29, 29 March 2015 (UTC)