Thanks so much for watching! If you are curious as to how long you can stay in a nuclear spent fuel pool, please check out: th-cam.com/video/diHG9W27XeU/w-d-xo.htmlsi=dpxPwZbkX_UrXDEb
It shouldn't matter if you're a foot away, ten feet away, or a mile away. That close to the sun, it's approximately a flat plane, and integrating the inverse square law for a plane actually gives you a constant value!
Not nearly but it is high. The gravity on the surface of the sun is about 28 Gee's. In comparison the gravity on the outer 'surface' of Jupiter is only 2.4 Gee's. 😄 (That's using the Newtonian eq A = GM / r^2.)
The question was to heat up because it was too cold in Colorado, so xkcd focused on temperature, not all the hazards on the surface of the sun. As you say, they are many variables to Calculate.
But without knowing all the dangers associated, some young impressionable folk may get the idea to actually teleport to the sun and back for a nanosecond.
Typically you're either going to explode or basically have your heart and entire nervous system fried. If you get very lucky and you have something that can actually draw the bulk of the electricity away from you then you might survive. Basically we're talking about people who I mean were technically in the wrong place but basically everything went perfectly in their favour. But now you can more often than not you're probably going to die if you get hit by lightning you'll probably just go deaf indoor blind and probably get hit by a decent dose of x-rays
No, no, light will always begin to enter your eyes regardless of the distance. The output stream is continuous. You don’t have a delay while light travels from the surface to your eyes.
@@TriangulumGalaxy_1 Not the point. Since the sun does not light up at the moment you appear, being 30cm away wouldn't mean you don't receive anything. You will always take in a nanosecond whatever radiation was already traveling and was within 30cm of where you appear.
Can you show us the exact calculation for the radiation dosage on the surface? I'm not convinced that it would be lethal, but maybe I'm wrong. Either way, I'd line to see you calculate it.
The sun actually doesn't emit any xrays itself. it follows the black body radiation. The inside if filled with xrays because thats the temperature they are created at. The photons are absorbed and reemitted at at lower and lower energy levels closer to the surface as it gets cooler. the photons slowly carry the energy to the surface but the only ones that come off it are only created in the surface itself. What actually comes off the sun as Xrays is all created in the corona from ejected particles at very high temperature. But the Corona is extremly thinn. the energy density compared to the inside or even the surface of the sun is VERY low. a nanosecod is not enough time for any real damge to be done, even by xrays as these make up a miniscule fraction of energy emitted by the sun.
We are talking about black body curve of 5800K for the exterior and it still emits a little xray and gamma naturally. But in deeper layers where the temperature gets ridiculously high the ionizing photons also get ridiculously high
@@rea1m_ The black body radiation for 5800K has nearly no emission below UV light, so the central assumption that a nanosecond there would be fatal due to gamma and x-ray radiation is flawed *if* you only consider that blackbody radiation.
Thank you!! Yeah, I was also thinking about the blackbody curve, which drops off exponentially with increasing photon energy. The photosphere is going to emit virtually no x-rays. Now, _some_ of the emission from the corona will propagate back down to the photosphere, but this video didn't break that out as a separate calculation. Even if the "millions of sieverts per second" (from the corona???) _is_ right, that's millisieverts per nanosecond - and a millisievert is not a big deal. One other nit about all of the "surface of the Sun" cases. The photosphere is not a sharp boundary on a human scale. One can pick the depth from which half of the visible light from the Sun is emitted, but the photosphere is about 300 miles thick. Face towards the Sun's core, face away from the Sun's core, one is immersed in somewhat hazy white hot gas for a nanosecond in either case.
@@jeffreysoreff9588 The 10 µJ/cm² reported in xkcd should include all ionizing radiation delivered from the solar surface in that nanosecond, so I don't understand the claim that the radiation dose would be high, either. I'm not an expert on this, but I ran some numbers and got a similar value of 6.3 µJ/cm². Assuming an average 62 kg adult with a crude estimate of .2m² facing the sun and all of that energy as ionizing radiation gives about 200 µGy delivered to a person. That should be 200 µSv for photons, but even assuming that is delivered with the highest multiplier (20 for neutrons), which we shouldn't, that is still only 4 mSv. Even with uneven penetration, I'm dubious we could get a factor of 250 to get it up to 1 sievert, and this is after being generous. Calcs: Solar constant max * (1 AU/radius of Sun)² * 1 ns = 63 mJ/m² energy delivered. 63 mJ/m² * .2 m² / 62 kg = .2 mGy.
Not all radiation is thermal radiation. I don't know how transparent the sun is to (hard) radiation, so I guess it's somewhere in the middle between - thermal radiation and some radiation created by nuclear stuff.
depends on how you get there. Get a small comet and put it on a Sun-grazing trajectory. Stay inside it until you get there. The Parker Solar Probe is expected to survive several passes within 6 million km of the Sun's surface. en.wikipedia.org/wiki/Parker_Solar_Probe
Yep - a Sievert is a unit of absorbed dose in J/kg. Simply by following the black body law there is much less UV/X Ray/Gamma Ray radiation than light and heat, and the UV would only give you sunburn. There are also the solar protons and such, but again, the length of time exposed is just so short.
Dude, there's no such thing as "one foot above the surface of the Sun". You're talking about kilometers of ever-thickening plasma. From 150,000,000km it may look solid, but it just ain't so.
I would expect the ratio of different EM frequencies at the surface of the sun to be similar to that in interplanetary space, as empty space does not diminish the radiation. As the most immediate risk of facing and staring directly at the sun without a visor is blindness, then I would contend that even at a closer distance and shorter time, that if it doesn't blind you, it won't significantly irradiate you.
@@Aarush.A.S6 million km! Bro the moon is approximately 300000km far away which is one-third of a million. The atmosphere ends way before that. Probably 10000-20000 km.
I'm trying to double-check whether you are actually right or wrong, but I couldn't find the data on how much ionising radiation does the Sun emit per square meter (or as a percentage of its total output) - and I'm also not sure if my idea on how to actually convert it to an approximate dose is correct. I'd love it if you did a follow-up video and actually did these calculations. Edit: I have found some articles by NASA that say the surface of the Sun doesn't actually emit a lot of X-rays, these are produced mostly by the corona. I think this question is actually much more complicated than it seems
@@katrinabryce the total energy output of the sun is something like 10^26 W, the surface area is 10^18 sq m, divided it gives you 10^8 Watts per meter, so a lot, but a nanosecond is 10^-9, so a meter of the Sun outputs 0.1j of energy in a nanosecond. Not sure how to convert it into a dose received by a human body though. If all of that energy would be evenly distributed throughout your body it's basically nothing, but calculating it this way doesn't seem right - but I do think xkcd is closer to the right answer
@@igorbednarski8048 If it was 100% ionising radiation, then the dose would be 100mSv, which the UK Health Security Agency describes as "Level at which changes in blood cells can be readily observed". That is an absolute worst-case scenario, and it is in reality going to be a lot lower than that.
@@katrinabryce 0.1J per square meter is not 0.1Sv, unless you're irradiating a really strange human that is a flat 1 square meter pancake that weighs exactly 1kg, they get uniformly irradiated with nothing but x-rays and absorb all of it evenly. Focus all of that 0.1J into your little finger in the form of high energy neutrons and you might end up getting 100 Sievers instead of 100 milisieverts.
It doesn't matter how far you are from the sun surface, the few feets wouldn't make a difference, because the radiation was there already. Sun didn't started radiating only when you teleported.
I'm curious what the justification for a high gamma flux at the surface of the sun is. At the photosphere you're pretty deep inside an atmosphere that can provide shielding and I wouldn't expect the temperatures to be high enough for there to be any locally produced gamma rays. There would definitely be tons of UV and some X-rays, though.
If we define the surface as that what is visible from earth, the ratio between visible light and gamma/x rays can't be much different than what we observe from here (above the atmosphere of course). Because there definitely isn't much in the way that could absorb it on the way. And the falloff from distance is the same of course.
Radiation from charged particles could be different, but i would like to see the justification for that. Its not like there should be any nuclear reactions taking place that far from the core.
I feel like the flaw here is that it refers to radiation that needs to reach you. What about the radiation that was already there? Wouldn't that burn you too?
That point about temperature and heat is demonstrated pretty well by the thermosphere. The average free energy per particle up there (The temperature) is very high, but there is very little of them so the total energy (Heat) per unit volume is minimal.
Was gonna say as a welder we get covered in sparks. Most of those are close to "sun heat" they just don't have enough mass to burn anything. So I figured time would do something similar
Made me think of Two shakes of a lamb's tail. I than found this on Reddit. This expression first appeared in publication in 1840, but is undoubtedly much older than that in vernacular usage. Interestingly if was originally paired with its opposite-“two shakes of a dead lamb’s tail”--which of course meant never. The expression is occasionally reduced to “two shakes,” meaning the same thing. “See you in two shakes” was a fairly common expression I would hear when I was growing up. The expression is gradually disappearing from written and spoken usage, one assumes because fewer and fewer people grow up actually seeing lambs wag their tails. The one place it will probably be preserved is among nuclear scientists, since during the Manhattan Project they adopted the term “shake” to denote 10 nanoseconds. Strange how words and expressions evolve! Language evolves. Not sure seeing lambs tails in action has any relevance really.
I just put in a suggestion to What If? about how a car engine that fused hydrogen instead of burning fuel would act, and I think that question, though unlikely to be answered by him, would definitely get your attention.
6:25 No actually gravity wouldn't be extreme at all. Gravitational force is inversely proportional to the square of the distance, and at the core that distance would be 0. So the Gravitational force would be 0 at that point. Now of course, the other factors would kill you but Gravity wont be one of those.
He didn't say that there's extreme gravity at the center or that it would kill you, did he? He only said that it induces fusion reaction with plasma, which is true.
True, it's pressure. Note that on the surface, other than your feet, you're safe (enough). The Ruler rule will protect the rest of your body (though it depends on your teleportation device. A cut-rate unit might not completely clear your landing zone of matter/energy)
Just wondering how we know the radiation environment at the surface? Is it data from the Parker Solar Probe? To me, an educated (but not in the nuclear field) person,I would say that X-rays and Gamma rays would be mostly confined to the Sun's core, and would be unable to escape the confines of the core due to the densely packed matter therein, it would be reradiated outwards as heat, light, radio waves, etc due to the heat from absorbing that radiation. I know that there's radioactive products in there that escaped the fusion process (e g. Tritium, Helium-3), and probably neutron activated radioactive materials as well, since the sun has some "metallicity" to it (I e. Things that aren't hydrogen and helium). I know that there's tiny amounts of material that get fused beyond helium, also... that become the main fusion source when the sun becomes a red giant later in its life.
Hmm... I didn't think the sun's photosphere had much in the way of x-rays & gamma rays. As the sun is quite optically dense, none of the light from the hotter regions could get there, so it should be very close to the standard black body radiation for that temperature. I suppose if it is like the picture, AJ will take some x-rays from behind, coming down from the quite hot corona. Alpha & alpha like radiation from the helium nuclei & protons shouldn't be an issue, as they won't have time to inflitrate the respiratory track & will be stopped by the top layers of the skin. Beta radiation might a concern with all the free electrons. I don't think they'll be able to penetrate down to the basal layer of the skin with the much lower kinetic energy they have relative to the beta radiation from decay that we usually refer to. That's a guess though.
This thought experiment has a huge implicit assumption that drastically changes outcomes: what happens to the matter and energy that was inside the space you teleported to? And what happens to it when you teleport back? If you completely replace all matter and energy in the spot you teleported to, 1 femtosecond in the core wouldn't be that bad since it'd only directly affect the outermost 300nm of your body (dead skin cells). Then, the way you return is important since those waves within your perimeter would do a lot of damage to your body if you bring them back with you.
Of considerable additional concern is that at the temperatures you experience inside the sun's core, all the atoms will be moving at a noticeable fraction of the speed of light. That means that a large fraction of the matter within one to several tens of nanometers will penetrate into the outer layers of your skin and will also come back with you. The sun is dense enough that even a modest fraction of a cubic millimeter* of material absorbed into the surface of the skin would be a very large number of lethal doses of radiation. When you are talking about how many alpha particles your body has absorbed you really do not want a number expressed in tens or hundreds of milligrams. *One square meter of surface area times 10 nanometers (3.3% light speed) would be 1/10th of a cubic millimeter at solar core density of high energy particles. I am not sure what speed is actually equivalent to 15 million Kelvin, but I imagine I am within an order of magnitude of correct.
I always think of a light nanosecond as a metric foot. It's not exactly a foot, but it's the same order of magnitude and closer in proportion than a metric ton is to an imperial ton (which are also the same orders of magnitude, such that measuring bridge strength in either gets you well within the safety margin in the other).
If you were only in the center of the sun for one femtosecond, the radiation, however intense, would only have time to heat up about 0.0003 mm of your exterior. Upon teleporting back to earth, could that tiny layer of plasma burn you? Would the radiation traveling through that layer be enough to ionize the rest of you in the nanosecond following the teleport? I assume the pressure, however intense, couldn't crush you quicker than the speed of light.
@3:50 is there a particular reason the radiation would be constrained to "up to" the surface of the sun? Cuz in my mind it would be, well, just radiating out, creating a stream of radiation. Hence if you dipped into that stream (much like a river) you'd still get wet, or in this case, radiated?
Light goes a foot per nanosecond is required lab knowledge. I mean those little coax cable are even labeled in 1/2, 1, 2, 4, 8… ns. They’re a bit shorter since the pulses go about 0.8 c.
The absorbed radiation dose is in the best case bad, but getting worse is highly dependent on which place you are in Sun's magnetic field. In the height of a sunspot season being in the middle of a a flare or protuberance is unimaginably bad for you. Any accelerated protons (be it the Sun's magnetic field or another cause) have a really steep Bragg peak, some of it inside you where it really counts.
Ionizing radiation would not be a concern at the surface of the sun as it is all absorbed internally and re-emitted at the surface as thermal radiation. I couldn’t find out how much of space radiation is from the sun vs from cosmic sources, but assuming 100 mSv/yr at Earth’s distance and assuming a spread by the inverse square law, the dose at the surface of the sun will be ~4,500 Sv/yr or about 143 femto-Sieverts in a nanosecond. This is a comically small amount of radiation and would have no real effect on your accumulated dose in a day.
We should all remember that the sun is big. - The core is twice as far from the surface as the Earth is from the Moon. Those gamma rays are not going to make it there..
"on the sun" "touching the sun" - there isn't any "surface" of the sun. It's a gradually thickening plasma. You'd have to specify a density level as the "surface"
Heat being a function of temper and time is also why the whole "if you drop a nuke there's a certain distance at which frozen pizzas will be perfectly cooked" theory is wrong. The heat will be there but the chemical reactions from that heat won't have time to go through.
What if you dropped a continous stream of nukes so as to ensure that the temperature is maintained over time? Surely then the frozen pizza would be perfectly cooked. Let's say the nukes are magically incapable of damaging each other so the first one doesn't destroy the incoming stream.
What's the difference between 'touching the surface' & being say 3 meters away? I thought the gas just goes transparent as it becomes more diffuse, cooler & less ionized at a certain point that we call the surface, but it's a smooth transition not a sharp boundary. Plus aren't the actual nuclear reactions all happening in the core?
I wonder, since the spectrum of radiation the sun produces at the surface is relatively low in penetrating ionizing radiation(pretty high in UV though) would the surface be producing enough x and gama to be lethal at 1 nanosecond despite being so close to the source. To be clear i dont have any schooling in this, i work IT and i just remember something about if we lost our atmosphere and magnetosphere we wouldnt all die of radiation sickness over night because the level of x and gama are so low compared to UV and visible.
Hello Mr. Folse. I've recenrly been made aware of Pebble Bed REactors. May I know you opinion on the design? Good but hard to convert to? No need to convert to? Cost prohibitive? There's quite a bit of positive takes on the matter so I'm just trying to get a more rounded view.
I thought a shake was a light-metre, about three nanoseconds. I think I might have learned that from “The Sum Of All Fears”, though. Maybe don’t learn nuclear physics from Tom Clancy. ;)
One foot: exactly 30.48 cm. One light-nanosecond: exactly 29.9792458 cm. Also, does it need explaining that specifically a foot from the sun is meaningless? Sunstorms and solar flairs emit x-rays and possibly gamma rays. As for fusion -- at the surface of the sun you are farther from fusion than the moon is from the earth. If you're within a 100 miles of the sun (I'm not even sure that that's meaningful) approximate the sun as an infinite plane. Then it doesn't matter how far you are. You get the same exposure to the sun. The comment of glaza4957, I find plausible. Inside the core, one has extreme radiation pressure, energy density, and temperature. The peak radiation is in the X-rays. How long does it take to vaporize? In real life Icarus would have gotten *cold* flying high. The folk who gave us that folk tale didn't know how far the sun is.
Gravity right in the middle of the sun would technically be 0. As the integral is 0 ("sum" of force in all directions) Gravity differential between parts of your body that are not right in the middle though (like pretty much your entire body), is not gonna be 0. You're probably gonna experience some tearing apart before the shockwave of the pressure (traveling at speed of sound in your body) reaches through your body entirely. You'll kinda look like a cavitation bubble, is what I'm trying to say
That's not quite right. Close to the center of the sun, where gravity isn't zero, it would still be small. As if only the part of the sun closer to the center than you existed.
@@Pablo360ableah but while gravity moves at the speed of light, pressure moves at the speed of the medium. I suppose we could calculate the average particle speed at that temperature to see how far the sun pushes into your space. Less than a foot, but possibly enough to pancake you all the same
I pointed out that the gravity gradient at the surface of the sun would also kill you, but some disagreed. I'm not a physicist, so I might be way wrong, though.
The surface gravity of the Sun is about 28 gees. Col. John Stapp showed that humans can survive that, for _very_ brief periods. en.wikipedia.org/wiki/Surface_gravity en.wikipedia.org/wiki/John_Stapp
your logic is flawed. the radiation doesn't start when you teleport to the sun. it is already coming for that spot. Also the sun doesn't really have a "surface" to begin with. certainly nothing well defined like the surface of a slightly damp rock, like earth.
Hummm.... Here is a quick back-of-the-envelope calculation from a non-expert. The irradiation received from a non-point source (like the Sun) is proportional to its visible angular surface which is, for spherical objects, expressed by the formula 2*pi*(1-cos(theta/2)) where theta is the object angular size (so its apparent surface on the virtual sphere of radius 1 around the observer). On the surface of the Sun, theta will be 180° (so half Sun and half sky) and its angular surface 2*pi*(1-cos(180°/2)) = 2*pi. From Earth, the Sun angular size is about 0.5° so its angular surface is 2*pi*(1-cos(0.5°/2)). From that, we can derive that on the Sun surface, the irradiation shall be 1/cos(0.5°/2) = 105,263 times greater than what we experience from Earth. We can double that value if we assume that the Sun atmosphere (i.e. a plasma?) is also radiating. The conclusion is that the amount of radiation received in 1 nanosecond (1/1,000,000,000 s) shall be roughly 0.00021 = 1/4800 times the amount during 1 second from Earth so quite negligible. We should also consider the protective effect of Earth atmosphere and of Earth magnetic field but that should not change the result by much. So my non-expert opinion is that XKCD is probably right.
What is the speed of light in meat? A nanosecond might not be enough time for the gammas to travel through the entire thing. Definitely not if the sun is to your side!
4:00 "what the dose rate is" Assuming you're standing, nothing from the sun's surface will get past your calves. You might lose your feet, but nothing else will be affected.
4:00 At radiation levels this high, wouldn't a lot of your living organic matter - if not all of it - get ionized into dead organic matter immediately?
Stefan Boltzmann law. But as a neat trick, the system reaches equilibrium once everything is at the same temperature, so you can instead ask what if the person was the one at 5000K, see how much energy they radiate, and divide it by 2 to account for the field of view.
A nanosecond considering the exposure of heat radiation tremendous solar wind which I'm sure at that distance would actually affect something as large as us and the intense gravity and other title forces even in the nanoseconds I'm pretty sure all of those forces would have enough time to act on you
Radiation poisoning? Uh, source? Yeah the sun is a huge fusion reactor- at the core. Yeah it’s huge, yeah the radiation is coming from all directions, but without doing any calculations I’m not sure what you are basing your claim on.
Sorry, you’ve made a few mistakes here: first, on the surface, there’s a fairly easy way to think about the radiation dose. The temperature is 5800K, the equilibrium temperature on Earth is around 300K, which is 1/20th, so by the T^4 rule, this is like spending 20^4 x 1ns = 160 microseconds, at Earths distance but completely unshielded by atmosphere. Even unshielded by atmosphere, this is very survivable. Sorry Nuclear Engineer, you have over-estimated how the ionising radiation at the suns surface, don’t know where you got your hundreds millions Sieverts from. Secondly, at the core, pressure is *not* going to be what kills you fastest. Go back to that “light travels 1 foot per nanosecond”. Any chance “contact effect” is transmitted by the interactions between adjacent particles, whether it’s atoms or plasma. The “speed at which force is transmitted” is the speed of sound in the medium, which is high but much lower than speed of light. It is a plasma, so the speed of sound is proportional to (kB x electron temperature / ion mass)^1/2 in natural units….ie relative to the speed of light. I think you can see where this is going - in any star, until the temperature is high enough that the electrons are relativistic, ie the star is held up by degeneracy pressure like a white dwarf, speed of sound is much less than speed of light. The suns core is mostly helium, although it isn’t fusing at those “low” temperatures. In actual numbers, with 15 million K core temperature, I get speed of sound = 0.0022c = 6000km/s. The conclusion of that, is that in one nanosecond, the pressure wave has only had time to “travel” less than a millimetre into the skin. The *photons* however, have already fully cooked you from front to back, since you are less than 30cm thick. However, you could say that it’s even a bit stranger than that. All the molecules in your body will have been fully ionised by the torrent of radiation, they won’t be “molecules” in any meaningful sense, as nothing will be bound. However, the incoming photons will be predominantly interacting with the electrons of your body, not the much heavier nuclei. The nuclei of all the molecules of your body will not yet have had time to start to physically move and fall apart, from having their electrons stripped from them. With a sufficiently detailed microscope, you’d look like a skeleton with all the “atoms” still in the correct places, even within their molecule, but fully ionised as a plasma!
sieverts is joules/kilogram. Even if you assume all the joules you receive are ionising radition, which they won't be, it is still going to be negligible.
In that nanosecond, the entire volume of your person would have absorbed all of the radiation in that space your body occupied plus the "moving radiation" you instantly occupied a space filled with a radiation, plus the radiation that hit you in that one nanosecond. Lethal Dose.
How ever you are 'transported' for just 1 nanosecond at the sun, your body will not fare well. Even travelling at the speed of light isnt good for your body (cause of the g-forces imparted on your body). I cant imagine arriving and then having to leave so abruptly is any good. Best to be 'transported' to the core. Make it swift so nothing has to come back.
The brain is like a cpu thats usually in idle mode the fact that the brain can replay your whole life in seconds or make time stand still tells you something of what it can do at full power.
![HIWWHEE | OHM [Official MV]](http://i.ytimg.com/vi/OJJLtTLqLag/mqdefault.jpg)
Thanks so much for watching! If you are curious as to how long you can stay in a nuclear spent fuel pool, please check out: th-cam.com/video/diHG9W27XeU/w-d-xo.htmlsi=dpxPwZbkX_UrXDEb
Anyone that suggests you do this at night is an idiot. Everyone know you only do this in winter.
What if you did it at night?
You just couldn't resist, could you?
🐱
Really?😂
You could survive at least five minutes
hahaha good one .... scary thing is there people out there who believes that
You know you’ve made some mistakes when the first question the nuclear engineer asks is “what’s gonna kill you first”
That’s how you know.
1. Solar luminosity (power output): 3.83E26 W
2. Solar surface area: 6.09E12 km^2 = 6.09E18 m^2
3. Solar surface luminosity: 3.83E26 W / 6.09E18 m^2 = 6.29E7 W/m^2
4. Ratio of hard radiation (X-ray, gamma): ~1%
5. Surface power of hard radiation: 6.29E7 W/m^2 * 1% = 6.29E5 W/m^2
6. Irradiated cross-section (human): ~1 m^2
7. Irradiated mass: ~100 kg
8. Absorbance: 100%
9. Dose rate: 6.29E5 W/m^2 * 1 m^2 / 100 kg * 100% = 6.29E3 Gy/s ≈ 6.29E3 Sv/s
10. Exposure time: 1E-9 s
11. Total dose: 6.29E3 Sv/s * 1E-9 s = 6.29E-6 Sv
12. Banana equivalent dose: 1E-7 Sv
13. Bananas needed for same dose: ~63
Bananas are scary.
This is definitely quite plausible.
So is this showing that the estimate of billions of sieverts per second was off? It's only a few thousand?
It shouldn't matter if you're a foot away, ten feet away, or a mile away. That close to the sun, it's approximately a flat plane, and integrating the inverse square law for a plane actually gives you a constant value!
Also the surface is not even well defined on a km scale.
Not nearly but it is high. The gravity on the surface of the sun is about 28 Gee's. In comparison the gravity on the outer 'surface' of Jupiter is only 2.4 Gee's. 😄 (That's using the Newtonian eq A = GM / r^2.)
@@Lucien86 28 G's for a nanosecond would not be noticeable. The human body can easily withstand massive G spikes in short bursts.
Fun fact: Grace Hopper, one of the Grand Old Women of Computer Science, used to hand out 30 cm bits of wire, as a handy illustration of nanoseconds.
That's Rear Admiral Grace Hopper 😊
And she said if you want some picoseconds get yourself a pepper mill 🙂
Interesting wavelength comparison. Radio waves in GHz and sound waves in kHz are roughly equal.
A blink, but in reverse? So a knilb?
Definitely
Agreed.
That's gunna be my next D&D goblin champion name. I'll definitely give them armor of blinking
That's as good as any other arbitrary word so sure.
Can we standardise pronunciation now though? "Nilb" or "Kuh-nilb"?
The question was to heat up because it was too cold in Colorado, so xkcd focused on temperature, not all the hazards on the surface of the sun. As you say, they are many variables to Calculate.
But without knowing all the dangers associated, some young impressionable folk may get the idea to actually teleport to the sun and back for a nanosecond.
@@alphazero924 maybe you are right, also there wasn't a disclaimer with "dont do this at home"
@@Nikolai_brrrr Well you can't do it at home, you do it at the sun. If you have the sun in your home there are other considerations.
You are reminding me that lightning is hotter than the sun but people can survive it if hit by
Minus the radiation dose, and this is for the surface of the sun not the core
not the centre of the sun it isn't
Typically you're either going to explode or basically have your heart and entire nervous system fried. If you get very lucky and you have something that can actually draw the bulk of the electricity away from you then you might survive. Basically we're talking about people who I mean were technically in the wrong place but basically everything went perfectly in their favour. But now you can more often than not you're probably going to die if you get hit by lightning you'll probably just go deaf indoor blind and probably get hit by a decent dose of x-rays
I think most survived lightning strikes are not direct hits, but unluckily getting some of the current as it dissipates in the ground.
@@Clayne151 indirect or the person got lucky and had just the right things on them to redirect the lightning away from their core or brain.
No, no, light will always begin to enter your eyes regardless of the distance.
The output stream is continuous. You don’t have a delay while light travels from the surface to your eyes.
It takes time to process the light
@@TriangulumGalaxy_1 Not the point. Since the sun does not light up at the moment you appear, being 30cm away wouldn't mean you don't receive anything. You will always take in a nanosecond whatever radiation was already traveling and was within 30cm of where you appear.
@@Brorebeth Ohh I see
He never suggested that the stream of light would not be contiuois, the 30 cm was just a side note...
You get 30cm worth of light waz the point
Can you show us the exact calculation for the radiation dosage on the surface? I'm not convinced that it would be lethal, but maybe I'm wrong. Either way, I'd line to see you calculate it.
I think this can only be resolved by conducting an experiment
There is a comment that calculates it. The dose is equivalent to eating 63 bananas (which are slightly radioactive). So you will probably live.
The sun actually doesn't emit any xrays itself. it follows the black body radiation. The inside if filled with xrays because thats the temperature they are created at. The photons are absorbed and reemitted at at lower and lower energy levels closer to the surface as it gets cooler. the photons slowly carry the energy to the surface but the only ones that come off it are only created in the surface itself.
What actually comes off the sun as Xrays is all created in the corona from ejected particles at very high temperature. But the Corona is extremly thinn. the energy density compared to the inside or even the surface of the sun is VERY low. a nanosecod is not enough time for any real damge to be done, even by xrays as these make up a miniscule fraction of energy emitted by the sun.
We are talking about black body curve of 5800K for the exterior and it still emits a little xray and gamma naturally. But in deeper layers where the temperature gets ridiculously high the ionizing photons also get ridiculously high
@@rea1m_ The black body radiation for 5800K has nearly no emission below UV light, so the central assumption that a nanosecond there would be fatal due to gamma and x-ray radiation is flawed *if* you only consider that blackbody radiation.
Thank you!! Yeah, I was also thinking about the blackbody curve, which drops off exponentially with increasing photon energy. The photosphere is going to emit virtually no x-rays. Now, _some_ of the emission from the corona will propagate back down to the photosphere, but this video didn't break that out as a separate calculation. Even if the "millions of sieverts per second" (from the corona???) _is_ right, that's millisieverts per nanosecond - and a millisievert is not a big deal.
One other nit about all of the "surface of the Sun" cases. The photosphere is not a sharp boundary on a human scale. One can pick the depth from which half of the visible light from the Sun is emitted, but the photosphere is about 300 miles thick. Face towards the Sun's core, face away from the Sun's core, one is immersed in somewhat hazy white hot gas for a nanosecond in either case.
@@jeffreysoreff9588 The 10 µJ/cm² reported in xkcd should include all ionizing radiation delivered from the solar surface in that nanosecond, so I don't understand the claim that the radiation dose would be high, either.
I'm not an expert on this, but I ran some numbers and got a similar value of 6.3 µJ/cm². Assuming an average 62 kg adult with a crude estimate of .2m² facing the sun and all of that energy as ionizing radiation gives about 200 µGy delivered to a person. That should be 200 µSv for photons, but even assuming that is delivered with the highest multiplier (20 for neutrons), which we shouldn't, that is still only 4 mSv. Even with uneven penetration, I'm dubious we could get a factor of 250 to get it up to 1 sievert, and this is after being generous.
Calcs: Solar constant max * (1 AU/radius of Sun)² * 1 ns = 63 mJ/m² energy delivered. 63 mJ/m² * .2 m² / 62 kg = .2 mGy.
Not all radiation is thermal radiation. I don't know how transparent the sun is to (hard) radiation, so I guess it's somewhere in the middle between - thermal radiation and some radiation created by nuclear stuff.
Something tells me the radiation dose is actually going to be less than the actual heat energy you absorb on your trip there.
depends on how you get there. Get a small comet and put it on a Sun-grazing trajectory. Stay inside it until you get there.
The Parker Solar Probe is expected to survive several passes within 6 million km of the Sun's surface. en.wikipedia.org/wiki/Parker_Solar_Probe
Yep - a Sievert is a unit of absorbed dose in J/kg. Simply by following the black body law there is much less UV/X Ray/Gamma Ray radiation than light and heat, and the UV would only give you sunburn.
There are also the solar protons and such, but again, the length of time exposed is just so short.
Dude, there's no such thing as "one foot above the surface of the Sun". You're talking about kilometers of ever-thickening plasma. From 150,000,000km it may look solid, but it just ain't so.
I would expect the ratio of different EM frequencies at the surface of the sun to be similar to that in interplanetary space, as empty space does not diminish the radiation. As the most immediate risk of facing and staring directly at the sun without a visor is blindness, then I would contend that even at a closer distance and shorter time, that if it doesn't blind you, it won't significantly irradiate you.
This is an excellent and very satisfying proof
The “surface” of the sun isn”t like the earth. You’re not just a solid 1 foot from the “surface”. It’s a wildly varying gas surface.
Yeah thats why he said its hard to define what temperature and rad dose youd get because it varies so much with this and solar flares
Yeah, it is basically the same as asking where the Earth's atmosphere ends too
@@GummieI there is no definition but maybe 6 million kilometers
@@Aarush.A.S6 million km! Bro the moon is approximately 300000km far away which is one-third of a million. The atmosphere ends way before that. Probably 10000-20000 km.
0:13 or at winter, when it’s colder
We're technically closer to the sun in winter😅
Sun's radiation is mostly thermal, high energy particles are not significant in energy output.
I'm trying to double-check whether you are actually right or wrong, but I couldn't find the data on how much ionising radiation does the Sun emit per square meter (or as a percentage of its total output) - and I'm also not sure if my idea on how to actually convert it to an approximate dose is correct. I'd love it if you did a follow-up video and actually did these calculations.
Edit: I have found some articles by NASA that say the surface of the Sun doesn't actually emit a lot of X-rays, these are produced mostly by the corona. I think this question is actually much more complicated than it seems
1 sievert is 1 joule/kilogram, so even if you assume all the energy you receive is ionising radiation, it is going to be negligible.
I mean… the only way to see if it’s true is to actually do it 🧐
Which ofc it won’t be done for a lot of reasons
@@katrinabryce the total energy output of the sun is something like 10^26 W, the surface area is 10^18 sq m, divided it gives you 10^8 Watts per meter, so a lot, but a nanosecond is 10^-9, so a meter of the Sun outputs 0.1j of energy in a nanosecond.
Not sure how to convert it into a dose received by a human body though. If all of that energy would be evenly distributed throughout your body it's basically nothing, but calculating it this way doesn't seem right - but I do think xkcd is closer to the right answer
@@igorbednarski8048 If it was 100% ionising radiation, then the dose would be 100mSv, which the UK Health Security Agency describes as "Level at which changes in blood cells can be readily observed". That is an absolute worst-case scenario, and it is in reality going to be a lot lower than that.
@@katrinabryce 0.1J per square meter is not 0.1Sv, unless you're irradiating a really strange human that is a flat 1 square meter pancake that weighs exactly 1kg, they get uniformly irradiated with nothing but x-rays and absorb all of it evenly.
Focus all of that 0.1J into your little finger in the form of high energy neutrons and you might end up getting 100 Sievers instead of 100 milisieverts.
People make,, by mass and even by volume, much more heat than the sun in average. People are indeed bright
It doesn't matter how far you are from the sun surface, the few feets wouldn't make a difference, because the radiation was there already. Sun didn't started radiating only when you teleported.
I wouldn't be surprised if he responded to this video with his own video doing the calculations of how high of a dose you would get.
I'm curious what the justification for a high gamma flux at the surface of the sun is. At the photosphere you're pretty deep inside an atmosphere that can provide shielding and I wouldn't expect the temperatures to be high enough for there to be any locally produced gamma rays. There would definitely be tons of UV and some X-rays, though.
If we define the surface as that what is visible from earth, the ratio between visible light and gamma/x rays can't be much different than what we observe from here (above the atmosphere of course).
Because there definitely isn't much in the way that could absorb it on the way.
And the falloff from distance is the same of course.
Radiation from charged particles could be different, but i would like to see the justification for that. Its not like there should be any nuclear reactions taking place that far from the core.
I feel like the flaw here is that it refers to radiation that needs to reach you. What about the radiation that was already there? Wouldn't that burn you too?
That point about temperature and heat is demonstrated pretty well by the thermosphere. The average free energy per particle up there (The temperature) is very high, but there is very little of them so the total energy (Heat) per unit volume is minimal.
Was gonna say as a welder we get covered in sparks. Most of those are close to "sun heat" they just don't have enough mass to burn anything. So I figured time would do something similar
7:02 ok I may have disturbed my neighbours with that laugh 🤣
What if you did it during a total solar eclipse?
I hope that's a sarcastic comment
@@sirynka the "What if you did it at night?" was already said, so I had to come up with something else.
Teleportation device would not send you back because of Moon.
- I may be wrong...
- please don't conduct an experiment
Made me think of Two shakes of a lamb's tail. I than found this on Reddit. This expression first appeared in publication in 1840, but is undoubtedly much older than that in vernacular usage. Interestingly if was originally paired with its opposite-“two shakes of a dead lamb’s tail”--which of course meant never. The expression is occasionally reduced to “two shakes,” meaning the same thing. “See you in two shakes” was a fairly common expression I would hear when I was growing up. The expression is gradually disappearing from written and spoken usage, one assumes because fewer and fewer people grow up actually seeing lambs wag their tails. The one place it will probably be preserved is among nuclear scientists, since during the Manhattan Project they adopted the term “shake” to denote 10 nanoseconds. Strange how words and expressions evolve! Language evolves. Not sure seeing lambs tails in action has any relevance really.
I just put in a suggestion to What If? about how a car engine that fused hydrogen instead of burning fuel would act, and I think that question, though unlikely to be answered by him, would definitely get your attention.
6:25 No actually gravity wouldn't be extreme at all. Gravitational force is inversely proportional to the square of the distance, and at the core that distance would be 0. So the Gravitational force would be 0 at that point. Now of course, the other factors would kill you but Gravity wont be one of those.
He didn't say that there's extreme gravity at the center or that it would kill you, did he? He only said that it induces fusion reaction with plasma, which is true.
Extreme pressure based on gravity pulling on the entire radius of the sun. 26 petapascals.
True, it's pressure. Note that on the surface, other than your feet, you're safe (enough). The Ruler rule will protect the rest of your body (though it depends on your teleportation device. A cut-rate unit might not completely clear your landing zone of matter/energy)
oh there isnt any gravity on you sure... but what about the entire mass and the resulting 250 billion bar of pressure created by the fusion?.
The gravity of the entire star causes pressure thay crushes you. It's semantics at this point.
Just wondering how we know the radiation environment at the surface? Is it data from the Parker Solar Probe? To me, an educated (but not in the nuclear field) person,I would say that X-rays and Gamma rays would be mostly confined to the Sun's core, and would be unable to escape the confines of the core due to the densely packed matter therein, it would be reradiated outwards as heat, light, radio waves, etc due to the heat from absorbing that radiation. I know that there's radioactive products in there that escaped the fusion process (e g. Tritium, Helium-3), and probably neutron activated radioactive materials as well, since the sun has some "metallicity" to it (I e. Things that aren't hydrogen and helium). I know that there's tiny amounts of material that get fused beyond helium, also... that become the main fusion source when the sun becomes a red giant later in its life.
Hmm... I didn't think the sun's photosphere had much in the way of x-rays & gamma rays. As the sun is quite optically dense, none of the light from the hotter regions could get there, so it should be very close to the standard black body radiation for that temperature. I suppose if it is like the picture, AJ will take some x-rays from behind, coming down from the quite hot corona.
Alpha & alpha like radiation from the helium nuclei & protons shouldn't be an issue, as they won't have time to inflitrate the respiratory track & will be stopped by the top layers of the skin.
Beta radiation might a concern with all the free electrons. I don't think they'll be able to penetrate down to the basal layer of the skin with the much lower kinetic energy they have relative to the beta radiation from decay that we usually refer to. That's a guess though.
This thought experiment has a huge implicit assumption that drastically changes outcomes: what happens to the matter and energy that was inside the space you teleported to? And what happens to it when you teleport back?
If you completely replace all matter and energy in the spot you teleported to, 1 femtosecond in the core wouldn't be that bad since it'd only directly affect the outermost 300nm of your body (dead skin cells). Then, the way you return is important since those waves within your perimeter would do a lot of damage to your body if you bring them back with you.
Of considerable additional concern is that at the temperatures you experience inside the sun's core, all the atoms will be moving at a noticeable fraction of the speed of light. That means that a large fraction of the matter within one to several tens of nanometers will penetrate into the outer layers of your skin and will also come back with you. The sun is dense enough that even a modest fraction of a cubic millimeter* of material absorbed into the surface of the skin would be a very large number of lethal doses of radiation. When you are talking about how many alpha particles your body has absorbed you really do not want a number expressed in tens or hundreds of milligrams.
*One square meter of surface area times 10 nanometers (3.3% light speed) would be 1/10th of a cubic millimeter at solar core density of high energy particles. I am not sure what speed is actually equivalent to 15 million Kelvin, but I imagine I am within an order of magnitude of correct.
Light travels 30cm/ns, so in 1ns, it is going to go either all of the way through or most of the way though if it hits you on the front or back.
I always think of a light nanosecond as a metric foot. It's not exactly a foot, but it's the same order of magnitude and closer in proportion than a metric ton is to an imperial ton (which are also the same orders of magnitude, such that measuring bridge strength in either gets you well within the safety margin in the other).
If you were only in the center of the sun for one femtosecond, the radiation, however intense, would only have time to heat up about 0.0003 mm of your exterior. Upon teleporting back to earth, could that tiny layer of plasma burn you? Would the radiation traveling through that layer be enough to ionize the rest of you in the nanosecond following the teleport? I assume the pressure, however intense, couldn't crush you quicker than the speed of light.
It's interesting that we call the extremities of the photosphere the surface of the sun considering it's less dense than the earth's atmosphere there.
So going to the sun for even a second makes you super dead. But what if you tried being a billionth of super dead?
@3:50 is there a particular reason the radiation would be constrained to "up to" the surface of the sun? Cuz in my mind it would be, well, just radiating out, creating a stream of radiation. Hence if you dipped into that stream (much like a river) you'd still get wet, or in this case, radiated?
6:10 I love how the question isn't the normal "huh, I wonder what would happen?", but "Lets see, what is going to kill you first?".
Light goes a foot per nanosecond is required lab knowledge. I mean those little coax cable are even labeled in 1/2, 1, 2, 4, 8… ns. They’re a bit shorter since the pulses go about 0.8 c.
The absorbed radiation dose is in the best case bad, but getting worse is highly dependent on which place you are in Sun's magnetic field. In the height of a sunspot season being in the middle of a a flare or protuberance is unimaginably bad for you. Any accelerated protons (be it the Sun's magnetic field or another cause) have a really steep Bragg peak, some of it inside you where it really counts.
That opening was hilarious
Ionizing radiation would not be a concern at the surface of the sun as it is all absorbed internally and re-emitted at the surface as thermal radiation.
I couldn’t find out how much of space radiation is from the sun vs from cosmic sources, but assuming 100 mSv/yr at Earth’s distance and assuming a spread by the inverse square law, the dose at the surface of the sun will be ~4,500 Sv/yr or about 143 femto-Sieverts in a nanosecond. This is a comically small amount of radiation and would have no real effect on your accumulated dose in a day.
We should all remember that the sun is big. - The core is twice as far from the surface as the Earth is from the Moon. Those gamma rays are not going to make it there..
Someone needs to test this. We can only speculate until then.
I'm curious where you get the ionization radiation numbers for the surface of the sun?
They don't consider that the radiation and light are constant if you teleported to the Sun there would already be light were you teleported.
"on the sun" "touching the sun" - there isn't any "surface" of the sun. It's a gradually thickening plasma. You'd have to specify a density level as the "surface"
Heat being a function of temper and time is also why the whole "if you drop a nuke there's a certain distance at which frozen pizzas will be perfectly cooked" theory is wrong. The heat will be there but the chemical reactions from that heat won't have time to go through.
What if you dropped a continous stream of nukes so as to ensure that the temperature is maintained over time? Surely then the frozen pizza would be perfectly cooked. Let's say the nukes are magically incapable of damaging each other so the first one doesn't destroy the incoming stream.
@@leobriccocola8141they're clearly must be an answer. It is after all the greatest question of our time.
@@leobriccocola8141 This is like that video where they cooked a chicken by slapping it taken to the extreme
What's the difference between 'touching the surface' & being say 3 meters away? I thought the gas just goes transparent as it becomes more diffuse, cooler & less ionized at a certain point that we call the surface, but it's a smooth transition not a sharp boundary. Plus aren't the actual nuclear reactions all happening in the core?
I wonder, since the spectrum of radiation the sun produces at the surface is relatively low in penetrating ionizing radiation(pretty high in UV though) would the surface be producing enough x and gama to be lethal at 1 nanosecond despite being so close to the source.
To be clear i dont have any schooling in this, i work IT and i just remember something about if we lost our atmosphere and magnetosphere we wouldnt all die of radiation sickness over night because the level of x and gama are so low compared to UV and visible.
Oh, that's a good WhatIf question. What if you transported a cubic mm of the sun's core to home plate at Yankee's Stadium?
Hello Mr. Folse. I've recenrly been made aware of Pebble Bed REactors. May I know you opinion on the design? Good but hard to convert to? No need to convert to? Cost prohibitive? There's quite a bit of positive takes on the matter so I'm just trying to get a more rounded view.
I watched they’re video when it was minutes old, I don’t even watch XKCD
I thought a shake was a light-metre, about three nanoseconds. I think I might have learned that from “The Sum Of All Fears”, though. Maybe don’t learn nuclear physics from Tom Clancy. ;)
One foot: exactly 30.48 cm. One light-nanosecond: exactly 29.9792458 cm. Also, does it need explaining that specifically a foot from the sun is meaningless?
Sunstorms and solar flairs emit x-rays and possibly gamma rays. As for fusion -- at the surface of the sun you are farther from fusion than the moon is from the earth.
If you're within a 100 miles of the sun (I'm not even sure that that's meaningful) approximate the sun as an infinite plane. Then it doesn't matter how far you are. You get the same exposure to the sun. The comment of glaza4957, I find plausible.
Inside the core, one has extreme radiation pressure, energy density, and temperature. The peak radiation is in the X-rays. How long does it take to vaporize?
In real life Icarus would have gotten *cold* flying high. The folk who gave us that folk tale didn't know how far the sun is.
ah, so that means the already extant gamma rays are filling that 30cm,- that you just beamed into. I think you'd explode
I wonder what he'd react to radiation based SCPs like SCP-003 or what he would say about Akiva radiation
Gravity right in the middle of the sun would technically be 0. As the integral is 0 ("sum" of force in all directions)
Gravity differential between parts of your body that are not right in the middle though (like pretty much your entire body), is not gonna be 0.
You're probably gonna experience some tearing apart before the shockwave of the pressure (traveling at speed of sound in your body) reaches through your body entirely.
You'll kinda look like a cavitation bubble, is what I'm trying to say
That's not quite right. Close to the center of the sun, where gravity isn't zero, it would still be small. As if only the part of the sun closer to the center than you existed.
To be clear, you would be crushed by the pressure, so gravity would still be one of the things that kills you. Just not directly.
@@Pablo360ableah but while gravity moves at the speed of light, pressure moves at the speed of the medium. I suppose we could calculate the average particle speed at that temperature to see how far the sun pushes into your space. Less than a foot, but possibly enough to pancake you all the same
@@mjp121 You're missing the point. The gravity close to the center of the sun is *low.*
I pointed out that the gravity gradient at the surface of the sun would also kill you, but some disagreed. I'm not a physicist, so I might be way wrong, though.
The surface gravity of the Sun is about 28 gees. Col. John Stapp showed that humans can survive that, for _very_ brief periods.
en.wikipedia.org/wiki/Surface_gravity
en.wikipedia.org/wiki/John_Stapp
your logic is flawed. the radiation doesn't start when you teleport to the sun. it is already coming for that spot. Also the sun doesn't really have a "surface" to begin with. certainly nothing well defined like the surface of a slightly damp rock, like earth.
I would like to believe a reverse blink is a blonk
Hummm.... Here is a quick back-of-the-envelope calculation from a non-expert. The irradiation received from a non-point source (like the Sun) is proportional to its visible angular surface which is, for spherical objects, expressed by the formula 2*pi*(1-cos(theta/2)) where theta is the object angular size (so its apparent surface on the virtual sphere of radius 1 around the observer). On the surface of the Sun, theta will be 180° (so half Sun and half sky) and its angular surface 2*pi*(1-cos(180°/2)) = 2*pi. From Earth, the Sun angular size is about 0.5° so its angular surface is 2*pi*(1-cos(0.5°/2)). From that, we can derive that on the Sun surface, the irradiation shall be 1/cos(0.5°/2) = 105,263 times greater than what we experience from Earth. We can double that value if we assume that the Sun atmosphere (i.e. a plasma?) is also radiating. The conclusion is that the amount of radiation received in 1 nanosecond (1/1,000,000,000 s) shall be roughly 0.00021 = 1/4800 times the amount during 1 second from Earth so quite negligible.
We should also consider the protective effect of Earth atmosphere and of Earth magnetic field but that should not change the result by much.
So my non-expert opinion is that XKCD is probably right.
Let's check that. Build a mission to Sun.
Also the sun doesn't really have an exact specific "surface"
I think I could survive a nanosecond on the Sun just fine. The extraction maneuver to get me out in under a nanosecond, however...
Doesn't gravity work at the speed of light. So even one micro second, if there was a surface of the sun, wouldn't he be a pan cake before he got back?
Excuse me sir I think it would be interesting to hear some of your reactions on some of radioactive Drew's videos.
What is the speed of light in meat? A nanosecond might not be enough time for the gammas to travel through the entire thing. Definitely not if the sun is to your side!
4:00 "what the dose rate is"
Assuming you're standing, nothing from the sun's surface will get past your calves. You might lose your feet, but nothing else will be affected.
4:00 At radiation levels this high, wouldn't a lot of your living organic matter - if not all of it - get ionized into dead organic matter immediately?
How long on the sun’s surface, or core for that matter would be safe then?
I enjoy the ASMR mic scratching
On the surface…it’s a good explanation
How long is a day on the sun? Asking for a friend.
The sidereal rotation period (sidereal day) is 25.05 earth days at the equator to 34.4 earth days at the poles.
Pretty sure it’s hotter than 15 million degrees bedside fusion occurs at 150 million degrees doesn’t it?
How is the energy rate from light calculated? He did say it took up almost 180 degree view that means light comes from everywhere on one side
Stefan Boltzmann law.
But as a neat trick, the system reaches equilibrium once everything is at the same temperature, so you can instead ask what if the person was the one at 5000K, see how much energy they radiate, and divide it by 2 to account for the field of view.
A nanosecond considering the exposure of heat radiation tremendous solar wind which I'm sure at that distance would actually affect something as large as us and the intense gravity and other title forces even in the nanoseconds I'm pretty sure all of those forces would have enough time to act on you
Okay I guess a nanosecond is shorter than I thought it was but I still don't think you would survive.
Also the x-rays
So the iron in our blood would remain? Would it become Fe18?
great take
Radiation poisoning? Uh, source? Yeah the sun is a huge fusion reactor- at the core. Yeah it’s huge, yeah the radiation is coming from all directions, but without doing any calculations I’m not sure what you are basing your claim on.
But, how many barns?
Not even a peco second!
Just turn the sun off first!
Wouldn’t this fall under the five second rule?
Sorry, you’ve made a few mistakes here: first, on the surface, there’s a fairly easy way to think about the radiation dose. The temperature is 5800K, the equilibrium temperature on Earth is around 300K, which is 1/20th, so by the T^4 rule, this is like spending 20^4 x 1ns = 160 microseconds, at Earths distance but completely unshielded by atmosphere. Even unshielded by atmosphere, this is very survivable. Sorry Nuclear Engineer, you have over-estimated how the ionising radiation at the suns surface, don’t know where you got your hundreds millions Sieverts from.
Secondly, at the core, pressure is *not* going to be what kills you fastest. Go back to that “light travels 1 foot per nanosecond”. Any chance “contact effect” is transmitted by the interactions between adjacent particles, whether it’s atoms or plasma. The “speed at which force is transmitted” is the speed of sound in the medium, which is high but much lower than speed of light. It is a plasma, so the speed of sound is proportional to (kB x electron temperature / ion mass)^1/2 in natural units….ie relative to the speed of light. I think you can see where this is going - in any star, until the temperature is high enough that the electrons are relativistic, ie the star is held up by degeneracy pressure like a white dwarf, speed of sound is much less than speed of light. The suns core is mostly helium, although it isn’t fusing at those “low” temperatures. In actual numbers, with 15 million K core temperature, I get speed of sound = 0.0022c = 6000km/s.
The conclusion of that, is that in one nanosecond, the pressure wave has only had time to “travel” less than a millimetre into the skin. The *photons* however, have already fully cooked you from front to back, since you are less than 30cm thick. However, you could say that it’s even a bit stranger than that. All the molecules in your body will have been fully ionised by the torrent of radiation, they won’t be “molecules” in any meaningful sense, as nothing will be bound. However, the incoming photons will be predominantly interacting with the electrons of your body, not the much heavier nuclei. The nuclei of all the molecules of your body will not yet have had time to start to physically move and fall apart, from having their electrons stripped from them.
With a sufficiently detailed microscope, you’d look like a skeleton with all the “atoms” still in the correct places, even within their molecule, but fully ionised as a plasma!
sieverts is joules/kilogram. Even if you assume all the joules you receive are ionising radition, which they won't be, it is still going to be negligible.
It is obvious, the radiation would be lethal...
Inverse blink= blip😅
Surface of the sun is not solid, is it? So where do you decide where surface is, think like ocean as the surface.
What's the shortest period of time a human could survive being teleported to the center of a neutron star?
As close to zero as possible
You talk like the surface of the sun as if it were a flat surface. It's actually a very deep layer of extreme turbulence.
cool
In that nanosecond, the entire volume of your person would have absorbed all of the radiation in that space your body occupied plus the "moving radiation" you instantly occupied a space filled with a radiation, plus the radiation that hit you in that one nanosecond. Lethal Dose.
Ah yes, our good old friend, radiation poisoning.
How ever you are 'transported' for just 1 nanosecond at the sun, your body will not fare well. Even travelling at the speed of light isnt good for your body (cause of the g-forces imparted on your body). I cant imagine arriving and then having to leave so abruptly is any good.
Best to be 'transported' to the core. Make it swift so nothing has to come back.
The brain is like a cpu thats usually in idle mode the fact that the brain can replay your whole life in seconds or make time stand still tells you something of what it can do at full power.
And yet people are still idiots.
hii