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September 10

What is the over/under for the closest a planet has been to Earth in the Phanerozoic? What about any planet pair?

2. Is the probabilistically most likely value for "what was the biggest Phanerozoic perturbation?" enough to cause detectable climate change if it magically happened now?

Planet means the kind we have 8 of. Sagittarian Milky Way (talk) 03:41, 10 September 2020 (UTC)[reply]

It is generally thought that the Solar System as a whole is unstable (see the article on Stability of the Solar System) and that the system of equations is is chaotic in the technical sense of mathematical chaos theory. On a time scale of a few million years nothing dramatic is to be expected, and the effects are too small to have been observed since Homo sapiens gazed up to the skies. But even the most precise long-term models for the orbital motion of the Solar System are not valid over more than a few tens of millions of years, let lone over the 500+ million years the Phanerozoic has lasted. Numerical integration of the differential equations on supercomputers suggests the possibility of collisions of Mercury, Mars or Venus with the Earth some 3.34 billion years into the future, but not sooner than within one billion years. Since the equations are time-symmetric, the models can equally be used to postdict the past planetary motions. I don't know if anyone has actually done that. It seems that for a time scale below a billion years into the past the result is likewise not likely to be dramatically different from the current orbits. But the lack of major upheaval for a long future time span is no guarantee of the same for the past. Until someone runs that simulation backwards the best answer to the question in the heading is that we don't know. As to the second question, current climate models have no contingency for a direct collision of Mars with Earth, but I think it is a fair bet the climate change would be detectable.^[1] --Lambiam 08:05, 10 September 2020 (UTC)[reply]

The closest planet to earth, on average, is Mercury (no really: [2], [3]). The closest planet to earth at the closest point on either of their orbits is Venus. The gravitational effect of Venus at its closest point is easy enough to calculate; it is neither close enough nor large enough for General Relativity to matter, so use F = G (M1 * M2)/(d^2). You should also do the calculation for the gravitational effect of the Sun on the earth too; I'd imagine the effect of Venus is well out of the range of the significant figures of our measurements. You may also want to do Jupiter, given the difference in masses, Jupiter may have a larger overall gravitational effect on Earth than Venus, but again, these numbers are going to be several orders of magnitude into the uncertainty range for the effect of both the Sun and the Moon, and at that level, basically meaningless as they would get lost in the normal expected variations of those bodies (basically just part of the "noise"). --Jayron 32 10:50, 10 September 2020 (UTC)[reply]

I've heard the Mercury thing before, apparently the math makes it not tied with Venus. Sagittarian Milky Way (talk) 01:09, 11 September 2020 (UTC)[reply]

One can estimate an upper bound on any perturbations of Earth's orbit by considering the hypothesis that the largest perturbations in the Earth's climate with unknown cause, was caused by a perturbations of the Earth's orbit. Now, going a bit beyond the Phanerozoic period, we can consider the snowball Earth period, and for that you actually need quite small perturbation of earth's orbit, way smaller than what can happen due to the effects mentioned by Lambian above. Count Iblis (talk) 15:03, 10 September 2020 (UTC)[reply]

September 11

Time it would take for Earth and Mars to collide

A friend brought up the question of how long it would take for the Earth and Mars to collide. There were several diversions but the core of the question is, if Earth and Mars were the only objects in space, how long would it take for them to collide. We're assuming the gravitational constant remains what it currently is. And no other bodies are affecting them. The distance would be, let's say a light year. And they start out stationary. I thought it would be fairly straightforward. Something along the lines of just using the two masses, gravity, and distance. But I'm not sure how that math problem would be set up. And my Google skills are failing me for what seems like a simple equation. Any help?

I know that school is just starting here in the US but I can assure you this isn't for a school project. I'm a dude in my 40s with no homework due any time soon. Thanks, †dismas†|^(talk) 02:28, 11 September 2020 (UTC)[reply]

Per Newton (not Einstein) and his law of universal gravitation

F=G{\frac {m_{1}m_{2}}{r^{2}}},

gives the force F pulling two objects together. Solve for

G = 6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻².‍^[1], m_earth = 5.97237×10²⁴ kg, m_mars =  6.4171×10²³ kg

Then Newton's 2nd Law of Motion a = F / m

gives the accelerations a_earth and a_mars of each planet towards the other. We can arbitrarily consider one of the planets as a stationary reference. Then the separation distance

s = 1 Light-year - radius_earth - radius_mars

 = 9.46 x 10¹² - 6371.0 - 3.39 km

will be closed by the other body accelerating at a = a_earth + a_mars

Remembering the equation of motion s = ut + ½ a t ² the actual calculation of

t=2sqrt(s/a)

is left to the reader. 84.209.119.241 (talk) 08:25, 11 September 2020 (UTC)[reply]

^ "2022 CODATA Value: Newtonian constant of gravitation". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.

I'm afraid that's wrong. Crucially, the gravitational force and hence the acceleration does not stay constant as the two bodies approach each other. I'm not going to attempt to solve the equation of motion (Newton's 2nd law), but merely point to the article free-fall time, which gives the solution. --Wrongfilter (talk) 08:34, 11 September 2020 (UTC)[reply]

(ec) To start, I'd make the problem more general. We start out with a Newtonian universe containing two stationary bodies, being point masses of magnitudes

m_{1}

and

m_{2}

at a given distance

d_{0}

. Obeying Newton's laws of motion they start moving toward each other at time

t=0

and their distance

d_{t}

will decrease as a function of

t

. Solve

d_{t}=0

for

t

. The centre of mass of the two-body system is stationary at time

t=0

, and Newton's laws (specifically the conservation of momentum) imply it will remain so, so that the bodies will meet there. Likewise, the conservation of angular momentum implies they will move along the straight line connecting them and passing through the centre, which we will take to be the origin. (Even without appeal to conservation of angular momentum, this already follows from elementary considerations of symmetry.) We can therefore think of the two bodies as being located at the positions

d_{1,t}={\tfrac {m_{2}}{m_{1}+m_{2}}}d_{t}

and

d_{2,t}=-{\tfrac {m_{1}}{m_{1}+m_{2}}}d_{t}

. To simplify the notation, we define

s=s_{t}=d_{1,t}

and

\mu ={\tfrac {m_{2}}{m_{1}+m_{2}}}

, so that

s=\mu d_{t}

(and therefore

d=d_{t}=\mu ^{{-}1}s

), and solve

s=0

. The forces attracting the bodies to each other are opposite in sign but equal in magnitude, which is given by

F=Gm_{1}m_{2}d^{{-}2}=

G\mu ^{2}m_{1}m_{2}s^{{-}2}=

\gamma m_{1}s^{{-}2}

, where

\gamma =G\mu ^{2}m_{2}

.

The acceleration on the first body, taking the proper sign so that the acceleration is towards the origin, is given by

{\ddot {s}}=a=-F/m_{1}=\gamma s^{{-}2}

.

I'll leave it to the two of you to solve this simple ODE and plug in the values for the variables in the model, but don’t be shy to tell us if you need more help. BTW, the planets you want to collide, in their regular orbits, are never more than 4.25 × 10^-5 light-year or 23 light-minutes apart, so putting them a light-year apart is a far stretch~~; in the universe as we currently understand it, you'd not be able to ignore relativistic effects~~. --Lambiam 10:41, 11 September 2020 (UTC)[reply]

From the energy conservation law it follows that

{\frac {\mu }{2}}\left({\frac {dr}{dt}}\right)-G{\frac {m_{1}m_{2}}{r}}=-G{\frac {m_{1}m_{2}}{r_{0}}}

,

where

r_{0}

is the initial distance,

\mu =m_{1}m_{2}/(m_{1}+m_{2})

. Then the time

\tau

to collision will be

\tau ={\sqrt {\frac {\mu }{2Gm_{1}m_{2}}}}\int \limits _{r_{E}+r_{M}}^{r_{0}}{\frac {dr}{\sqrt {{\frac {1}{r}}-{\frac {1}{r_{0}}}}}}={\sqrt {\frac {\mu r_{0}^{3}}{2Gm_{1}m_{2}}}}\int \limits _{(r_{E}+r_{M})/r_{0}}^{1}{\frac {{\sqrt {t}}dt}{\sqrt {1-t}}}

.

You can calculate the integral yourself. Ruslik_Zero 20:41, 11 September 2020 (UTC)[reply]

I have a faint memory of reading the assertion (possibly due to Newton) that if an orbiting body were to stop suddenly, relative to its primary, it would fall to the barycenter in 1/√32 of its former orbital period. If that's accurate, you need only work out what the period of the Earth-Mars binary would be at that distance, and divide by the appropriate constant. —Tamfang (talk) 00:24, 15 September 2020 (UTC)[reply]

It is true and can be derived from the above formula:

\tau ={\sqrt {\frac {\mu r_{0}^{3}}{2Gm_{1}m_{2}}}}\left({\frac {\pi }{2}}+{\sqrt {a(1-a)}}+\operatorname {asin} {\sqrt {a}}\right)={\sqrt {\frac {\pi ^{2}\mu r_{0}^{3}}{8Gm_{1}m_{2}}}}

,

(where

a=r/r_{0}

) when r=0. Ruslik_Zero 18:47, 16 September 2020 (UTC)[reply]

Does milk slosh in a sealed tetrapak?

When I shake a new sealed tetrapak of milk, it always seemed like the milk doesn't slosh. Once I open it and remove the seal it does. Is something about the sealing process preventing the milk from sloshing (even though surely it's still got space to move around)? Or is it because the milk is sloshing, but the sound can't travel through the vacuum? (Which would be very cool, like having a tiny microcosm of space in your milk carton.) --2A01:4C8:64:898F:1:2:A027:D730 (talk) 16:46, 11 September 2020 (UTC)[reply]

Milk only sloshes at a phase boundary. If there is no air in the tetrapak, there is no phase boundary against which the milk can slosh. Once opened, air gets in, and now you have the ability to slosh. --Jayron 32 17:52, 11 September 2020 (UTC)[reply]

So a question that remains, for me, is whether the milk gets mixed when it is shaken, or does one need to open and seal it again before shaking in order to mix the milk?--Shantavira|^{feed me} 18:23, 11 September 2020 (UTC)[reply]

If there is no air inside – whether a tetrapack, bottle or can – shaking has no or very little effect. Turning the pack on different sides and slowly keeping turning may help to mix if separated components (e.g. milk and cream) have different densities; this requires no vigorous anything but just patience to let gravity do its thing. --Lambiam 23:10, 11 September 2020 (UTC)[reply]

Tetrapaks aren't used here, so I have no familiarity with opening them, but if there was no air inside, I can't see how there wouldn't be milk spilling out whenever you do. However, there might be just a small bubble of air that could move around without "sloshing". --174.88.168.23 (talk) 01:19, 12 September 2020 (UTC)[reply]

Assuming the container isn't rigid, it can slosh even without air by deforming the containiner itself. I just tried a "tall rectangular half-gallon" and shaking it up and down caused it to bulge alternately on the bottom vs top of the sides. That means the liquid is able to move inside even if there is no non-liquid space. DMacks (talk) 20:28, 12 September 2020 (UTC)[reply]

An additional consideration is that cow milk froths pretty readily. Depending on how much air is in there (assuming there is some amount), you might not hear much sloshing because the foam is a) deadening the sound and b) filling the previous air gap with a structure of milk foam. Matt Deres (talk) 12:40, 12 September 2020 (UTC)[reply]

As to the question about whether the sound can travel through the vacuum, the answer is that it doesn't need to. If you take a Vacuum flask and half fill it with water and then slosh it about, you'll be able to hear the liquid perfectly well despite in that case most of the surrounding space beyond the interior wall being vacuum. The sound waves can pass out by vibrations in the glass of the wall that then get transmitted to the exterior atmosphere. Michael D. Turnbull (talk) 12:52, 13 September 2020 (UTC)[reply]

LED Lightbulb Problem

I bought a new LED light bulb for a 3-way lamp, labeled 50-100-150W, which means that its output is equivalent to that of an old 50-100-150W incandescent three-way lightbulb. When I put the bulb into the lamp and cycled the switch, it went 0-50-100-50. I couldn't get it to produce what was supposed to be full illumination at full power. So then I thought that maybe there was something wrong with the lamp, so I switched this new bulb and another existing bulb between two similar lamps. The bulb still switched 0-50-100-50, and did not go up to 150. My question is: What is wrong? It isn't that one of the two sets of light-emitting diodes doesn't work. Each of the sets of diodes works. They just don't work at the same time. Why doesn't the higher-power set of diodes work when I have switched the lamp to energize both sets of diodes? Robert McClenon (talk) 17:50, 11 September 2020 (UTC)[reply]

This page talks about incompatibility of using incandescent lamp dimmers for LEDS. It might help, i don't really know much about the topic tbh. Zindor (talk) 18:02, 11 September 2020 (UTC)[reply]

Dimmers are irrelevant. The traditional 3-way light is a bulb with two separate filaments powered independently, and does not involve a dimmer. I don't know about Robert's problem but I would suggest contacting the manufacturer. --174.88.168.23 (talk) 01:22, 12 September 2020 (UTC)[reply]

Yes, that is a little bit weird. Robert, if you have a replacement incandescent bulb, maybe you can try that to see if the 3-way switch works properly. Also, examine the contacts on your LED bulb to see that they are in the right places. A 3-way LED bulb is kind of a hack, though. LED's are dimmable, though the dimming electronics are different from old-fashioned incandescent (triac) dimmers. Places like homedepot.com will have tons of stuff like that. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 02:25, 12 September 2020 (UTC)[reply]

I switched the bulb in question with a working 3-way incandescent bulb. The incandescent bulb worked fine. The questionable LED bulb continued to work 0-50-100-50 in another similar lamp. One comment I got is that maybe the complicated electronics in the bulb is trying to control the current to the diode assemblies, and that there is some error in the "intelligence". That is, if you put artificial intelligence in something, you risk putting artificial madness in it. So I have a bulb that doesn't work the way it is supposed to work. Robert McClenon (talk) 04:53, 12 September 2020 (UTC)[reply]

"if you put artificial intelligence in something, you risk putting artificial madness in it" sounds like the tagline to a great cyber-dystopia film. Flesh it out a bit with computer screens that cast glowing text on people's faces and maybe some boobs and you're got a hit on your hands. Let me know how to invest in this. Matt Deres (talk) 12:47, 12 September 2020 (UTC) [reply]

The phrase should be voiced by Jeff Goldblum, hopefully in a more pithy formulation. --Lambiam 20:26, 12 September 2020 (UTC)[reply]

Though this be artificial madness, yet there is an algorithm in 't. --Lambiam 20:30, 12 September 2020 (UTC)[reply]

It's worth closely reading the packaging to check whether it actually promises that it lights all the LEDs with both live contacts powered, or if it just promises to be "compatible" with 3-way lamp fixtures. Every standard bulb is compatible with a 3-way fixture because the fixture was designed to be "backwards-compatible"; there's just a second hot contact for the second filament in an incandescent. In a "3-way LED lamp", the power controller has to monitor both hot contacts and switch on all the LEDs when they're both powered. A chintzy bulb that was designed to be super-cheap while fooling you could shave off components. (Electronics people: would the bulb work as described if it just connected both hots to a common input? That would explain why it doesn't have a "150W" mode.) --47.146.63.87 (talk) 23:25, 12 September 2020 (UTC)[reply]

Robert, what is the brand and model of your particular bulb? -- ToE 16:08, 13 September 2020 (UTC)[reply]

User:Thinking of England - General Electric Soft-White 3-way LED bulb. Robert McClenon (talk) 05:10, 14 September 2020 (UTC)[reply]

I still suggest contacting the manufacturer. --174.88.168.23 (talk) 05:51, 16 September 2020 (UTC)[reply]

Reduction potential at mercury electrode

Hydrogen has high over-potential at mercury electrode. But I could not find the value when I searched Google. What are the reduction potentials of hydrogen, sodium, water and oxygen at mercury electrode in alkaline media? Thanks. Horus1927 20:21, 11 September 2020 (UTC) — Preceding unsigned comment added by Horus1927 (talk • contribs)

Many are tabulated at Standard electrode potential (data page). Usually, hydrogen is set at zero and any other is measured relative to that. Michael D. Turnbull (talk) 15:43, 13 September 2020 (UTC)[reply]

September 12

Ventilation fan

I'm wondering why typical household ventilation fans are as small as they are. My reasoning is as follows (tell me if I'm wrong somewhere). Ventilation requirements are basically in air changes per hour which means for a given room size you need a given cfm (cubic feet per minute) of air, and fans are rated in CFM. 1000 cfm would be a typical small indoor fan. Basically the fan pushes out a cylinder of air the same diameter as the fan blades, and its velocity (per minute) is the cfm divided by the area swept by the blades. From the molar weight of air and the ideal gas law, you can find the mass of 1000 cfm of air. The kinetic energy per minute is 1/2·m·v² which after some unit conversions gives you the electric power in watts needed to run the fan (there is a small loss factor but this wattage really does seem to be reflected in the fan specs). The thing here is, that the cfm stays constant but the airspeed (and therefore power) decreases with the square of the diameter. So to minimize your electric bill, you really should want the largest diameter fan you can get.

This shows up in big ceiling fans, but right now I'm using a 20 inch box fan, it is pretty noisy and power hungry, and I'm wondering why I couldn't get one that was say 4 feet in diameter. Any idea? Am I missing something? Thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 02:31, 12 September 2020 (UTC)[reply]

Additional question: I have the 20" box fan blowing through a furnace filter to help clean the air coming out.[4] The filter slows down the airflow somewhat, so that would be one advantage of a smaller (higher velocity) fan. Is there a reasonable low-tech way to gauge the airspeed change? Say, by seeing how much a piece of paper (maybe weighted by a penny) hung in front of it gets deflected? I could try to concoct something like this. I have a voltmeter so in principle I could measure the electrical power going into the fan, but I'm not really concerned about the power bill right now. I mostly want to compute the cfm getting through the filter. Thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 02:38, 12 September 2020 (UTC)[reply]

The something that you're missing is practicality. A large fan on the ceiling is acceptable, but consumers are concerned with factors in addition to efficiency, such as: How would a 4-foot fan fit with my decor; and where can I find space for such a thing? -- They don't make 4 ft. residential fans because nobody would buy them (with one exception?); but, if you really want to, you can buy one: [5] —2606:A000:1126:28D:FDD2:BED1:91BF:2ACB (talk) 05:32, 12 September 2020 (UTC)[reply]

That fan you linked is a huge, expensive, noisy, 1.5hp (1.1kw) industrial fan. I was imagining something that ran at much lower speed. very quiet, using little energy. If not 48", then 30" or so seems practical for many settings such as in a wide window. Right now my 20" fan is sitting on a chair in the middle of the room, which is terrible decor by any stretch, but it's helping the air and a 30" would fit about as well. I also have to wonder why anyone uses 12" or 16" fans when they can use 20". Finally I'm wondering whether this relationship between diameter and power efficiency (watts per cfm) is well known--or alternatively, whether it is mistaken. It does make me more interested in getting a large ceiling fan for hot days. Home Depot has some residential ones up to 7 feet across, and industrial ones (appropriately called "big ass fan"[6]) up to 14 feet (750 watts, they don't even tell you the cfm but I think I can calculate it from the diameter and power). Anyway, thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 08:15, 12 September 2020 (UTC)[reply]

I would refer you to Big Ass Fans for all your Big Ass Fan needs.... - Nunh-huh 05:15, 14 September 2020 (UTC)[reply]

A larger box fan might also run into safety concerns, especially as regards children and small animals around it. --Khajidha (talk) 14:16, 13 September 2020 (UTC)[reply]

Optically hide a big fan in new construction with a dropped ceiling, pitch black tallest open-ended honeycomb that's worth it ceiling grate for the fake lower ceiling, pitch black entire ventilation system and largest gap between top of grate and bottom of fan that's worth it. You can paint the bottom of the grate any color you want, in fact a light color might camo the fan more. Put as many lights as you can that still seems homey, all under the grate with caps to prevent direct grate illumination and evenly distributed in area and brightness, summing to a typical total room illumination level in all. Well that's the best I could come up with. Sagittarian Milky Way (talk) 13:16, 12 September 2020 (UTC)[reply]

Or slowly and slightly turn the top half or more of a very tall but otherwise standard square subway grate-type grate, you can make it geometrically impossible to see the fan no matter where you put your eyeball. The top of the grate cells should be pointing up and to the left or right or front or back, whichever is downwind in the exhaust system. Sagittarian Milky Way (talk) 17:32, 12 September 2020 (UTC)[reply]

If I got a ceiling fan I wouldn't want it to be hidden. They look cool as hell imho. But, part of that is they look a bit retro since the modern era generally uses a/c instead of fans. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 20:12, 12 September 2020 (UTC)[reply]

I may be mistaken, but I think most of the energy expenditure goes into the mechanical friction of the motor plus electrical losses because electrical motors are not 100% efficient even disregarding friction. The energy saved by using a fan with a large diameter may be disappointing. Noise production costs energy, so go for the less noisy fans. Commercial ceiling fans have anywhere from two to ten blades, sometimes curved like the wooden propellors of early planes, sometimes plain and flat. I think the curved ones are that way because it looks slick, and the flat ones because it is cheaper to manufacture. I doubt the areodynamic properties are a serious factor entering the considerations of the manufacturers. --Lambiam 20:22, 12 September 2020 (UTC)[reply]

I haven't done the math for very large or small fans, but for midsized ones (and for vacuum cleaners), the energy really does seem to mostly go into moving the air. See air watt. I believe my 20" box fan uses around 100W (haven't measured) and it definitely makes noise, but think of how loud a 100W stereo would be. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 20:45, 12 September 2020 (UTC)[reply]

The Dyson C47 has reportedly a suction power of 180 airwatts, but an input power of 1210 W.^[7], which is an efficiency of 15%. --Lambiam 16:25, 13 September 2020 (UTC)[reply]

What makes radar so valuable?

I read about radars, and it seems that except for special beyond-horizon radars; most radars can detect only objects which are in a sight line with the object. Meaning that theoretically, can been seen with a naked eye(with binoculars I guess?). So why then the radar makes such a different? why nobody uses a visible light for example?--Exx8 (talk) 18:47, 12 September 2020 (UTC)[reply]

A fighter jet at radar limit is too small for binoculars. In daytime. Sagittarian Milky Way (talk) 19:16, 12 September 2020 (UTC)[reply]

See Chain Home for the first early warning radar, which had a detection range of 160km (100 miles) back in 1936. Mikenorton (talk) 19:29, 12 September 2020 (UTC)[reply]

Another factor is that radar can "see through" clouds and other meteorological phenomena. See Radar#Principles. AndrewWTaylor (talk) 19:34, 12 September 2020 (UTC)[reply]

In addition to the points above, even in fair weather you'd need to keep scanning around in a 360° sweep to notice vehicles, possibly several, approaching from unpredictable directions. With radar you need to look merely at the screen. --Lambiam 19:57, 12 September 2020 (UTC)[reply]

@Sagittarian Milky Way, why? what makes radar able to detect thing in a distance which can't be observed with visible light?--Exx8 (talk) 20:25, 12 September 2020 (UTC)[reply]

Well, at night it is dark, so you can't observe anything with visible light unless it has lights on or you light it up somehow, in which case black paint would do a good job of stopping reflections. During the day your sensors will be flooded with sunlight because it is daytime so you have to somehow resolve a tiny dark speck. Maybe you could do that with a high resolution wide angle camera these days but that takes fairly modern tech. With radar, you illuminate the target and it's enough to notice a very faint reflection since there is not much background noise at radar frequencies. It's like being able to see a distant star at night because of the faint light reaching you. Finally, you can measure the doppler shift in the reflected radar signal and that tells you the velocity of a moving target. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 20:51, 12 September 2020 (UTC)[reply]

At night, if you don't have too much light pollution, you can look for artificial satellites passing overhead. Except, you don't get to look at any one for very long, because they'll pass through your line of sight in tens of seconds. Things in orbit are moving quite a bit faster than aircraft, but still, this gives you an idea. Now imagine trying to pinpoint something in the daytime moving fast at 50,000 feet, and keep your bearing on it while it moves, and trying to identify it. And then do that for possibly dozens of aircraft at a time. --47.146.63.87 (talk) 23:31, 12 September 2020 (UTC)[reply]

RADAR = "RAdio Detection And Ranging". In addition to the detection and determination of azimuth, which radar does better than visual sighting as described above, it also gives accurate ranging to aid in tracking or targeting. -- ToE 00:16, 13 September 2020 (UTC)[reply]

I read an article recently about Joan Curran who invented Radar chaff. Was very interesting to learn about her work. Zindor (talk) 12:53, 13 September 2020 (UTC)[reply]

This book (which made its author famous) may be of interest [8]. 2A02:C7F:D63F:AF00:48A1:AB9F:4B06:6D65 (talk) 18:04, 14 September 2020 (UTC)[reply]

September 13

The speed of electricity

How fast does electricity move? I've heard it said that electricity moves at the speed of light. If this is the case then surely there is something else, other than light, that moves at the speed of light. Does this not contravene the rules of physics? Thank you — Preceding unsigned comment added by 2A00:23C6:6884:6200:75D7:FCD:5331:8FC7 (talk) 09:02, 13 September 2020 (UTC)[reply]

Have you looked at Speed of electricity? When people say that electric current moves at the speed of light (even though that is not entirely accurate), what they mean is that the wave/signal travels at that speed; the electrons or ions that transport the electric charges move at a much slower rate. Anything with a non-zero mass that moves at the speed of light would violate the laws of physics, but electric current has no inherent mass, so it does not violate any physical law, even if it would move at the speed of light. - Lindert (talk) 11:21, 13 September 2020 (UTC)[reply]

It's the speed of light, in the conductor (if they weren't opaque). The linked article Velocity factors gives typical speeds. For copper it's around 2/3rds of the speed of light in a vacuum, or 200,000 km/sec. LongHairedFop (talk) 13:14, 13 September 2020 (UTC)[reply]

The "speed of light" is not inherently about light at all, which makes the name somewhat poor. (Like many names of things in science, it caught on by happenstance, and now we just have to put up with it.) Special relativity asserts that no two points in spacetime can exchange information faster than c. It just so happens that c is also the speed at which electromagnetic waves (including light) travel in a vacuum. Since "electricity" is an electromagnetic wave (or at least can be modeled as one), it will travel at the "speed of light" in whatever thing it's moving through; this is always less than c if it's not a vacuum. Your confusion is common and completely understandable, resulting from the somewhat misleading name. I suggest this video for more detail. --47.146.63.87 (talk) 07:02, 14 September 2020 (UTC)[reply]

black hole collision

[9] If two black holes collide and merge into a bigger black hole, during the approach does that at least temporarily create a region from which no light can escape, but whose shape is something like a figure 8 or dumbbell? And is that in conflict with the no-hair theorem that says black holes are always spherical? Thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 15:27, 13 September 2020 (UTC)[reply]

Yes, it creates a toroid [10]. As for whether that violates the no hair theorem, I don't know. Handschuh-^{talk to me} 00:53, 14 September 2020 (UTC)[reply]

Only non-spinning black holes have a spherical event horizon. The no-hair theorem doesn't assert "black holes are spherical"; it just asserts "stable" black holes can be fully described by eleven parameters. Also, again, it only applies to "stable" black holes. --47.146.63.87 (talk) 06:48, 14 September 2020 (UTC)[reply]

Does this mean that the Einstein–Maxwell equations of gravitation and electromagnetism in general relativity only apply to stationary situations? If so, the text of the articles concerned should make that clear. --Lambiam 07:53, 14 September 2020 (UTC)[reply]

The equations apply to any situation, of course, stationary or not. But the idea was that the no-hair "theorem" (it's a conjecture, strictly speaking) only applies to stationary solutions of the equations. I thought so until five minutes ago, but apparently that's not correct. This paper says that the frequencies and decay rates of the overtone modes of the ringdowns signal are uniquely determined by the final hole’s mass and spin magnitude. Hence, while you can talk about non-spherical shapes during ringdown that does not give you any new information beyond mass and spin. The situation already applies to the stable Kerr black hole: the event horizon is non-spherical, so you can measure an ellipticity of the black hole. But again, this tells you nothing new as the ellipticity is uniquely determined by the spin. In that sense the no-hair theorem applies to the ringdown phase as well. --Wrongfilter (talk) 08:51, 14 September 2020 (UTC)[reply]

Continuity considerations suggest compellingly that there has to be a transitional phase between the event horizons of the merging black holes being a pair of perfect spheroids before the merger, and the event horizon being a single spheroid after the merger. This is, however, excluded by the formulation given by the No-hair theorem article. So which one is it: continuity but additional conditions needed for the theorem to apply as stated, or a physical discontinuity? --Lambiam 15:31, 14 September 2020 (UTC)[reply]

The transitional phase is called ringdown, and from the cited paper I gather that the way ringdown proceeds is uniquely determined by total mass and total spin (I wonder whether the mass ratio shouldn't have an influence, though). But nobody said anything about discontinuity. --Wrongfilter (talk) 16:56, 14 September 2020 (UTC)[reply]

Presumably nobody said anything about discontinuity because there is none. But if during the ringdown period, however brief, "any distortion in the shape is dissipated as more gravitational waves",^[11] the no-hair theorem is not unconditionally true in its stated form. (Some black holes do ring, others with exactly the same mass, electric charge, and angular momentum do not.) --Lambiam 17:35, 14 September 2020 (UTC)[reply]

Well, I only cited a peer-reviewed paper, so... --Wrongfilter (talk) 17:52, 14 September 2020 (UTC)[reply]

And yet another way to look at the ring down is that it is just an echo of the gravitational wave lensed by the black hole. You can expect infinite echos, getting weaker and weaker as light or gravitation waves passing very close in get bent more and more, and exceed 360°, 720° and even more extreme spiralling around the final black hole. I may have written the bit Lambiam quoted, but it may not be the right way to look at ringdown. Note that you will never observe an event horizon or its formation, and that the black hole (s) have not yet actually been formed, as they only do so in our infinite future. Timescales are seriously distorted. Graeme Bartlett (talk) 01:07, 18 September 2020 (UTC)[reply]

(Yes, you did). We lack the terminology to properly discuss black holes in English, so any such exposition going into detail is in some way wrong or misleading. I maintain that the idea that a pair of perfect spheroids "instantaneously" merges into one perfect spheroid is more misleading than others. This idea stems from a popular formulation of the no-hair theorem, which therefore is apparently also misleading. But all will become clear in our infinite future. --Lambiam 07:19, 18 September 2020 (UTC)[reply]

September 14

Effects of Gravitational Waves

The gravitational wave sources detected by LIGO and other experiments are so far away that the amplitude of the waves are tiny by the time they reach earth. The article section Gravitational_wave#Effects_of_passing gives a description of how it would affect some particles on a plane, but I'm interested to know what the practical effects of a wave would be on material close to an intense source. Would it distort materials so much it destroys them? Would it injure organisms? Or does a ripple in the spacetime curvature not really affect things too much? If a tertiary system existed consisting of two black holes and a third massive object, could the merger of the two black holes have some substantial effect on the third object? Say the third object was a star, could the distortion affect the rate of fusion in the star's core, or trigger a gravitational collapse of the star, or anything like that? Handschuh-^{talk to me} 01:13, 14 September 2020 (UTC)[reply]

Not an answer, but Scott Manley recently put up a video, 'Impossible' Black Hole Created by Largest Gravitational Wave Event [YouTube, 10:51], about GW190521, the gravitational wave signal detected on 21 May 2019 of what appears to have been the merger of two black holes of 85 and 66 solar masses, resulting in a single black hole of 142 solar masses, where the "remaining 9 solar masses were radiated as energy in the form of gravitational waves"! -- ToE 01:45, 14 September 2020 (UTC)[reply]

Yes, it was reading about the 9 solar masses of energy radiated out into the universe that go me thinking of what kind of effect it might have on something/someone close to the source, and also, the coinciding light flash (for which astronomers have already proposed an alternative explanation) got me thinking about how the GW might affect nearby stars. Handschuh-^{talk to me} 02:26, 14 September 2020 (UTC)[reply]

Also not a full answer, but the frequency of the first gravitational wave ever observed went up to 250 Hz. That means a wavelength of more than 1000 km. The effect on a much smaller body being sloshed around in a wave of an extreme amplitude is much less violent than that of waves with ripples commensurate with the size of the body. Warning. This does not mean it is safe to observe a black hole merger from up close. Always keep a safe distance. --Lambiam 07:45, 14 September 2020 (UTC)[reply]

If you take the observed strain and use that it decays inversely proportional to the distance fro the source, then you find that at a distance of 1 AU from the source, the strain would have been just about detectable by our seismometers. Count Iblis (talk) 11:28, 16 September 2020 (UTC)[reply]

That interesting...so at a very short cosmological distance the effects are hardly earth shattering. Where did you get the observed strain from? It doesn't seem to be in the GW190521 article. Handschuh-^{talk to me} 23:39, 16 September 2020 (UTC)[reply]

I used the strain sensitivity of

10^{-23}/{\sqrt {\text{Hz}}}

of LIGO, the observed strain is then an order of magnitude larger, you can then multiply this by the distance to the source divided by the distance of some hypothetical planet close to the source to obtain the strain at that planet. Count Iblis (talk) 22:46, 17 September 2020 (UTC)[reply]

Marsh and swamp

In Polunin and Walters A Guide to Vegetation of Britain and Europe there is a different definition of marsh and swamp from Wikipedia (i.e. a marsh is a wetland dominated by grassland and swamp a forested wetland): a marsh is an area where at least in summer the water level is below the soil surface, whereas a swamp is permanently saturated with water. Could this distinction be defined with other words than marsh and swamp?--Carnby (talk) 16:02, 14 September 2020 (UTC)[reply]

I'm going to go on a bit of a side rant here; I have long suspected that scientists create word distinctions among synonyms that didn't used to exist, but try to make them so, and sometimes different groups of scientists will create different distinctions than each other because they never got together and got their story straight. The English Language has a surprising variety of words for (well anything, but in this case...) areas of non-quite-land-not-quite-water, and I'm not sure the distinction between those words as you note existed as long as the different words themselves. I suspect such word difference were initially geographic rather than in quality or character, so that people from one area called such land a "swamp" or a "bog" or a "marsh" or a "bayou" or a "mire" or a "fen" or a... whatever. People just named the wetland by their local name, and later the distinctions between such things as whether it had trees or grasses or whatnot was imposed by people trying to make sense of the system. The problem you note, where two different supposedly expert sources have two mutually exclusive and entirely non-compatible systems for drawing such distinctions, is a symptom of that history. In the means of actually providing sources to my side-rant, consider the etymology of various terms, swamp, marsh, fen, bog, they all have, in their origins going back to the 1500s or earlier, the same basic meaning, that being "wet spongy ground". That being said, by 1775, the distinction between swamp and marsh was drawn, as noted in the link I provided for swamp, "[B]y swamps then in general is to be understood any low grounds subject to inundations, distinguished from marshes, in having a large growth of timber, and much underwood, canes, reeds, wythes, vines, briers, and such like, so matted together, that they are in a great measure impenetrable to man or beast .... [Bernard Romans, "A Concise History of East and West Florida," 1775]" So take that what you will. --Jayron 32 11:38, 15 September 2020 (UTC)[reply]

To complicate things even more, which I assume everyone is begging for, it appears plausible (at least to me) that different branches of physical geography that study wetlands from different angles (e.g. ecology versus hydrology) use different classifications. Some fields (e.g. limnology) were developed originally by non-English scientists, and there may not be a clear correspondence beteen the French and German terms and the English ones. For the variety of types of wetlands, just look at Category:Wetlands – good luck developing an unambiguous classification for these. I noticed that Moorland is not in that category and not mentioned in the wetlands article, although some moors definitely are and many (most?) that are not started out that way, as mires. --Lambiam 16:02, 15 September 2020 (UTC)[reply]

" I have long suspected that scientists create word distinctions among synonyms that didn't used to exist" I think we can safely say that suspicion is more than justified. Speed and velocity being a typical example (and even acceleration or it's latin root probably had a similar meaning, lacking the technical distinction we now impose upon it). Handschuh-^{talk to me} 23:47, 16 September 2020 (UTC)[reply]

Similarly "Mass" and "Weight". --Jayron 32 15:02, 17 September 2020 (UTC)[reply]

The existing English word mass was repurposed to translate Newton's repurposing of the Latin massa for a concept that did not yet exist; the same happened for acceleration and inertia. Obviously, these terms did not have the technical physics meanings they have today before someone introduced the concepts they name. If these terms had synonyms (inertia = sloth), the technical sense was not simultaneously imparted on the synonyms. The earlier, everyday meaning of mass(a) was "lump (of dough)". --Lambiam 20:38, 17 September 2020 (UTC)[reply]

Yeah, but there's a few bad teaching practices that try to explain that "weight" only refers to gravitational mass while "mass" only refers to inertial mass, that's basically bullshit because we've known the two concepts are the same thing due to the equivalence principle. Even during the time of Newton there was suspicion they were because any measurement of one resulted in an equivalent measure of the other, but such conjectures were not proven until about 100 years ago. There's some other texts and things that try to say that weight is only referring to gravitational force (i.e. Newtons) while mass is "mass" (i.e. kilograms), but I don't know any actual physicists that make such distinctions and legitimately say things like "Such and such weights X Newtons". If one means force, they say force. If one means weight, the use kilograms just fine. --Jayron 32 13:03, 18 September 2020 (UTC)[reply]

I'd like to add "morass" to the discussion. HiLo48 (talk) 04:30, 17 September 2020 (UTC)[reply]

as in "I've been eating a lot during this quarantine and I have morass than I used to.--Jayron 32 07:35, 17 September 2020 (UTC)[reply]

Let's not get bogged down in trivia. ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:36, 17 September 2020 (UTC)[reply]

September 17

Teach flight tricks to parrot?

Is it possible to teach a parrot (goffin) to do loops and barrel rolls in flight? Or things like flying upside down or backwards? Can't find much online. Thanks. 146.200.128.134 (talk) 04:22, 17 September 2020 (UTC)[reply]

Despite their intelligence, goffins are not built for acrobatics, and I doubt whether any bird would be able to fly upside down or backwards.--Shantavira|^{feed me} 08:57, 17 September 2020 (UTC)[reply]

Hummingbirds can fly backwards and upsidedown. --TrogWoolley (talk) 09:32, 17 September 2020 (UTC)[reply]

I've seen Hummingbirds vanish into thin air, then reappear a moment later 30 metres away. Amazing birds. Try typing 'do a barrel roll' into your parrot, might work. Zindor (talk) 11:26, 17 September 2020 (UTC)[reply]

Seagulls can fly (or glide) backwards. 146.200.128.134 (talk) 17:13, 17 September 2020 (UTC)[reply]

But can they do it in high heels? —Tamfang (talk) 01:59, 18 September 2020 (UTC)[reply]

September 18

Energy from rusting

Roughly speaking how much energy, in Joules, is given off by completely rusting 1Kg of Iron? 2A01:E34:EF5E:4640:6149:84E9:11CE:A892 (talk) 09:47, 18 September 2020 (UTC)[reply]

The article on energy density has a handy table. Rust can have different compositions, but the "burned to iron(iii) oxide" row should be a decent enough approximation. 85.76.15.6 (talk) 10:52, 18 September 2020 (UTC)[reply]

Why are Pnictogen hydrides so poisonous?

Why are Pnictogen hydrides like ammonia,phosphine and arsine so poisonous?Acidic Carbon (Corrode) 13:23, 18 September 2020 (UTC)[reply]

Should there be a redirect: Pnictogen hydrides → Pnictogen hydride ? --2606:A000:1126:28D:31A8:66E0:89AA:D98E (talk) 16:48, 18 September 2020 (UTC)[reply]

Each one's article talks about its specific toxicity profile. At least phosphine#Toxicity, arsine#Toxicology, and stibine#Toxicology are quite explicit about what their mode of action is. DMacks (talk) 17:05, 18 September 2020 (UTC)[reply]

Missing a hurricane

Hurricane Alice (December 1954) says that two cyclones, including a Category 2 hurricane, operationally went undetected in 1954. In a time before satellite pictures, how would meteorologists learn that they hadn't noticed a storm? 2601:5C6:8081:35C0:2C6E:461:D260:3BF (talk) 22:26, 18 September 2020 (UTC)[reply]

No source is cited for that specific statement, but José A. Colón's paper cited as a reference from the paragraph refers to another undetected hurricane in January 1951. Basically the answer to "how did they learn" seems to be that they reanalyzed the data they had already collected, when the storms were over with and they had time to do it. --174.88.168.23 (talk) 23:08, 18 September 2020 (UTC)[reply]

[physconst-G-1] "2022 CODATA Value: Newtonian constant of gravitation". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.

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