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Many people with hypertonia are susceptible to meteorological changes. Perhaps (as stated above) the absence of a limb amplifies such problems?

Negative below, I'm sure you meant. And, of course, someone stated earlier in the thread, that it could be bedrock, found only on the bottom four layers of the minecraft map, that has a stupendous density (explaining, why it's completely unbreakable) that would handwave away the dubious cause of uniform gravity when burying deep into the ground (also it would allow the rest of the ground to have a sane density).

Ha! Good one! So what-if.xkcd has done this using the european way of representing fuel consumption (liters per 100km, or a unit of area). That surface area is basically the answer to the question "if instead of burning fuel a car leaves a stream of consumed fuel behind it, what would the cross section of that stream be?". So I tried to consider, what significance does the inverse quantity have. Imagine, you have an infinitely long channel of fuel along the roadside with a cross-section of unit area. On the road you have several cars which have no fuel tanks, but instead have scoops picking up fuel from this channel. This inverse area is "how many cars will this channel be able to sustain". Crazy thought experiments

PS, uh yeah, double-checked the numbers. Minecraft is 1 quadrillion blocks (or square meters) in surface, or one billion square km. The numbers in the picture at the start of the thread are wrong (someone threw in too many zeros) That's on the same scale as the surface area of earth (perhaps, twice as much). PPS, realized, that Neptune isn't that much larger than Earth.

For all intents and purposes we can consider the minecraft world to be an infinite (in lateral dimensions) slab, as minecraft's world is vastly greater in lateral dimensions than in thickness. Let's say, it's about equal amounts of air and ground (128 blocks up is air, 128 blocks down is ground). In that case all we need is mass per unit of area. By analogy to Gauss' law for a charged plane we can easily calculate the gravitational field for a massive plane: g=2ÀÃÂsqγ, where ÃÂsq is mass of slab per unit area, γ is the gravitational constant. That would yield: On the other hand, mass of slab per unit area is average density of minecraft world * 128 meters of thickness (I mean, of course just the ground, not counting the air), therefore g=128m*2ÀÃÂγ, where àis the proper density of minecraft ground. g=2 * 3.14 * 128m * ÃÂγ. However, we do not know neither the density, nor the gravitational constant of minecraft. Although, as has been stated earlier, we know that the g in minecraft is 23 m/s2. Therefore the product ÃÂγ is a mere 0.0286s-2. On the other hand, in the real world àis of the order of thousands of kg per m3, while γ is 6.67*10-11m3 kg-1 s-2. So a first approximation gives us either stupefying densities or a world in which gravity is quite a strong a force as electrostatic or even nuclear interactions. Additionally, it is quite clear, that the gravitational force in minecraft does not change much (or, for that matter, noticeably) as you go deeper into the ground. That means, that the mass of dirt and rock above the bedrock is negligible, compared to the bedrock itself and whatever may hypothetically lie beneath it. This returns us to somewhat more realistic densities/gravitational constants. For instance, if we take a density of 4 tons per cubic meter and the real world gravitational constant, we would find the real thickness of the minecraft world (including a hypothetical magma layer beneath the bedrock) to be about 14 thousand km. Funny enough, that's on the same scale as earth's diameter. Going pack to the first post though... 32 million blocks in length? That's only 32 000 km. Imagine a square 32000 by 32000 km. You could lay out a box of roughly 3x3 earths on this square. So minecraft is 3 earths long, 3 earths wide and 1 earth thick. Y'know, quite consistent with the real world. <wrong>Nothing like the surface area of Neptune</wrong> One 8th the surface area of Neptune.

Ek=kQ2(1/r1-1/r2)=mpv2/2. v=(2kQ2/mp(1/r1-1/r2))1/2=2.75*104m/s That's if the absolutely fictitious idea of having a "fixed" proton holds.

Firstly, you let some typos creep into your calculations. It should be 7.2*10^21 J = 7.2*10^12 GJ for the KE of this rock. As far as the rest of the discussion goes, you haven't specified an inertial frame of reference to talk about 20 km/s. Without it there is no point in discussing the energy output of the Tsar Bomb and comparing it to the energy of the asteroid in an arbitrary reference frame. What also is needed, is a mechanism through which the energy of the bomb works "against" the kinetic energy of the asteroid. X-ray radiation for instance would hardly make a dent, if that's all of the energy output of the bomb - it would be along the lines of 10 um/s delta-v. PS Guys, I can see your point on superscripts... but symbols? That sweet little multiplication cross, for instance. I don't see a character table for commonly used math symbols here in the forum editor, do you really expect someone to dive into the built-in Windows Character Map every time he needs to post anything more than a smiley with "10char" attached? The tl;dr is, that WYSIWYG editors for math notation drive me into a state of rage (especially when they're not really suited to the task), but I promise to start using proper superscripts 'n' all as soon the forum starts parsing LaTeX input :-)

I prefer to believe, that some things (scaled down by factor 10 universe) are truly no more than game-balancing decisions, rather than having a specific storyline behind it.

Nah, that's just ludicrous. That fraction of particles would be on the order of e^(-1000000/6000), if you plug in the numbers, then the expected quantity of such energetic particles in the entire sun would be around 10^(-15). On the other hand, how do you define the temperature of the corona? The mean free path for a particle in a gas is around 1km (for a particle density of 10^15 per cubic meter, as it is in the corona). Accelerated by the sun's electromagnetic field, perhaps? A mean electric field of 0.1 V/m would be enough to impart this kind of energy to an elementary charge over these lengths. EDIT: The question about defining temperature is rhetoric, of course. I was hinting, that more conventional methods of heat transfer hardly would play a major role in this problem as was stated by the OP.

I may have missed something, but the paper in question discusses a crystal with spin-polarised He*-He* bonds, not He*-He. Something akin to metallic hydrogen.

Just under 1.43*10^26 Joules

Wow. Thanks. Makes for some nice bedtime reading.

Since the thread has already been long ago derailed... K^2, I was also surprised to hear, that we already have a "theory of everything" (unified field theory, or whatever you prefer to call it), so could you perhaps elaborate the fine points of not having a renormalizeable theory, but having an "Effective Theory" that is renormalizeable above the Planck scale? To avoid unnecessary embarrassment, I've got no background in cosmology, just a diploma in condensed matter physics :-)

I wonder, if this concept could be doable at higher elevations (e.g. http://en.wikipedia.org/wiki/Qinghai%E2%80%93Tibet_Railway - place a linac up there).

To quote Jostein Gaarder, "Life is like a lottery in which only the winning numbers are shown".

...and having done all that you should be well on the way to finishing your doctoral thesis in theoretical physics:sticktongue:

Sure! The difference is that for a VTOL aircraft the velocity that matters is relative to the atmosphere which is normally more or less matched with the landing pad (exception - strong, but intermittent or frequently changing winds), whereas in space the trajectory of a spacecraft going at the same speed as the docking port at a given moment may deviate from the trajectory of the docking port itself quite rapidly even if the space station is truly huge. In fact, while the rpms of a large station may be quite small and you'll be able not to worry about the docking port turning away by 90 or so degrees while you're struggling to get yourself in an "orbit" around the station, the actual velocity of the docking port relative to the rotational axis is comparably larger and that leads to a bigger difference in the orbit of the approaching spacecraft as compared to the orbit of the station.

Yeah, that's quite clear. Coriolis would be noticeable if you were to dock at the rotational axis and then move radially all the way to the outer edge of you station. If the docking bay is small relative to the diameter of your structure, Coriolis effects would also be negligible. The tougher part would be to match velocities with the docking port (which puts you on a very different trajectory compared to the station as a whole) and then as soon as you are aligned with the docking port you need to start accelerating towards the center of the station at 1g just to appear to be stationary in the co-rotating frame of reference.

Yeah, the comparison with VTOL craft is nice, but they do generally use air-breathing engines, right? 1000 m/s "spare" dV in space and on earth are two entirely different things. Naturally, a person moving from the center of rotation to the outer edge of our spinning construction would experience quite a sizeable Coriolis effect. Ah! My bad, this has already been mentioned. Now that I think of it, Dweller also wondered about the motion of air in such a construction. Now wouldn't that be very similar to the motion of free electrons in thermal equilibrium in a solid under a uniform magnetic field directed along the axis of rotation?

This may have been already considered or dismissed, but why not just dock to a structure at the axis of rotation of the station?

Ah, so here you finally state the answer to my recent question! We are interested in the energy increase at apoapsis (please correct me if I misunderstood). Indeed, I didn't ever consider what happens to the rocket after finite time, rather I was considering what happens immediately after a burn, but of course of course by the time the rocket gets from periapsis to apoapsis in a moving frame of reference the change in energy of the parent body will be non-negligible.

In principle, the notebook is already "molten". It's made up of organic polymers that are all tangled up with each other (i.e. it's not a crystalline solid). When you heat it up, rather than untangling the polymers, you'll sooner break the intra-molecular chemical bonds and your "melt" would actually be a vapor chemically very different from the initial notebook. This is very much unlike simpler compounds which when heated first break their crystal structure, but each molecule in a melt and further on in a vapor retains the original chemical composition. You can also heat up some thermoplastic plastics and see, that they don't really have a fixed melting point, they just gradually get softer as they warm up.

I'll answer briefly now and go into more detail later, when I'm not at work :-) dU(x, y)/dt = ∂U/∂x dx/dt + ∂U/∂y dy/dt = -Fx vx - Fx vy dE/dt = m (ax vx + ay vy) - Fx vx - Fx vy = 0 The red "x"s should of course be "y"s. And you are absolutely right (as I also mentioned in my previous post) that in the general case ∂U/∂t is non-zero. However, at periapsis or apoapsis the ship's motion is perpendicular to the gravitational force acting upon it and is in fact parallel to one of the lines of constant potential. At that moment ∂U/∂t will undoubtedly be zero. As the simplest description of the Oberth effect is "burn at maximum velocity" and maximum velocity is achieved at periapsis, I feel that this is the most important case to consider.

Thanks very much for this video. This actually clears a lot up. I guess, the long part with "let go - no, i won't - no you have to" kinda' blurred out the drama of this event.

@K^2 Nice problem, thanks. I didn't consider that. That does not, by the way, prove, that energy is not conserved, as pointed out in the link AeroEnergy posted. Now about the energy of the parent body (does it remain constant or not). The change of energy should be taken into account... but only when the rocket is not at apoapsis or periapsis. If we are in a frame of reference tied to the parent body, its change in velocity is negligible (m_rocket/(m_planet+m_rocket) << 1). If we work with differentials, then the change of energy would not only be proportional to this tiny factor, it would also be proportional to the square of dV and that can certainly be neglected. If we were to go into a frame of reference tied to the rocket, the change of energy of the parent body is no longer negligible in the general case. However, at apoapsis or periapsis the planet is moving perpendicular to the line connecting it to the rocket, thus the gravitational force from the rocket acting on the planet is not performing work (well it is, but it is proportional to dt squared and thus also negligible). Similarly, the change of the potential energy is also proportional to the second-order differential and is also negligible. EDIT The tl;dr of AeroEnergy's link is that when the cart rolls down the hill, there is also some recoil acting on the hill that accelerates it backwards. Of course if we say the hill is much heavier than the cart, the change in the kinetic energy of the hill is negligible. However, in the moving frame of reference that you suggested, the change of energy of the hill is no longer negligible and is actually equal to double mgh, exactly the energy that seems to have "vanished".