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March 4

Why is Nyiragongo a stratovolcano?

Given the fluidity of its lava, why isn't it a shield volcano? ZFT (talk) 04:27, 4 March 2021 (UTC)

That's a good question, the neighbouring Nyamuragira is a shield volcano, although it erupts high potassium basalts. I've looked through this thorough review and I can't find a clear explanation - apart from the earliest stage eruptions, the volcano is dominated by lava flows. Mikenorton (talk) 09:04, 4 March 2021 (UTC)

I've seen a discussion a while ago on another volcano that the eventual slope of a volcano depends on the median lava flow length, which in turn is a function of the median volume of each eruption. So even a volcano with very liquid lavas will grow steep if the median lava flow is short. Jo-Jo Eumerus (talk) 11:19, 5 March 2021 (UTC)

Do astronomical observatories really shoot laser into sky like this?

Do astronomical observatories really shoot laser into sky like this? Here is the picture taken from Optical amplifier Rizosome (talk) 06:28, 4 March 2021 (UTC)

See laser guide star for an explanation of the technique - why would you doubt it? Mikenorton (talk) 08:44, 4 March 2021 (UTC)

In usual terminology, space observatories are observatories located in outer space. Observatories observing space (usually from the ground, but sometimes from space or even a high-altitude balloon, aircraft or from below the Earth's surface) are called astronomical observatories. And yes, sometimes they shoot a laser into the sky. Apart from lunar ranging, it's sometimes done for adaptive optics. Adaptive optics uses a deformable mirror to correct for diffraction of light in moving bubbles of air of different density. To measure the required corrections, the system has to observe a sufficiently bright reference star (much brighter than the object you're interested in). If no sufficiently bright star is available close enough to the object you want to observe, a laser can be used to project one in the higher parts of the atmosphere. Of course, they have to make sure they don't hit an aircraft passing over, as those lasers can be very annoying to pilots (just as aircraft can be annoying to astronomers). PiusImpavidus (talk) 09:02, 4 March 2021 (UTC) @PiusImpavidus: I changed my title, still your answer supports my question? Rizosome (talk) 13:46, 4 March 2021 (UTC)

Yes, that was already the question I answered. The laser-equipped observatories are always ground-based, not in the air, in space or below ground. Except gravitational wave observatories, which are laser-equipped and can be below ground, but don't shoot their laser beams into the sky. PiusImpavidus (talk) 16:35, 4 March 2021 (UTC)

The Royal Observatory, Greenwich has a similar laser, The Meridian Laser, which only serves to show the location of the Greenwich Meridian and has no scientific purpose beyond education. It's more of a museum than a working observatory, but one of their historic telescopes has been brought back into use recently. Alansplodge (talk) 15:30, 4 March 2021 (UTC)

Yes, I once saw that, when leaving the Maritime Museum on a winter afternoon/evening (after the fire alarm went off). PiusImpavidus (talk) 16:35, 4 March 2021 (UTC)

March 5

Seabird IDs

In this video, what are the cliff-nesting birds first seen at 23 seconds, and the tern-like birds at 3:30? --24.43.123.79 (talk) 00:19, 5 March 2021 (UTC)

The birds at 23 seconds are probably Bermuda petrels, a deduction based on the fact that the list of CornellLab's bird cams shows they have a couple of cams covering this species. PaleCloudedWhite (talk) 07:46, 5 March 2021 (UTC)

My best guess for the tern-like birds - based not on personal knowledge, but on perusing lots of google searches - is red-billed tropicbird. The images at our article seem to have very red bills - maybe too red for the birds in the video - though the images of the same species at this google image search seem to fit better. PaleCloudedWhite (talk) 09:10, 5 March 2021 (UTC)

Why Space X didn't reveal the reason behind SN 10 failure?

Why Space X still didn't reveal the reason behind SN 10 failure? Rizosome (talk) 02:49, 5 March 2021 (UTC)

Is that the one that blew up within the last day or so? ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:26, 5 March 2021 (UTC)

Maybe because they aren't yet completely sure what the reason was? Our article already says "due to a suspected methane leak." In any case, as a privately owned company conducting tests on prototypes, they are under no obligation to reveal their findings immediately, and will doubtless only do so when and if it becomes advantageous for them. {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 06:20, 5 March 2021 (UTC)

• It's been a day and a half, what did you expect? It could be easily observed that at least some of the landing legs didn't lock into place, so that might have something to do with it. However, SpaceX (Elon Musk's twitter probably) will let us know when they worked it out, they always have in the past (cf the Starlink 19 landing failure already having been explained). Fgf10 (talk) 09:34, 5 March 2021 (UTC)
• (ec) Usually they provide a full explanation for such failures, but the investigation needed for a full explanation takes time. If there was a methane leak, then the question is, how come? A leak can perhaps explain the explosion, but there should have been no leak in the first place. Did something rupture? If so what, how, and why? There is also some chatter about the landing legs not properly deploying,^[1] which may have been a factor in a chain of mishaps. It is wonderful how the Space X people (that is, people of Space X, not X-Men from space) describe the incident in a self-congratulory text on their website:

 As if the flight test was not exciting enough, SN10 experienced a rapid unscheduled disassembly shortly after landing. 

Fortunately there was no one on board, or they too would have experienced a rapid unscheduled disassembly.  --Lambiam 09:45, 5 March 2021 (UTC)

Reports yesterday said initially it's suspected there was a leak. Hence the kaboom. That term "unscheduled disassembly" reminds me of when Challenger exploded in mid-air, and the NASA announcer describing things while they were happening called it "a major malfunction". ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:57, 5 March 2021 (UTC)

First, the announcer who said that didn't have access to live video of the flight. Second, there was no explosion; quoting from memory, what people misperceived as an explosion was only "a fast, almost explosive fire". Third, if by the Challenger you mean the orbiter, it not only didn't explode, it came away from that event intact, only to be destroyed on impact with the sea surface. --142.112.149.107 (talk) 21:54, 5 March 2021 (UTC)

How do you know he didn't have the live video feed? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:11, 5 March 2021 (UTC)

He "was reading routine values off gauges at his console". 39 seconds later he reported that the flight dynamics officer had reported an explosion. --142.112.149.107 (talk) 23:44, 6 March 2021 (UTC)

• It can take months of investigation to understand the causes of such disasters. The Challenger disaster investigation took a long time; the Rogers Commission Report was released in June of 1986, the Challenger disaster happened in January of that year. The Columbia disaster investigation took a similar amount of time, the Columbia Accident Investigation Board released a report in August, 2003 when the accident happened in February. The VSS Enterprise crash investigation took even longer; the final findings were released in July 2015, nine months after the October 2014 crash. I wouldn't expect anything for some time. --Jayron32 12:58, 5 March 2021 (UTC)

  SpaceX doesn't operate like that in most cases. If the cause is obvious either they or Elon Musk tends to release it within days. That's what has happened with previous Starship tests and Falcon 9 landing failures. The exceptions being the two failures on launches for external customers . Fgf10 (talk) 19:24, 5 March 2021 (UTC)

  But what if they don't know? The OPs question of "Why hasn't SpaceX said why the rocket failed?" has an obvious answer. "They don't know yet". --Jayron32 20:39, 5 March 2021 (UTC)

  These are tests that are conducted with the express goal of finding out what failures may happen so that the design can be improved. Everything is heavily instrumented and recorded to make it possible to quickly get at the bottom.  --Lambiam 20:46, 5 March 2021 (UTC)

Elon Musk:

Thrust was low despite being commanded high for reasons unknown at present, hence hard touchdown. We’ve never seen this before.

Next time, min two engines all the way to the ground & restart engine 3 if engine 1 or 2 have issues.

Tweeted 2010-02-06 01:51 UTC.

Remember that the SpaceX Raptor engine is still being developed, tested, and refined. -- ToE 02:18, 6 March 2021 (UTC)

Another Musk Tweet:

SN10 engine was low on thrust due (probably) to partial helium ingestion from fuel header tank. Impact of 10m/s crushed legs & part of skirt. Multiple fixes in work for SN11.

Q from NASASpaceFlight's Chris Bergin:

This is a tricky one given that I believe said helium pressurization was added to the CH4 header tank to mitigate what happened with SN8. That's why it's a test program, of course.

Musk:

Fair point. If autogenous pressurization had been used, CH4 bubbles would most likely have reverted to liquid.

Helium in header was used to prevent ullage collapse from slosh, which happened in prior flight. My fault for approving. Sounded good at the time.

-- ToE 21:00, 9 March 2021 (UTC)

Role of anaerobic metabolism in the global carbon cycle

Has anyone ever tried to quantitatively estimate the role of anaerobic metabolism in the global carbon cycle? Jo-Jo Eumerus (talk) 11:17, 5 March 2021 (UTC)

The citation quoted in anaerobic metabolism that's most likely to have a good estimate is (on Pubmed) [2] which has a "simliar articles" section that could also point to useful items. Ironically, at first sight one might think that organisms which take carbon dioxide out of the atmosphere, as some anaerobes do, would be good to mitigate global warming. Unfortunately, they turn it into methane, which is an even worse greenhouse gas! Mike Turnbull (talk) 14:10, 5 March 2021 (UTC)

Unfortunately, that source is focused on man-made emissions, but I need information on all anaerobic throughput. Jo-Jo Eumerus (talk) 10:51, 6 March 2021 (UTC)

@Jo-Jo Eumerus:: I know of one book that covers the GCC pretty comprehensively, and does mention anaerobes. The book is Mannion, A. M. (12 January 2006). Carbon and Its Domestication. ISBN 1402039565., with e-book at ISBN=1402039581. Looking at my copy, on page 53 it has a carbon budget in units of 10⁹ tC yr^-1 that talks about a "missing sink" — maybe that could be where anaerobes have a role? The book refers to the IPCC's "Climate Change 2001. The Scientific Basis", which has its preface/foreword "here" (PDF).. Later reports may be even more relevant. Mike Turnbull (talk) 13:14, 6 March 2021 (UTC)

March 6

Equation regarding amount of heat in a certain amount of matter.

Hi, I was looking for an equation which tells me how much a certain amount of heat in a certain amount of matter will translate into.

As a reference, it has been said that the head of a pin as hot as the surface of the sun (10,000 Fahrenheit) could kill a person from 90 miles away.

I was looking for a specific equation though - for example, assuming no other variables - If I was standing 90 miles away from a pinhead (let's say 1.5 mm diameter) as hot as the sun's surface (10,000 F) - how hot would it be where I'm standing? Thanks.--IBBishops (talk) 00:04, 6 March 2021 (UTC)

Simple answer: The Sun subtends an angle of approximately 0.52° as observed from here on Earth. Since a pinhead at 90 miles will appear much, much smaller, you would feel much, much less radiant heat from it than you feel from the Sun.

If your pinhead has a diameter of 1.5 mm, then at 90 miles it subtends only 5.9×10^-7 °. Since you are interested in the solid angle of the heat source, you want the square of the ratio of the subtended angles: Thus the received radiant heat from the pinhead will be (5.9×10^-7 ° / 0.52°)² = 1.3×10^-12 that of the sun. Inversely, the sun will feel 770 billion (7.7×10¹¹) times warmer to you. -- ToE 03:50, 6 March 2021 (UTC)

Nearly as simple answer: Per Solar radius: 1 R_☉ = 695,700 km, so your pinhead is 0.75 mm / 695,700 (km/R_☉) = 1.1×10^-12 R_☉ in radius.

Per Astronomical unit: 1 AU = 92955807 miles, so your pinhead is 90 mi / 92955807 (mi/AU) = 9.7×10^-7 AU away.

Given its temperature, the amount of heat radiated by the sun (or your pinhead-sun) is a function of its surface area, which varies with its radius squared. The portion of that heat your body intercepts is inversely proportional to your distance squared from the heat source. So the heat you receive from your pinhead will be (1.1×10^-12 / 9.7×10^-7)² = 1.3×10^-12 that of the sun. Inversely, the sun will feel 770 billion (7.7×10¹¹) times warmer to you. -- ToE 04:43, 6 March 2021 (UTC)

These two back-of-the-envelope calculations match because they are in essence the same geometrical argument.

But they both ignore the practical issue that if you were to be in 90 mi line-of-site of your pinhead, its radiant heat would be further attenuated by passing through much, much more atmosphere than that of the mid-day sun. -- ToE 04:43, 6 March 2021 (UTC)

But you want a black-body radiation calculation? Stefan–Boltzmann will oblige.

The total power emitted by your pinhead is 4πr²σT⁴ = 4π(0.00075 m)²(5.67×10−8 W⋅m⁻²⋅K⁻⁴)(5800 K)⁴ = 454 watts.

(That's slightly less than 1/3 that of a typical 1500 W electric space heater!)

Divide that be the area of a 90 mi radius sphere = 4π(90 mi)² = 2.63×10¹¹ m² to determine the irradiance of 1.72×10^-9 W/m².

The solar constant is 1361 W/m², which is 791 billion times greater than the calculated irradiance of the pin, which is a close enough match given the precision of these calculations.

(Both the solar constant and the calculated irradiance of the pin do not consider any atmospheric attenuation.) -- ToE 05:48, 6 March 2021 (UTC)

But you asked "how hot would it be where I'm standing?" and I instead answered what the radiant flux density would be.

We could instead calculate your planetary equilibrium temperature as if you were a rotating (or heat conductive) black body in a universe empty of everything else except your 5800 K pinhead 90 miles away. How warm would it keep you?

$T_{\rm {you}}={\sqrt[{4}]{{\frac {A_{\rm {abs}}}{A_{\rm {rad}}}}{\frac {1}{\sigma }}{\frac {L}{4\pi D^{2}}}}}$

Where:

A_abs/A_rad is the ratio of your absorption area to your radiation area. In this case it is 1/4, which is the ratio of the area of the disk, πr², you present perpendicular to the pinheadlight to that of your entire surface area, 4πr², with which you reradiate that energy via your own black body radiation. (We are determining your equilibrium temperature where those energy fluxes balance.)

L/(4πD²) is the irradiance we calculated above, with L being the total luminosity of your pinhead and D the 90 mi between you two.

The 1/σ and fourth root are from us solving the Stefan–Boltzmann law for temperature.

And for a black body albedo a = 0 & emissivity ε = 1 so those terms don't appear.

Thus T_you = ( (1/4)(1/(5.67×10−8 W⋅m−2⋅K−4))(1.72×10-9 W/m2) )^1/4 = 0.295 K.

So as hot as that pinhead is, it does almost nothing to keep you warm. (Though there are colder spots in Florida.)

If you wish to recalculate for objects of different sizes and distance, it's much better to substitute in our luminance formula and simplify.

$T_{\rm {you}}={\sqrt[{4}]{{\frac {1}{4}}{\frac {1}{\sigma }}{\frac {4\pi R^{2}{\sigma }T^{4}}{4\pi D^{2}}}}}={\sqrt[{4}]{{\frac {1}{4}}{\frac {R^{2}}{D^{2}}}T^{4}}}={\sqrt {\frac {R}{2D}}}T$

Redoing our calculation:

T_you = ( 0.00075 m / (2 ⋅ 144841 m) )^1/2 ⋅ 5800 K = 0.295 K.

Más sabe el diablo por viejo que por diablo, and the parallel moral here is that the power of the sun comes more from its enormous size than simply from its temperature. -- ToE 14:53, 6 March 2021 (UTC)

None of the above calculations consider how your pinhead manages to hold together at such a temperature.

Nor do they consider the 454 watt power source which maintains it temperature.

Without power input it will quickly dim due to its tiny surface-to-volume ratio.

V = (4/3)π(0.075 cm)³ = 1.77×10^-3 cm³.

If we assign it the density and specific heat capacity of steel, roughly ρ = 7.87 g/cm³ and c = 0.466 J/(g⋅K),

and we assume that specific heat capacity remains constant across a wide range of temperatures and there are no phase changes to deal with,

then your pinhead's heat capacity will be 1.77×10^-3 cm³ ⋅ 7.87 g/cm³ ⋅ 0.466 J/(g⋅K) = 6.48×10^-3 J/K, and even multiplying that by the full 5800 K gives us only 37.6 J. So it clearly can't maintain it's 454 watt luminous output for but a fraction of a second, and it will rapidly dim.

This is no different than a pinhead of steel being blasted by the 40,000° F jet from a plasma cutter. It will be bright but ephemeral. If you are in the workshop, then you should be wearing welding goggles, but 90 miles away, you won't even be aware of its existence.

Compare this to the sun where, other than the sudden drop in neutrino flux, we wouldn't notice anything for 10,000 years were its fusion to somehow suddenly stop. -- ToE 15:58, 6 March 2021 (UTC)

As pointed out by ToE, the solid angle has to be similar to that of the Sun. This means that at a large distance the object needs to be huge to lead to enough radiant flux. Take e.g. a fireball from a large impact. Or large gas fires: "The fires from the wells and the oil and gas lines (all of which ruptured, one by one) had produced flames with a height of about 200 metres and a peak rate of energy consumption of ~100 gigawatts, three times the rate of UK total energy consumption." In this case the radiant heat was so large that it was felt inside rescue helicopters at a distance of 1 km away. Count Iblis (talk) 02:59, 7 March 2021 (UTC)

RBMK part 2

So I've read that paper about the Chernobyl disaster which was linked in response to my previous questions about the RBMK reactor (where the hell did the article go?!), and while its main hypothesis is not plausible (it claims there was an actual nuclear explosion during the disaster, among other implausible claims), it does bring up at least one salient point: it says that the RBMK main circulation pumps have a built-in low-voltage/low-frequency trip (which would explain why they wound down so fast all 3 times after the turbine was tripped), and that this, rather than the graphite tips of the control rods, was the reason for the thermal runaway at Chernobyl. So my question is: (1) is it true that the pumps trip automatically due to low voltage and/or frequency, and (2) if so, is it plausible that at least part of the role which the INSAG-7 report attributes to the graphite tips in causing the reactivity spike was actually played by the abrupt stoppage of the circulation pumps (as the paper alleges)? 2601:646:8A01:B180:7446:DB1C:3033:BF8E (talk) 03:22, 6 March 2021 (UTC)

To answer your first question, see RBMK. Since the "R" part of the acronym stands for "Reaktor", I guess calling it the "RBMK reactor" or "High-Power Channel-type Reactor reactor" is considered tautological. {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 05:32, 6 March 2021 (UTC)

• But it would make sense for RBMK reactor to be a redirect to RBMK. Anyone feel like creating it? --142.112.149.107 (talk) 23:57, 6 March 2021 (UTC)

• Redirect now done. Alansplodge (talk) 13:34, 7 March 2021 (UTC)

• Thanks. --142.112.149.107 (talk) 08:43, 8 March 2021 (UTC)

We have an article on that: RAS syndrome. -- ToE 06:08, 6 March 2021 (UTC)

Sorry, the article does not answer my first question -- it says that the coolant pumps are prone to cavitation (which I already knew from reading said article, and which was specifically mentioned in both INSAG-1 and INSAG-7 as a contributing factor to the disaster), but it says nothing about the power being cut to said pumps! 2601:646:8A01:B180:7446:DB1C:3033:BF8E (talk) 07:57, 6 March 2021 (UTC)

The first question you asked in this thread seems to be "where the hell did the article go?!". It may not have been numbered, but since it was the first question you asked, it seems reasonable to call it "your first question". The article is called RBMK not RBMK reactor for the reasons stated which is an answer to that question. (It doesn't look like it was ever called anything else [3] although I don't know if any redirects were deleted. It looks like it was only ever wikilinked under RMBK in Wikipedia:Reference desk/Archives/Science/2021 February 22#RBMK reactor.) No one said the article answers any of your other questions AFAICT. To avoid confusion, it might be better to ensure you number all your questions, or don't number any. Nil Einne (talk) 14:28, 6 March 2021 (UTC)

I did number all my questions: (1) is it true that the pumps trip automatically due to low voltage and/or frequency, and (2) if so, is it plausible that at least part of the role which the INSAG-7 report attributes to the graphite tips in causing the reactivity spike was actually played by the abrupt stoppage of the circulation pumps (as the paper alleges)? 2601:646:8A01:B180:21E9:FB1D:6AAA:EDD3 (talk) 08:12, 7 March 2021 (UTC)

Where I'm from, "where the hell did the article go?" is a question, numbered or not. Should we refer to it as your zeroth question?  --Lambiam 11:23, 7 March 2021 (UTC)

It's worth asking the question why is the claim that there was "an actual nuclear explosion" implausible? It's not a great practice in science to just think something is implausible because "it seems patently absurd," or some such. Why is it actually implausible? --OuroborosCobra (talk) 18:44, 7 March 2021 (UTC)

Because for one to occur, it is not enough for the reactor merely to become prompt-critical -- it has to be prompt-critical by a very wide margin so that there is an energy spike of trillions of joules within a small fraction of a second, and the power output at 1:23:46 on that morning was nowhere even close to that (not even by the highest estimates!) Anyway, the question was about the pumps, not about what exactly exploded, right? 2601:646:8A01:B180:DDEB:56CE:4137:67EA (talk) 08:53, 8 March 2021 (UTC)

What if it fizzled? I'm not saying it was a full on, successful nuclear bomb. --OuroborosCobra (talk) 19:07, 8 March 2021 (UTC)

March 7

Falcon 9 launch

Falcon 9 trail

File:Falcon 9 launch 3-4-21.jpg

Falcon 9 launch

In the wee hours of March 4, I was photographing the launch of a Falcon 9 from Cape Canaveral, from 180 miles north of there. About 5-6 minutes after launch, it became visible to me again and I took this 39-second exposure. I could see only a faint dot of the flame from the engines, but this photo shows some sort of purplish gas or something below it. I've gotten something like it once before and I've seen other photos showing it. What is it? Bubba73 ^{You talkin' to me?} 02:25, 7 March 2021 (UTC)

Twilight phenomenon perhaps? --jpgordon^{𝄢𝄆 𝄐𝄇} 06:01, 7 March 2021 (UTC)

Perhaps, but this happens when it is not too long after sunset or too long before sunrise (30-60 minutes), when sunlight is shining on it. This launch was at 3:24AM and sunrise was at 6:45AM.

I'm guessing that it is (1) dumping fuel or some gas, or (2) it is ionizing the gas in the ionosphere. But I'd like to know. Bubba73 ^{You talkin' to me?} 06:36, 7 March 2021 (UTC)

Falcon 9 launches (and first stage landings) are continuously filmed from various cameras on the ground and several mounted on/in the first and second stages, and shown (initially live) on SpaceX's YouTube Channel, where they remain available thereafter (as well as on other space-related channels). Ordinarily it would be easy for you to watch the appropriate video and see what events and activities correlate with your exposure, but as it happens the Stage 1 downlink apparently failed this time around: however, you could watch some earlier Starlink missions to see similar Stage 1 events.

As a personal observation: the exhaust gases from such rocket launches spread more broadly as the rockets get higher because of the reducing atmospheric pressures they are expelled into, and in twilight as seen from those altitudes (per Jpgorden), the Sun's oblique illumination combined with a perspective unusual to ground watchers can give rise to some otherwise rarely seen phenomena.

Falcon 9 lanches do not generally involve 'dumping fuel', but after the Second stage has separated (around 2:40 from lift-off and 70km altitude on this occasion) and its single engine started (a few seconds later at about 75km altitude), the now-ballistic First stage adjusts its attitude using inert pressurised nitrogen jets and later (6:29 from lift-off at 55km altitude this time) re-ignites three of its nine engines in a re-entry burn (of about 20 seconds) to slow it down (and again for a low-altitude landing burn later). {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 08:09, 7 March 2021 (UTC)

I've added a 198-second exposure, taken before the first one. It was taken about 180 miles north of the Cape. It took more than 1 minute for it to become visible over the horizon. It must to get to be hundreds of miles away in these photos and the cloud is perhaps miles wide. Bubba73 ^{You talkin' to me?} 00:06, 8 March 2021 (UTC)

This was with a 20mm lens on a full-frame camera, so calculations can be done. Estimating that it is 500 miles away in the top photograph (5-6 minutes after launch from 180 miles away), that puts the width of the visible purple area about 5-6 miles wide. Bubba73 ^{You talkin' to me?} 02:54, 8 March 2021 (UTC)

Clearly that bottom photo shows first stage (booster) flight through MECO and stage separation, followed by second stage flight.

Regarding the top photo, when the burn became visible to you again, did it last for about 20 seconds?

If so, that (and the ~ 6 min past launch timing) strongly suggests the second stage entry burn, as mentioned above.

(The stage one entry burn start-up and shutdown callouts were at T+6:28 and T+6:49 as heard on that video. The launch timeline on the SpaceX page (archived here) gave "1st stage entry burn complete" at T+6:43, but didn't give a start-up time; the burn probably started at about T+6:23.)

You say that the second exposure was for 39 seconds. Presumably you started after it became visible. Did you continue for another 20 seconds or so after the dot of flame disappeared? -- ToE 15:58, 8 March 2021 (UTC)

As it was too early for the sun to be illuminating the exhaust plumes, I assume they were lit by the engine itself. The Merlin 1D when operating in a vacuum has a specific impulse of 310 s, which when given as an effective exhaust velocity is 3.0 km/s, so it will span your 5-6 miles width within seconds. Note that at the far right of the trail, where the photo started with the entry burn in progress, you can see an arc of illuminated exhaust plume which has already streaked away from where the stage had passed.

The three engines of the entry burn and the lower altitude explain its greater visibly compared to the fainter plume from the higher, single-engine second stage flight.

What I don't understand is why it is visible only below the line of flight. -- ToE 15:58, 8 March 2021 (UTC)

I can't recall about the 20 seconds. After MECO, I lost sight of it but I kept the shutter of the camera open. After some time (I don't know how long) I spotted it again. After it crossed over the palm tree, I knew it was out of the camera's field of view, so I swung the camera around for the photo at the top. Knowing the details of the camera, measuring the photo, and estimating that it was 500 miles away, I calculated that it was 5-6 miles wide. It seems to be that it would be extremely thin when spread out that much. But also in the bottom photo (the first chronologically), it is getting wider as time goes by, perhaps because of lower atmospheric pressure? Would the gas be glowing? Bubba73 ^{You talkin' to me?} 00:32, 9 March 2021 (UTC)

• BTW, the top photo is the full frame. The bottom photo is cropped, so it is enlarged with respect to the top one. Bubba73 ^{You talkin' to me?} 00:34, 9 March 2021 (UTC)

Can heavy water be used as a food preservative?

The stronger hydrogen bonds of deuterium compared to hydrogen derails biochemical processes, causing organism that take in too much heavy water to die. Can this be exploited to inhibit bacterial and fungal growth after harvesting of crops by watering crops with heavy water some time before harvesting? Count Iblis (talk) 03:08, 7 March 2021 (UTC)

That would depend on the relative toxicity of heavy water to bacteria and funguses vs. that to humans -- but even if this proves possible, it would still be far more economical to use salt for the same purpose (heavy water is very expensive!) 2601:646:8A01:B180:21E9:FB1D:6AAA:EDD3 (talk) 08:15, 7 March 2021 (UTC)

[Edit Conflict] Possibly, although its effects on the crop's ripening would also have to be taken into account. (I presume you've read Heavy water#Effect on biological systems and the succeeding sections?) However, since (from cursory web searching) heavy water costs several tens of US dollars per pound to produce, I'd think it would be prohibitively expensive even if it did work. {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 08:21, 7 March 2021 (UTC)

Swapping H for D in biological systems does not really "derail" processes, if by that you mean "stops them in their tracks". The kinetic isotope effect just means that the rate of reaction is (usually) slowed. This can have real consequences, as discussed in "this paper quoted in deuterium oxide article" (PDF).. One thought experiment (there are likely to be actual examples but I've not tried to find the refs) is to replace one or a few H for D in a drug or agrochemical with the intention of extending its half-life by altering the metabolic processes acting on it. For example, in a phenyl ring one point of oxidative metabolic attack is in the positions of highest electron density. Putting a D instead of H there would likely extend the useful lifetime. Or a methyl group subject to oxidation could be swapped for a CD₃ group. The costs of doing that would not be prohibitive if the active ingredient was thereby given some useful improvement. Such tactics are well known where fluorine is the atom substituted for the H but one problem is that other properties of the fluorinated material may differ too much from the original material. Mike Turnbull (talk) 12:17, 7 March 2021 (UTC)

Found a reference: "The development of deuterium-containing drugs". Mike Turnbull (talk) — Preceding undated comment added 12:49, 7 March 2021 (UTC)

You might be interested in Di-deuterated linoleic acid ethyl ester. This has primarily been studied as a drug, though. --47.152.93.24 (talk) 16:57, 10 March 2021 (UTC)

Charge a black hole

I understand that having a black hole with a significant charge is p~0 in practice. But suppose that you could dump somehow a few stellar masses of electrons into a black hole. Could you by adding enough charge overcome the black hole's gravitational cohesion? 93.136.167.229 (talk) 05:17, 7 March 2021 (UTC)

I don't think so. Gravity overcomes the strong force in black holes, so it should easily overcome the electromagnetic force. Mike Turnbull (talk) 12:39, 7 March 2021 (UTC)

Black holes become electrically charged when charged objects fall into them falls into them. Physicists calculate that black holes have an “extremal limit,” a saturation point where they store as much electric charge as possible for their size. When a charged black hole evaporates and shrinks in the manner described by Hawking, it will eventually reach this extremal limit. It’s then as small as it can get, given how charged it is. It can’t evaporate further. JumboWriter (talk) 12:53, 7 March 2021 (UTC)

References

Very interesting article! So it's conceivable that by adding enough electrons (which have Q > M) we might be able to create a Q > M black hole which wouldn't evaporate (at least via Hawking radiation), or we might not even be able to stuff those electrons into the hole? 93.136.3.148 (talk) 09:50, 8 March 2021 (UTC)

According to Reissner–Nordström_metric there is the maximum possible charge above which the black hole cannot exist. Ruslik_Zero 18:07, 7 March 2021 (UTC)

If I have made no mistake, that maximum capacity is given by:

$\left|{\frac {Q}{M}}\right|\leq \left({\frac {G}{K}}\right)^{\frac {1}{2}}\approx 8.6175\cdot 10^{{-}11}\,\mathrm {kg} ^{{-}1}\mathrm {C} ,$

where $G$ is the gravitational constant and $K$ is the Coulomb constant. So suppose now we have managed to charge a black hole right to its negative limit, and try to dump one more electron in it. If that dump cannot succeed, how does this impossibility manifest itself? As some kind of resistive force? Or is it possible (as a gedankenexperiment) but does the event-horizon singularity then cease to be? What does theory tells us here?  --Lambiam 11:30, 8 March 2021 (UTC)

Wow, that's a very small number - going from the data at the electron article, the electron carries about 176 GC/kg. That would mean we only need to dilute the black hole with 1 in 2 × 10²¹ parts by mass of electrons to get to this limit. If I didn't make any errors that'd be a million tons of electrons for every solar mass. 93.136.3.148 (talk) 11:50, 8 March 2021 (UTC)

For an electron $e/m\sim 10^{12}$ and therefore the electrical repulsion from the critical black hole will far exceed the gravitational attraction. So, it is impossible to add any electron to such a black hole. Moreover the article says: "Objects with a charge greater than their mass can exist in nature, but they can not collapse down to a black hole, and if they could, they would display a naked singularity". Ruslik_Zero 18:23, 9 March 2021 (UTC)

Does the electric force appear to tend to infinity (for an outside observer) as the electron approaches the event horizon? Otherwise, one could try to aim a high-velocity electron at the black hole.  --Lambiam 23:46, 9 March 2021 (UTC)

I suppose you can but the ratio of the charge to the relativistic mass of the electron will be much smaller than 10¹². The end result will be that the charge-to-mass ratio of the black hole will actually decrease after such a fast electron is added. Ruslik_Zero 12:43, 10 March 2021 (UTC)

Why is the outline of heatshield for Perseverance (rover) look so odd?

I am asking outline of heatshield for Perseverance (rover) look so odd:

IMAGE

Odd means it's outline of heatshield look like old retro photo look. Rizosome (talk) 08:43, 7 March 2021 (UTC)

What's the provenance of the image? Where does it come from? --Jayron32 19:24, 8 March 2021 (UTC)

Additionally, what is it supposed to look like? --OuroborosCobra (talk) 19:31, 8 March 2021 (UTC)

It is a still from a video released by NASA, which can be seen here. The still is a frame at 00:20⁺ into the video.  --Lambiam 21:31, 8 March 2021 (UTC)

If I read the article correctly, they used some off-the-shelf sports cameras (read GoPros). The graininess of the image could be explained by any number of ways then including the normal graininess one gets when doing image capture off of a video. --Jayron32 15:43, 9 March 2021 (UTC)

Yeah, seems to be the case. A few things to note. First, the view of the heat shield is the inside of the heat shield (facing "up" during entry), and not the ablative side. That would explain the aluminum foil looking material and what look like some sort of wiring. Second, as you said, the camera was an off-the-shelf sports camera, like a GoPro. Based on the footage, it looks like it was a camera intended to be used just once during the mission, to film the footage that we saw. Given that, NASA/JPL wasn't likely to spend the money or the mass on a really high quality camera. Why bother if the footage you get from a cheap one is good enough, and the thing only has to last for a single use? --OuroborosCobra (talk) 19:17, 9 March 2021 (UTC)

It looks like an almost-in-focus image of a high-contrast object against an out-of-focus, low-contrast background.

Here are some images of that heat shield on Earth: [4] [5] [6]

Here is a similar photo, but unlike the one linked in the question which is a still from the Rover Descent Camera, this is an image from the B&W Lander Vision System Camera (part of the Terrain-Relative Navigation system), which was hand colored by NASA's Doug Ellison. (That tweet also shows the B&W original.) With a wider angle and being focused on the terrain (in order to do its job), it gives a good view of the Jezero Crater that they were descending into.

For a description of the cameras, see The Mars 2020 Engineering Cameras and Microphone on the Perseverance Rover: A Next-Generation Imaging System for Mars Exploration. -- ToE 20:24, 9 March 2021 (UTC)

human / plant symbiosis

Does the symbiosis between humans and plants have a term? Nothing about symbiosis is mentioned in Agriculture, also the symbiosis is older than agriculture. How would one find papers about the subject (as seen from the biology side)? --SCIdude (talk) 09:17, 7 March 2021 (UTC)

There is a notion of "service-resource mutualism"^[7] in which the symbiont species can exist independently but benefit from the interaction. This is more general than an animal–plant symbiosis, though; cleaner fish and the fish they clean are an example of animal–animal mutualism. Fungi are not plants but can be cultivated in a similar way; a non-vertebrate animal–cultivee example is ant–fungus mutualism. This is more generally referred to as "cultivation mutualism".^[8] What type of symbiosis is the human–plant symbiosis older than agriculture you refer to? Which plant species were involved? Some nomadic peoples practice a limited kind of agriculture, planting or sowing in order to harvest on a later return. This practice is probably older than sedentary agriculture, but still is a form of agriculture.  --Lambiam 11:08, 7 March 2021 (UTC)

It's worth noting that, within agriculture, the interaction between humans and plants takes a specific form: namely, humans nurture and protect particular genetic lineages, rather than individual plants (we only protect individual plants up to the point where we eat them, so at an individual level, the relationship could be viewed as parasitic, but at a genetic level, it could be seen as mutually beneficial). PaleCloudedWhite (talk) 11:30, 7 March 2021 (UTC)

Still it's the specific plant species that, in whole, gets the benefits from e.g. transportation, because those lineages can escape control after they are transported. I also don't think fungus cultivation is equivalent because we can live without fungi but not without plants. --SCIdude (talk) 15:20, 7 March 2021 (UTC)

Fungi are great nutrient recyclers and have very close relationships with plants, and without fungi I would expect most productive ecosystems to cease functioning as they do now, so in that indirect sense we probably can't live without fungi. PaleCloudedWhite (talk) 15:33, 7 March 2021 (UTC)

We certainly couldn't live without Saccharomyces, which is responsible for beer, and beer is necessary for life! --Jayron32 19:32, 8 March 2021 (UTC)

March 8

chemistry

heating ammonia on a bunson burner makes what poison? — Preceding unsigned comment added by 173.49.56.54 (talk) 13:31, 8 March 2021 (UTC)

Ammonia can be used as a fuel (see Ammonia#Uses) and when oxidised will give the oxides of nitrogen, mainly nitric oxide or nitrogen dioxide as well as water, depending on how much oxygen was available. However, assuming that you had an aqueous solution of ammonia (which is the form most readily available), the main result of heating it would be to generate gaseous ammonia and give you a nasty, choking smell and eye irritation. So, overall, don't try this at home! Mike Turnbull (talk) 17:01, 8 March 2021 (UTC)

Speaking as someone who used to work with ammonia on an industrial scale, I couldn't agree more with your last point. On one occasion, workers on the plant got severe headaches and nosebleeds from having the wrong filters in their respirators (for dust rather than vapours) while they were handling multiple kilograms of ammonia. Rhythdybiau (talk) 19:24, 8 March 2021 (UTC)

I #3 this -- I experienced it the hard way by accidentally inhaling ammonia vapor, and although it was only a small amount I burned the inside of my nose quite severely and lost all sense of smell for several weeks! 2601:646:8A01:B180:C98E:59B1:C877:CE68 (talk) 06:27, 9 March 2021 (UTC)

Back in 2009, a natural gas line at a ConAgra plant making Slim Jims exploded about 2 miles from my house. The explosion ruptured an ammonia gas pipe; compressed ammonia is a common refrigerant. It was a Big Deal. Several tons of ammonia gas were leaked in a short period of time. read about it here. Ammonia is all kinds of useful, but also all kinds of dangerous. IIRC, it wasn't the natural gas explosion that caused most of the injuries from the accident, it was the ammonia leak. --Jayron32 13:03, 9 March 2021 (UTC)

In standard model of particle physics, what needed to be discovered?

From thorough research, I find Graviton still needed to be discovered. So anything else needed to be discovered in standard model table? Rizosome (talk) 13:36, 8 March 2021 (UTC)

The graviton is not part of the standard model. The Standard Model does not include gravity as among its forces. The most recent, and arguably last necessary, discovery to complete the Standard Model was the Higgs boson, discovered in 2012. Which is not to say that the Standard Model is a "theory of everything". It is not; it is only complete over its own domain with regard to particle physics, and then only in the sense that the fundamental particles it predicts have essentially all been discovered. It doesn't include gravity because it wasn't ever really designed to include gravity. It is self-consistent in the sense that it has no internal holes (i.e. nothing that it actually predicts is wanting or undiscovered or contradictory), but it is not, nor does it try, to account for all physical phenomena. There are a LARGE number of unsolved physics problems that the Standard Model does not address, and which are still largely unsolved. You can read more about this in several places. First of all, in the Standard Model article, in the lead section, the entire second paragraph deals with various aspects of physics the Standard Model leaves unanswered. Secondly, in the same article, if you go down to the "Challenges" section, you'll find some elaboration the same topic. Thirdly, there are several other Wikipedia articles you can read to learn more, including Physics beyond the Standard Model, and List of unsolved problems in physics. --Jayron32 19:21, 8 March 2021 (UTC)

Backwards time travel during black holes merger

In this video Neil deGrasse Tyson claims that before the event horizons overlap there's a trajectory that results in backwards time travel. Is there any serious, cite-able source for this? אילן שמעוני (talk) 19:56, 8 March 2021 (UTC)

Timestamp? --Amble (talk) 21:12, 8 March 2021 (UTC)

What? אילן שמעוני (talk) 08:41, 9 March 2021 (UTC)

"Timestamp" presuambly means at what point in that 17 minute video? Context is everything, but we don't want to have to watch the whole thing.--Shantavira|^{feed me} 09:01, 9 March 2021 (UTC)

The link was wrong somehow. Here's the right one that already contains timestamp: https://www.youtube.com/watch?v=iLKTZr00xBg&t=139s

אילן שמעוני (talk) 09:30, 9 March 2021 (UTC)

Sounds like bunkum to me. When I wrote the binary black hole article I consulted around 100 academic papers on the topic, and I cannot recall any such mention. If backwards time travel was in there I would certainly have noticed and written about it. Even more surprising than what NdGT mentioned was masses of black holes in a merger 1+1≠2 instead 1+1≈1.9 as mass is lost due to gravitational waves. Also when the gravitational waves pass, space is permanently deformed. And since gravitational waves carry momentum as well as energy, the merged black hole can shoot off at high velocity. Graeme Bartlett (talk) 11:00, 9 March 2021 (UTC)

Velocity at the expanse of mass? אילן שמעוני (talk) 12:49, 9 March 2021 (UTC)

Such things don't even require two black holes. Just one rotating black hole is required. See Penrose process. --Jayron32 15:36, 9 March 2021 (UTC)

• It sounds like he's oversimplifying the concept of a Closed timelike curve, but this is a well-trodden area of theoretical physics and General relativity. For most standard solutions, the spacetime path of a closed timelike curve ends up inside the event horizon of a black hole, or some other thing which makes them essentially useless for practical time travel; indeed one of the axioms of physics is causality, insofar as even if in some way time travel were possible, it wouldn't be possible to violate causality; this is called the Chronology protection conjecture, and basically says that once you actually work out any spacetime path that would result in actual time travel (except in the case of virtual particles and things like that where information can't actually be transmitted backwards in time, even if the particles themselves can travel that way) ends up not working because some other aspect of physics gets in the way. What NdGT is talking about, from my understanding, is that in the rare case of a pair of colliding black holes, there is a spacetime path that does not cross the event horizon of either black hole and thus would be theoretically possible. These kinds of bizarre solutions have been known about for some time, indeed Hawking proposed the chronology protection conjecture to deal with them; IIRC his solution was that something we haven't yet discovered about quantum gravity would turn out to fix the problem that NdGT is depending on to make his statement. --Jayron32 12:52, 9 March 2021 (UTC)

Sounds plausible enough. Whatever this specific solution is - he can't be the source for it - he himself states that GR math is not his field. Question is where he got this from. אילן שמעוני (talk) 15:22, 9 March 2021 (UTC)

The answer is likely "from something he vaguely remembered about closed timelike curves". It's "right enough" for the discussion he's having with the child in question, but of course lacks nuance. I couldn't find a specific example, but I didn't look too hard. If you google "closed timelike curve merging black holes" or something like that, you may find the specific scenario in question. --Jayron32 15:33, 9 March 2021 (UTC)

I did this search of course, found nothing relevant. אילן שמעוני (talk) 22:43, 9 March 2021 (UTC)

Thanks for the corrected link. As Jayron points out, Tyson is describing a closed timelike curve (CTC). There are many papers that include a review of spacetimes in GR that contain CTCs: [9], [10], [11]. If someone had shown that mergers (or near-miss encounters) of black holes gave rise to CTCs that don't cross an event horizon, that would be an essential result that would answer some of the open questions raised by that literature -- definitely important enough to be mentioned in these reviews (so that Graeme Bartlett would have read about it hundreds of times by now). So what is Tyson talking about here? Here are two possibilities that I think are reasonable:

1. He's talking about the Gott solution [12], which is one of the standard examples listed in the review articles and involves two objects passing by each other in a way very similar to what Tyson describes. However, it requires the objects to be infinitely long cosmic strings, not black holes. It sounds to me like this is what Tyson has in mind, and he's giving it as an example of the kind of thing that could happen with strongly curved spacetime, while leaving out the caveats (or else misremembering them).
2. He's talking about wormholes in the Kerr metric. This does relate to black holes, although it doesn't need two black holes; one is enough. This is another standard example listed in the review papers. It contains CTCs, but none that are entirely outside the event horizon, as Tyson seems to imply, but doesn't quite say. When he says "go around", he might be referring to paths that go through the middle of a ring-shaped singularity.

My best guess is that he's really talking about these types of scenarios in general, especially the Gott and Kerr solutions which are well known, and not trying to describe a specific solution (which might not be remembered in detail off the top of one's head). I don't think there's a known solution where encounters of two black holes produce CTCs that don't cross the horizons, since this would be an important result that would be mentioned in all the review articles, but we don't find it there. --Amble (talk) 18:07, 9 March 2021 (UTC)

March 9

Is it possible to perform heart surgery with silverware

Hello wikipedians I`m posting a question concerning whether or not it is possible to perform heart surgery with silverware in other words kitchenknives, forks, even spoons etc? If so what is the chance of success of this surgery? THANKS — Preceding unsigned comment added by 98.113.197.52 (talk) 23:05, 9 March 2021 (UTC)

Have you stabbed someone in the heart while they were in the midst of a heart attack, and are now hoping to repair the damage?

Time is of the essence and you shouldn't be waiting for an answer from us, even if it were our habit to give medical advice, which it is not.

Do please drop us a note to tell what you chose to do and how it worked out. -- ToE 23:32, 9 March 2021 (UTC)

maybe the OP will get back to us in another 6 weeks or so. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:30, 10 March 2021 (UTC)

As with writing Wikipedia articles, the answer is that competence is required. Mike Turnbull (talk) 10:35, 10 March 2021 (UTC)

True. For example, using a butter knife as a scalpel is not likely to work. My money would be on a Spork. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:35, 10 March 2021 (UTC)

Many things are possible, despite not being advisable. Surgical scalpels are much sharper than silverware. As for the latter question, please refer to the page header: "We don't answer requests for opinions, predictions or debate." --47.152.93.24 (talk) 17:16, 10 March 2021 (UTC)

We used to say "A good medic is an improvising medic", and that goes to any desperate medical situation. You can do surgery with the lid of a canned food, having the risk of tearing tissue, if that's the only alternative to certain death. Also, this. אילן שמעוני (talk) 19:45, 10 March 2021 (UTC)

Well, sure, if you've got no choice. Like "MacGyver". But since the OP first broached this subject several weeks ago, there doesn't seem to be any urgency. Maybe he's working on a new entry in the book series, The Worst-Case Scenario. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:38, 10 March 2021 (UTC)

March 11

Cure to aging question about women

If we will discover a successful cure to aging and manage to make an old post-menopausal woman young again, will this cure also result in this woman once again becoming fertile or would she actually need the implantation of new ovaries for that? Futurist110 (talk) 05:48, 11 March 2021 (UTC)

This is quite a hypothetical situation. If a woman was made young again you should expect everything in her to work the same way as a young woman, so yes she would likely be fertile. However if it was just a successful anti wrinkle treatment or a refresh of telomeres, or a cure for arthritis, or a clean up of cellular senescence then the answer would be no. Given your user name, you should be answering this! Graeme Bartlett (talk) 06:15, 11 March 2021 (UTC)

While the science is not as yet settled, there are suggestions that renewed fertility would likely not result from general rejuvenation.

Fertility depends on an ovarian follicle in one or other of the ovaries producing a mature ovum. Although female humans are born with millions of primordial follicles with the potential to mature, the number of these declines steeply with age, with only 10% remaining at around age 30. It appears that by around 55 there are few to none left, so subsequent rejuvenation by some as-yet-undiscovered means might restore the levels or hormones (whose decline cause the primary symptoms of the menopause) required to cause ovum maturation, but there would be no ovarian follicles left for them to work on.

Some additional means would have to be found either to cause the generation of new follicles, which has reportedly been observed in mice but not (yet) in humans, or to generate them in vitro from stem cells and implant them into the ovaries. If that were the procedure adopted, it would probably be more efficient to take the further step of generating mature ova in vitro and fertilise them ditto before implanting them into the womb as is already routinely done. {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 10:21, 11 March 2021 (UTC)