sevenperforce

Excellent question! This has actually already been asked (and answered) by xkcd! https://what-if.xkcd.com/115/ For some values of "answered" because there the question was a nanosecond, not a microsecond. Here is the relationship between how long you stay there and how much thermal energy your body absorbs (assuming you are an adult human with a body surface area of around 1.7 m2): 1 nanosecond: 0.17 J 500 nanoseconds: 85 J 1 microsecond: 170 J 500 microseconds: 85 kJ 1 millisecond: 170 kJ 1/100 second: 1.7 MJ That last bit is where things definitely get fun. 1.7 megajoules is the energy contained in about half a kilogram of TNT. So spending 1/100 of a second on the surface of the sun would be the equivalent of being painted with a thin layer of high explosive and then having it detonated. You will not have a good day. But let's look at those lower numbers. Your brain can only react to things on the order of a millisecond, anyway, so whatever heating effects you have are going to be effectively spread out over a millisecond. This allows us to get an idea of the effective heat flux. We assume you are somewhere cold and unpleasant and you are teleported to the surface of the sun and back for the specified period. What do you feel? 1 nanosecond. You experience an odd, dim flash of light in your eyes but you don't feel any warmer. 10 nanoseconds. The flash is bright. You experience the momentary sensation of being outside on a warm, sunny day but the feeling fades. 25 nanoseconds. You feel hot for a moment, like someone just opened a giant oven door right next to you but then closed it. 50 nanoseconds. You feel uncomfortably hot but it's not painful, just unpleasant, and the sensation fades quickly. 500 nanoseconds. The pain receptors are triggered and you recoil, but the pain is mild and doesn't last. 1 microsecond. A sudden burst of searing pain across your entire body. You find yourself screaming but look down at your body and see no burns or injuries. 10 microseconds. Once you stop screaming, you look down at your skin and find that you have a mild sunburn. 100 microseconds. Your entire body is sunburned and you will need hospitalization because of reduced skin respiration and other general injuries. 500 microseconds. Second-degree burns across your entire body. Very low chance of survival. 1 millisecond. You experience an odd, dim flash of light in your eyes but you don't feel anything else. Ever again.

Well, the energy-to-mass relationship is definitely dependent on neutrino mass. That's basic. But it's the graph above, giving the oscillation frequency in units of distance/energy, which seems to be the key. If the oscillation likelihood is observably sinusoidal with respect to distance/energy, then that means there is an underlying oscillation frequency which is sinusoidal with respect to proper time, independent of particle energy. That seems like it would be a big deal. But it's not stated explicitly anywhere.

Tagging @K^2 in this because his physics-fu may be sufficiently advanced from mine that he'll have the answer.......but anyone else is also more than welcome to chime in. I've been working on updating the above infographic and it's coming along quite well (here's a nice picture of the progress): I was trying to look up the neutrino oscillation frequency, thinking it could be a good analogue for particle lifetime. However, I couldn't find any mention of oscillation frequency or period anywhere. What I did find, on Wikipedia, were sine-wave probability distributions given with an x-axis in units of distance over energy, since the oscillation wavelength depends on particle energy. (since neutrinos have a range of energies when they are detected): I also found another page which provided example conversions between neutrino energy and neutrino velocity. Obviously, velocity impacts distance traveled, which would impact the wavelength. So I wondered what the relationship was like. Did more energetic neutrinos have a faster oscillation period or a slower oscillation period? I did the math, and it appeared to indicate that more energetic neutrinos had longer oscillation periods. A 10 eV neutrino would have an oscillation period of about 3.6e-10 s, a 1 keV neutrino would have an oscillation period of about 3.6e-8 s, a 1 MeV neutrino would have an oscillation period of about 3.6e-5 s, and so forth. This seemed odd. Usually, more massive particles tend to decay/oscillate faster...though I thought that perhaps this was only applicable to rest mass, not relativistic mass. However, then I realized that just as with the Frisch-Smith experiment that confirmed time dilation and special relativity, I would need to calculate the period in proper time, not coordinate time. So I solved for the Lorentz factor for each of the example neutrinos and divided. To my surprise, I got the exact same period for each: 5.01e-12 seconds. Did I just discover a constant neutrino oscillation frequency?? I looked rather frantically at a bunch of different research and couldn't find anything that talked about neutrino oscillation frequency/period. That would seem to be a pretty straightforward bit of information, but I couldn't find it listed anywhere. Help!

Parachutes would make Starship far more susceptible to wind conditions and thus make it harder to hit the landing net. The fairings did not have control surfaces, and so there was no way to guide them down to the net without a steerable chute. The steerable chute allowed them to guide the fairings into the net...but also allowed the fairings to touch down gently in the water, so they realized they didn't need the net at all. Starship has the flaperons for control so they can probably hit the net every time without much trouble. Hugely concerned. I can't imagine that contact with anything at 67 m/s -- no matter how much give it has -- will be good for heat shield tiles. IIRC the ballpark for landing propellant is on the order of 10-20 tonnes. Even if it was twice that much, it wouldn't have any meaningful impact on terminal velocity.

I want to replace the current diagram on the Wikipedia page for the standard model, which is hopelessly useless.

Having zero near one end but not quite at one end allows me to conveniently fudge the edge of the graph to add an infinity term, so that I can readily incorporate the electron and the up quark. Otherwise they would be off the chart by a factor of three.

That's the entire basis of my favorite tattoo. What are your thoughts on wrapping each group of fermions in an interaction overlay? Exactly. eV is the measured total energy of the particle, and so dividing by c2 turns it into a measure of mass because e = mc2. Does it make more sense on the more complex version to dispense with the whole "eV/c2" term and add that to a legend as well? I have to deal with the electron/positron and the +/- up quark, both of which have lifetimes far longer than the age of the universe. Speaking of +/-, is it readily ascertainable from the graphic that the parenthetical exponents after each particle signify the antiparticle numbers? Or do I need to add a legend for that as well? The photon, Higgs, and Z boson are all their own antiparticle so they get the (0) exponent, while the charge-carrying fermions and W boson all have electric charge and get the (+/-) exponent, and the stinking gluon has 8 freaking flavors and gets the (8) exponent. We don't really know whether there is such a thing as an antineutrino or if all the neutrinos just remain in a three-way (0) superposition so I used (?) for them.

Edited: Any thoughts on the color scheme? I made the weak nuclear interaction blue because it is technically part of the electroweak interaction which is a component of the electromagnetic force, and spin is a magnetic moment. But if this is confusing or distracting I can make it green (which is typically the preferred color scheme for depicting electroweak interactions) and tint the spin signifier as well. Decay paths and flavor oscillations are shown in a dark red, but would it make sense for them to match the color of the weak interaction since particle decay is mediated by the weak force? Another way to depict weak, QED, and color force interactions more cleanly would be to wrap each fermion group in an interaction overlay. E.g., a band of grey for the Higgs field, a rainbow band for the strong force, and a blue (or whatever other color) band for the weak interaction. Quarks would have all three, charged leptons would only have QED and weak, and neutrinos would just have the weak interaction. Would that be more illustrative? And is it clear from the diagram that the neutral leptons (the electron neutrino, muon neutrino, and tau neutrino) have 1/2 spin? I painted them that way but I don't know if it's readily visible.

Which lines are difficult to see? I'm not asking to be snarky; I'm asking because I don't know. I've been making it myself so it's hard to figure out what the "right" version would look like. Whoops, no, you caught an error. It should be either 10n with a negative scale, or 10-n with a positive scale. My double negative is entirely wrong. A top quark lives for 10-25 seconds, not 1025 seconds. It is the shortest-lived of all fundamental particles and is the only color-interacting fermion which can be directly observed, since it decays faster than the timescale of strong nuclear interactions that would otherwise fold it into a hadron. Thanks for catching the mistake. To that point, what's more intuitive? 10n sec with a negative scale or 10-n sec with a positive scale? I need feedback from people who aren't steeped in physics. Also, is it clear from the diagram that the masses of all the particles are to scale (with a volumetric constant-density representation)? Should I add a legend element to explain that? Here's the video I tried to link in the last post which somehow didn't make it.

Both involve a very convenient violation of the rocket equation, which is admittedly very nice. But Starship? Aish. The mechanics are fairly straightforward. A big net with a robust pully system. Maybe use some water displacement. But at 67 m/s bellyflop descent speed and a putative deceleration limit of 2 gees, you need 114 meters of braking distance. If you are generous and allow 3 gees, that's still 76 meters of braking distance. That's...frankly...insane. I know, right? I have always been a proponent of "use the legs on Mars that you must use on Earth because anything else is STUPID." But I suppose if you have 100 Earth landings for every 1 Mars landing then perhaps making single-use sacrificial Mars landing legs becomes attractive.

I appreciate that. I may or may not be in law school, so........ The OST doesn't prohibit the release of radioactive material into space; it just prohibits, substantively, (a) the placement of "nuclear weapons or other weapons of mass destruction in orbit" and says that member States (b) "shall avoid harmful contamination of space and celestial bodies" (emphasis mine). So the OST doesn't prohibit the atmospheric or exoatmospheric release of nuclear byproducts as long as no nuclear weapon was placed in orbit and any associated contamination is harmless (which this would presumably be). The "territorial limits" verbiage in the 1963 Limited Test Ban Treaty is the problem. There have been several instances in which the US or Russia (or, formerly, U.S.S.R.) performed underground tests which accidentally released radioactive byproducts into the atmosphere, which subsequently crossed international boundaries in minute quantities. In each case the opposite side complained bitterly but nothing was done, presumably because it was accidental. It's unclear whether the deliberate release of radioactive byproducts from a nuclear test (however peaceful) would be treated similarly. Although I suppose the signatories could readily enough ratify a codicil specifically excusing a one-time peaceful nuclear test, financed by Elon, under international supervision. It's not like the U.N. has any enforcement capabilities anyway. But I think the NPT would still prohibit Elon from purchasing a nuke outright.

All that makes good sense. Catching Starship? I'll have to see it to believe it.

Here's what's happening. In my own words. And, all things being considered........I'm fairly qualified. On a related note, what do you all think of this adaptation of the Standard Model Diagram? And here's a simpler, less-information-dense version: Honestly I think this is MUCH more descriptive than what is currently used on Wikipedia. The particles are all shown to scale (constant density representation, etc.). Spin and charge are shown visually. Lifetimes are to scale. The four particles we most commonly interact with are helpfully highlighted. The parenthetical exponent signifies antimatter variations in a pretty intuitive way. Thoughts? Is this fairly descriptive? How can I adjust?

Yeah, there's nothing in this article that says what they are doing, only that they claim to do it.

We already knew this: But now this. The man is nuts. But in a good way. I think?

The primary limitation on the sale of nuclear weapons is the Treaty on the Non-Proliferation of Nuclear Weapons (NPT), which absolutely prohibits nuclear powers from "transfer[ing] to any recipient whatsoever nuclear weapons or other nuclear explosive devices." However, Article V of the treaty permits "peaceful applications of nuclear explosions" to be "made available to non-nuclear-weapon States" which have ratified the treaty provided that there is "appropriate international observation" and that "the charge to such Parties for the explosive devices used will be as low as possible." The trouble, presumably, is that Elon Musk is not a "State" under the purposes of the NPT and cannot sign the treaty. Nuclear-weapon states are not permitted to make nuclear weapons available to peaceful purposes to non-nuclear-weapon states which have not ratified the NPT. So unless Elon could be defined as a state then he can't buy one. However, if Elon can't buy a nuke outright, could he pay the US or Great Britain or Russia or China to put one in a specifically-designed test chamber and trigger it? The Comprehensive Nuclear-Test-Ban Treaty has not yet come into force, but if it ever does, then the prohibition on "causing, encouraging, or in any way participating in the carrying out of any nuclear weapon test explosion or any other nuclear explosion" would also be a bar. The 1963 Limited Test Ban Treaty requires its signatories to "prohibit, to prevent, and not carry out" any nuclear explosion, including peaceful explosions, in the atmosphere, in outer space, or in water. It further prohibits nuclear explosions which would cause "radioactive debris to be present outside the territorial limits of the State under whose jurisdiction or control such explosion is conducted." So there's no prohibition on releasing radioactive debris; you just have to make sure said radioactive debris remains inside your territorial limits. But presumably, blowing up a nuke at the bottom of a tube in order to yeet a manhole cover out of Earth's gravity well would tend to send at least some debris high enough that it would drift outside of territorial limits.

What an absolutely spectacular view of that landing burn all the way down.

Oooooh, I see. I thought you were trying to do a best-case scenario, not a worst-case scenario. Missed that part. I guess we just need Elon to buy a Russian nuke and test it.

This is exactly the analysis I was looking for. **screams in joy** But in thinking about this...how sure are you that the force has its maximum at the center and drops to zero at the edge? Why wouldn't it be the other way around? Think about what we know from re-entry characteristics of a blunt body. Obviously we're talking about massively higher velocities here, but still. Maximum plasma temperatures are at the edge, not the center, because plasma temperature is a function of the compression ratio, and compression is highest at the edges. In the very center, the compressed air cannot escape and builds up, expanding both in the direction of travel and perpendicular to it, rarifying the air immediately in front of the vehicle. Thus the highest pressure would be on the edge, not the center. Intuitively, we're still looking at a pressure gradient, not absolute pressure, so would it even make a difference? Well, there are two more considerations. First, the gradient will be distributed differently if the peak pressure is at the edge and decreases toward the center. More importantly, though, this pressure differential is certainly going to be more than what the rigidity of the steel can handle, and so it will turn the disc into a mushroom, almost like how the shockwave forces on a shaped charge mold it into a dart. The disc will be shaped more and more into a dart until the difference in forces is less than what its rigidity can handle, at which point it will punch through the remainder of the atmosphere like the tungsten telephone poles from Project Thor. The most recent interview with Brownlee seems to be this one with Tech Insider in 2016. They make it seem like Brownlee absolutely confirmed the tale and believes it left the atmosphere. They also seem to indicate that the final speed estimate was done based on the footage.

The impact depth approximation provides a good rule of thumb for many cases but it is primarily applicable to high-speed impacts into soft but solid targets. For air impacts it is more of an order-of-magnitude estimate. It wouldn't be hard at all for the cap to end up stabilized edge-on. The shockwave and shock heating effects would likely have an impact (no pun intended) on penetration dynamics.

The vacuum Raptor uses the same turbopump assembly as the SL Raptor but the engine bell is larger. I think there are changes to the regenerative cooling loops, too, so that they cover a larger area.

Bingo. I just knew it was the CH4 turbopump that ultimately failed and precipitated the RUD. I wonder what their fix is, though.

Yes, hence Newton’s approximation. Depending on the tumbling characteristics it may or may not have been able to punch through the atmosphere. It’s hard to even know where to start to guess at the combination of forces in that kind of situation. Good catch, I totally miscounted the zeroes.

Let’s try going from first principles. The actually mass of the cap was 900 kg, not 2 tonnes as sometimes reported. With a 300 tonne TNT equivalent yield, that’s 1.26e12 J. 100% conversion of yield energy into kinetic energy (which is of course impossible) would yield: KE = 1/2 * m * v^2 v = sqrt(2*KE/m) v = sqrt(2.8e10) m/s v = 167.3 km/s I think my math is correct. That’s far, far more than the 66 km/s quoted, so I don’t think we have any concerns there. Also, remember that the cap was welded in place, so you would have had to build up pressure behind it....not that THAT would have taken very long. I am more curious about whether it would survive. You might assume that the drag involved would be sufficient to obliterate literally anything, but if the nuclear blast wave was able to accelerate it to that speed even faster, then that would tend to suggest otherwise. Ordinary drag mechanics don’t really apply, I don’t think. It would have been tumbling, but how many times could it realistically have tumbled before crossing the majority of the atmosphere? It would have been in near-empty space in less than 1/3 of a second. 66 km/s. Unfortunately we don’t know the height of the image in frame. The primary mode of heating for a blunt body passing through the atmosphere at hypersonic speeds is going to be convective. But convective heating between the rapidly-formed plasma and a steel plate is going to require time. It seems that there simply wouldn’t have been enough time for that convective heating to take place.

I am sure most of the people on this forum know about the manhole cover launch off the top of a nuclear test at 66 km/s in 1957. https://vm.tiktok.com/ZMePHarH7/ There is some debate about whether the manhole cover actually made it to space. At that velocity, it would have left the atmosphere in less than a second, so it would not have had time to melt. On the other hand, it well could have disintegrated from the forces...except for the fact that (a) it didn’t disintegrate when it was hit by the force that launched it and (b) it’s a one-time chunk of steel so if anything can survive, that would be it. I would use Newton’s approximation but I feel like at 66 km/s it probably breaks down. I have seen several conflicting statements from Brownlee about it. What do you think? Would a one-tonne steel disc 4” thick, launched from a Nevada desert at 66 km/s, be able to reach space? Or would it disintegrate?