K^2

The post I quoted contains the 50 meV rest mass estimate. I don't claim to read absolutely everything, but I at least take time to read things I quote. The 100 μeV is the typical kinetic energy of background neutrons. That would put total energy at 50.1meV. This is standard notation, too. If you look at any literature on background neutrinos, you will see people talking about them being in 10-6 - 10-4 eV ranges. In general, when we talk about particles at non-relativistic velocities, kinetic energy is the value that's being quoted. Always. Including rest mass in energy is pointless when talking about things happening at small fractions of c. The electrons in a CRT are 5keV, not 516keV, because including 511keV in the energy is not useful. And neutrinos in question are 100 μeV. Look, I've been doing particle physics for a living for a number of years. You're not going to catch me making dumb mistakes here. But at least, you can tell the difference between meV and μeV. I do appreciate that. P.S. dropping c² is pretty typical. Particle physics is usually done in "natural units", where c = 1 and ℏ = 1. It's a trivial matter to insert it back into results if ever we need the SI/cgs numbers. But otherwise, absolutely everything is measured in either eV or eV-1. Energy, mass, distances, and time.

How about 100 μeV neutrinos present in cosmic background radiation?

There are some high altitude, nearly sub-orbital aircraft proposals for which space sickness isn't an issue. BFR as transport will have to be ballistic. Ballistic suborbital trajectories, by necessity, include extended coasting times in which passengers experience microgravity. You're going to get at least half an hour of free fall, which is more than enough to develop space sickness in susceptible individuals. It's not as bad as start-and-go of vomit comets, but it is longer duration. A significant fraction of passengers will require medication, at very least, to avoid losing their previous meals.

I have no idea if any holes even are supposed to be drilled by hand. I was addressing the suggestion that somebody's drill slipped while drilling something near this area, and I can't see it happening. Unless you were supposed to make that specific hole in that specific place, a drill wouldn't be kept in general vicinity. I'm more familiar with how these procedures were followed in military, working on ICBMs. I've seen manuals I probably wasn't supposed to. But that's where Soviet space program takes its root, including Soyuz spacecraft specifically, and very little has changed in the way the rockets are built in Russia. I can try and see if any of my old school friends who went into aerospace still work on anything Soyuz related to try and confirm it, but I highly doubt that the procedure changed fundamentally.

The manufacturing process involves instructions that state which instrument has to be picked up from which bin, used in what order, and where it should be placed afterwards. Never mind the exact placement of every hole drilled. If procedure was otherwise followed, even accidental damage is impossible. The hole was almost certainly drilled deliberately, but it need not have been malicious. A plausible hypothesis was mentioned elsewhere, that the hole might have been necessary to patch through some wires used in testing, diagnosing a problem, or fixing it. Naturally, that would have been done in total violation of procedure, but I can see someone choosing to do that to avoid putting the construction out of schedule. Alternatively, there might be some portion of the process that regularly creates the difficulty in temporary wiring, and somebody probably ought to inspect other craft for similar shortcut holes. Definitely not an excuse, but a hole like that, properly patched, won't present danger to the ship most of the time. So it's anybody's guess how many before it have flown with similar damage without it ever becoming known.

Yes, because that would be direct confirmation of zero mass. Something physicists really like to have. Nonetheless, flavor oscillation is impossible under standard model if they are massless. And if they have a mass, they can travel at absolutely any speed. If you are not convinced by this argument, you should be arguing that neutrinos don't exist. We cannot measure them directly, just showers of particles they produce when they interact with matter. This is also why they have to be traveling at damn-near speed of light to be detected. Saying, "I believe they exist, but I don't believe they can travel at 10 miles per hour," is the kind of idiocy I generally expect from flat-earthers and creationists. They also like to selectively ignore the evidence. The fix I gave you avoided allocation all together, because std::vector::assign does not allocate if resize isn't required. I didn't realize that you remained clueless about that after I gave you the correct way to handle that loop. Moreover, assign itself will compile to a loop with just a few instructions and be inlined. There isn't an actual function call when you do std::vector::assign with an integral type. It's faster than memcpy. I'm not even going to respond to the type portion, because you obviously never learned how to use a disassembler, if you still think STL sizes return ints on x64. Please, please, just promise me you'll never come work in the Valley, and I'll leave your code alone.

You could try reading the paragraph above. No neutrino has ever been registered directly, period. Their very existence is inference on indirect measurements. Indirect measurements say walking pace neutrinos exist. So yes, you are still absolutely wrong when you say neutrinos travel at (nearly) lightspeed. And maybe they teach you not to use the words like "nearly" and "exactly" when talking about physical quantities at the same institution that taught you not to write clean code for a short demo. If you ever actually had taken even a survey course in physics or mathematics from a reputable university, they would teach you better. The fact that you are completely wrong regardless of this is an excellent indication of why nuances are important. You can't divide a real number by zero, but you can divide by nearly zero all day long. And if you don't think that's important, I would recommend a career in arts, far away from anything where precision actually matters. Again, I'd let it slide if you were simply wrong on the basis of missing that "nearly", but the fact that you're still under a miraculous impression that your claim was right after being told by somebody who studied particle physics for nearly a decade that it's not even remotely close, is the exact reason why its actually critical in highlighting your ignorance of the subject. You systematically replace facts by fiction, and if you simply answered questions outside of your competence, which is common enough on this forum, that wouldn't be a problem. The problem is that when you are given the exact reason why it is wrong, you don't even understand enough to get the explanation. And rather than say, "well, I don't understand that," you double down on your misinformation and pretend that anything you don't understand does not exist. I'm surprised, you acknowledge existence of subatomic particles at all, at this rate.

Distance 100km. Flight time of an aircraft? I'll let you figure out what the critical bit of information you're missing relevant to an elementary particle. Luminal speed has implications. It says, explicitly, that worldline is on a light cone. And that means that the particle is either massless or off the shell. Since a free particle has to be on-shell, and neutrinos exhibit behavior inconsistent with massless particles, clearly 'lightspeed' is always wrong. But you can't even claim an approximation. Everything we know about physics says that walking-pace neutrinos aren't just possible, they're all around us. It's another matter that we can only detect neutrinos possessing enough energy to trip a sensor. So a detectable neutrino travels very close to light speed. Saying that all neutrinos travel at lightspeed, even if you just forgot to use the word "nearly" in there, is flat wrong. And I'm well aware of the difference between identity (тождество) and equality (равенство). I'm also well aware that neither of these imply an approximation, so either you need to brush up on these yourself, or I have no idea where you're going with these.

Vaguely relevant article, but that last paragraph is a gem that seems worth sharing. Air Force Not Sure What To Do About Elon Musk Smoking Weed - GIZMODO

Yes, because we've already established that when you don't understand existing experiments or their implication, it means that the phenomenon is unknown to science. Yup, that's exactly how science works.

Almost certainly not.

Yeah, which is why I've been consistently referring to sidereal rotational period. Which is proper period of rotation of the body with respect to distant stars, regardless of its orbital motion. That is the quantity of interest if you are talking about solar day, because we then only need to find relative motion of the Sun, and we're in business. And of course, if you want very precise duration of the day for any given location, you need to do precise math on relative sun position, and that involves dramatically more math. Fortunately, that level of precision is rarely warranted if you're looking at duration only, and mean day is quite sufficient.

Sidereal rotation is own rotation. That's taken into account. Axis tilt is not accounted for, but so long as it's small with respect to plain of planet's orbit, mean solar day is not impacted on most of the surface. Only polar regions are going to be off. The size of these regions depends on the degree of the tilt. So again, small tilt means small regions where this doesn't apply. I've mentioned small tilt ('roughly perpendicular' bit) in my first post. Orbital period and inclination of the moon's orbit is entirely irrelevant. It has zero impact on duration of mean solar day. So no, tidal locking is not necessary and is explicitly not an assumption.

The response to the first season was great. We're ret-conning it to a bi-elliptic transfer.

Orbital time of moon around planet is irrelevant. It creates a small adjustment to relative position of the Sun, but it's periodic, so it has no impact on mean solar day. If you wanted a precise sunrise/sunset times, you'd have to take it into account. If you just want mean solar day, all you need to know is how long it takes you to go all the way around the Sun, which is planet's orbital period, and how long it takes the moon to rotate relative to the distant stars, which is sidereal rotational period. And there are no assumptions about tidal lock here.

The only necessary assumptions are that parent object (planet) is in near-circular orbit, and the moon's axis of rotation is always roughly perpendicular to direction towards the star, id est, inclination of moon's axis is small with respect to plain of orbit of the planet; and moon's orbit is relatively small compared to distance from the star. In that case, you have a very simple formula. Tday = 1 / (1 / Trotation - 1 / Tyear) Here: Tday - mean solar day on the moon Tyear - sidereal orbital period of the planet Trotation - sidereal rotational period of the moon Negative values can be used to denote retrograde motion of either body. This gives you mean solar day. The exact time of sunsets and sunrises will drift a bit either seasonally or monthly, depending on configuration, so even sunrise-sunrise time isn't going to be exactly constant, but it will not vary wildly, and will average to the above. If you need exact sunrise/sunset times, that's where things get a little complicated. There's nothing like a clean formula, but if you really need it, I can write you a script in C#, JS, or whatever. It will have a LOT of input parameters.

You drop the fully fueled rocket from great height, and see how fast it moves when it hits the ground. Or model the same on the computer, which tends to be safer. For a simple, rough estimate, you can use a model where air resistance is FD = (1/2) ρAv2. Technically, there is going to be a drag coefficient in there, but taking it to be 1 puts you in the ball park. Here, ρ is density of air near the surface, A is area of relevant cross-section of the rocket, and v is velocity. At terminal velocity, this is equal to weight of the rocket. In other words, just solve ρAvt2 = 2mg, where m is mass of the rocket, and g is surface gravity again. Keep in mind that this formula doesn't work great for speeds approaching speed of sound, which is why the whole thing is a very rough estimate for real rockets. If you are doing this in KSP, aerodynamics formula is the same, but area computations are a little wonky. I'm also not sure what it uses for air density and drag coefficients of different parts. My advice, go back to plan one and drop the rocket from great height nose first. See how fast it's going before it impacts the terrain, and use that. Better accuracy and more fun.

Nonetheless, certain topics are forbidden on this forum. Primarily conspiracy theories, but several pseudo-scientific loads of nonsense have been stomped on before, because they inherently do not lead to any constructive discussion. And administrators of this forum absolutely have the right to decide what does and does not get posted. This isn't a public space in any legal sense.

A very rough estimate for aerodynamic and gravity losses on ascent is 4gH/vt, where H is scale height, g is surface gravity, and vt is terminal velocity of your rocket near the surface. This is a semi-empirical result based on vertical ascent. It ignores many nuances of actual aerodynamics and gravity turn, but in practice works better than one ought to expect. At any rate, I'm not aware of any better estimate you can get from a simple formula. If you want more precise results, you'll need to do a lot of numerical work. Consequently, a very rough estimate of total delta-V is sqrt(GM/r) + 4gH/vt - v0, where v0 is your initial velocity due to rotation of the planet if you launch from equator. This will give you a good estimate for KSP, and a very rough estimate for real world. Still better than just the initial sqrt(GM/r), though.

The thing about cryo storage is that if you found a way to freeze and thaw the person safely, the answer is yes. Organics can be preserved for tens of thousands of years. Recently, some 40k year old worms were revived in a lab. Humans are a bit more complex, to put it mildly, but these worms weren't frozen under ideal conditions exactly. Keeping human body frozen at low temperature for 100k years shouldn't be a problem. Possibly a lot longer. The problem is entirely with freezing and to lesser degree thawing. The onset of freeze has to be incredibly fast in order to prevent damage on cellular levels. It's no good having flesh perfectly preserved, if majority of cells are dead due to damage to cell walls and various internal systems caused by ice formation. Neurons are particularly delicate, as usual, so you have to ensure freezing conditions that let almost all of the neurons survive. We know it can be done on small samples, but you have to chill the entire body rapidly. And this isn't just difficult. There is no known way to extract heat from the depths of the body fast enough to achieve it. Even dropping the body into liquid helium isn't fast enough, as heat conduction from inner body to the exterior is too slow. You'd perfectly preserve the skin and some muscle, but kill the brain. And that's not good. Thawing is the flip side of this. Fortunately, there are methods for delivering heat deeper into the body, which can allow restoring the circulation and warming the body up rapidly enough. The only time scale you are playing against here is getting oxygen delivered to tissues once they thaw. That means restoring circulation and normal body temperature in about 20-30 seconds. Still not trivial, but I can at least think of ways we'd do it with modern technology.

If we're talking really long timeframes, even sublimation and diffusions are factors. You'll be loosing matter from surface due to random vibrations in the lattice. Likewise, various critical impurities will be traveling through any circuitry rendering it inert. If you let the ship cool to temperature of the background, which is a few Kelvin, these processes might actually be too slow to cause problems even across billions of years. But then you're not bringing that ship back to life without outside help.

A submarine already can use its fins to go up and down without making alterations to its ballast. And there are recreational minisubs that rely just on that. But it's not a great option for a military submarine, as these things need capability of turning pretty much everything off and staying put. Having to keep moving to stay underwater would be a great disadvantage.

Wasn't superglue vapor and sand one of proposed ways of rapid construction on Mars? If it works on Mars, it might work in total vacuum to keep a rock pile asteroid together. You'll need a lot of superglue, though.

That really raises an important question. What is the G-tolerance of a common fly? I would imagine it to be pretty high. Low pressure in the wake won't be a problem, though. Flies are pretty resilient to near vacuum. Video behind spoiler tag.

Earthquakes do have a tendency to cluster. Usually, a stronger quake will have both preceding and following weaker quakes. The ground is settling, and it's rarely a singular event. Whether it means it's going to get worse or better, that's the bit that's impossible to tell. Sometimes you have a pretty powerful quake, and it turns out that it's just one of the "smaller ones" preceding a really big one. But these really big quakes are rare to begin with. I live in part of California that probably won't get thoroughly destroyed by a quake, but will definitely be left without power and water when The Big One hits. It's something you make preparations for, because eventually it will happen, and you will have to deal with it. In my case, dealing with it will probably involve grabbing provisions, getting onto a motorcycle, and going to Nevada.