K^2

Coriolis effect is proportional to velocity. It does not affect hover or any sufficiently slow motion on sufficiently large station. In fact, station has to be large enough so that the Coriolis effect on the person is pretty much negligible. This is why you can't build a station with artificial gravity 10m across. Basically, any station large enough for you to be landing a shuttle on, you won't be needing to worry about Coriolis effect. Seriously, it's not hard to approach a spinning station and hover to a landing. I have no idea what you are basing your conjectures on. It's not physics or experience. Physics says it's going to be almost identical to hovering in Earth's gravity, and experience with simulations confirms that.

We only chain tanks in KSP because of the way that the fuel logic works. With a real rocket, you are much better off having a manifold sitting atop the engine's main pump, and running a fuel line from manifold to each tank. Then simply switch which of these lines are feeding the manifold, and drop tanks as they are expended. There is no need to move fuel/oxidizer from tank to tank.

Yes, typos. Sorry about that. Yes, ∂U/∂t is zero at apsides in a frame co-moving with the rocket at either of these points. But then the rocket keeps moving along the elliptic path, and your coordinate system keeps moving relative to barycenter along the straight line, and you end up with energy of the ship changing. So by the time it travels from periapsis to apoapsis, how much energy was lost or gained? Just because you gained most energy at periapsis, if that energy was lost or gained, and that loss/gain was trajectory-dependent, as it is, then you don't know whether you ended up with optimal apoapsis. That's your entire problem. You maximize rocket's energy gain in the rocket's rest frame, but that does not result in optimal apoapsis. How do I know that? Because I know that rocket's energy is conserved in barycenter frame, I do know that you gain most energy when exhaust velocity matches rocket's velocity, and I do know that most energy gain results in highest apoapsis. You end up with a different result, and I tell you why. It's a wrong frame of reference for the problem for all of the reasons listed above.

You do have to have a pump on each tank. In order to drop a tank, you need to have a valve you can shut off. That puts some limitations on feasible diameter. At the same time, you have to have a huge mass flow to the engine. That means you have to take the fuel/oxidizer and accelerate them to very high speed before they even pass through the tank separation valve. There is no way to do that without a turbopump.

Ah. In that case, no dice. Anything in atmo that can serve as an efficient reducing agent would already have reduced something. The only known exception is Earth, and the only reason we have free oxygen in atmosphere is because of plants.

Something like Sabre, wouldn't do you a whole lot of good on, say, Mars. A turboprop with really large blades, kind of along the lines of V-22, should be able to fly very efficiently there, however. I think that's the sort of thing that OP is talking about.

Not really. I mean, at the face value, yes. But if you match angular velocity of the station as you pass underneath, which is really easy to do, then all you have to do to keep constant position relative to station is thrust towards the center of the station. Think about it from inside the station. You have a ship already sitting on the floor. You get in and tun on the lifting engines. They lift you off the floor. Did you suddenly start having control issues? No. From your perspective, you are just affected by gravity. The ship is already moving with the station, and keeping its orientation with respect to station. All you have to do is manage thrust to get hover. It's no harder than flying a VTOL jet on Earth. Ok, that takes training too, but presumably, we are talking about a trained pilot. Same thing as you approach. From the Station's frame of reference, the approaching ship is moving "up", but in effective free-fall. So all the ship has to do is start hovering when at apex, and it's going to maintain position with respect to station. Like I said, it's very easy to test out in Space Engineers. If you like KSP, you'd probably enjoy SE as well, so consider that a recommendation. In principle, you should be able to do this in KSP as well, but such a station seems like it'd be a major pain to launch.

There should be recoil towards the station due to tension in the ropes. But it's way too much in the movie.

Straight up jet propulsion would be pointless. A turbojet minus the atmospheric oxygen is just a rocket. A high bypass turbofan, however, can be way more fuel-efficient. If you want to build something like that in an atmosphere that does not have oxidizer, you would build your turbine to run completely on the on-board fuel and oxidizer, but you'd still have that drive a compressor fan that run on exterior atmo. Or, you can run a turboprop. Either way, it would be way more efficient at slow speeds. As for being able to build a turbine that runs on an oxidizer you brought with you, yeah, it's definitely doable. Turbopumps on most conventional rockets already work almost like that. You'd just have the turbine run a compressor or a gearbox to the propeller.

You can speed up to match station's velocity and then use your thrusters to "hover" in. You can try it out in Space Engineers. It's a bit tricky, but totally works. But yeah, if you are not moving with respect to station's center, the whole thing will be moving past you at quite some speed, regardless of whether you are in or out.

No, it is not. Change in both of these energies is negative. You are still mixing up signs. What you should have picked up from the link you posted is that work done against the cart is negative. So the fact that cart's energy goes down to zero makes sense. But the question of where that energy went remains. Precisely. But this only works out after you consider actual pair of action-reaction forces. If you approach the problem naively, and just say, "Well, the cart had kinetic energy mv²/2 and then it got mgh from going down hill, it's now traveling at root(2)v by conservation of energy." This is wrong way to do the problem. But this is only a wrong way to do this problem when the potential is time dependent. Lets say that the hill has a constant incline angle -θ. Then we can treat this as a 1D problem. The potential, in this case, is U(x) = mgh - mgx Sin(θ). Here, h is just a parameter, way to set U(0), rather than elevation. We can either say that at the bottom of the hill, where U(x) = 0, the potential energy has converted to kinetic and the velocity is now Sqrt(2gh), or we can compute acceleration. F = -∂U(x)/∂x = mg Sin(θ). Acceleration is then g Sin(θ). Distance the cart has traveled is d such that U(d) = 0 -> d Sin(θ) = h. And we get final velocity v = Sqrt(2ad) = Sqrt(2gh). Now, we go into coordinate system moving to the right at velocity u. The potential is now time-dependent. U(x,t) = U(x-ut) = mgh - mg(x - ut) Sin(θ). The cool thing about this is that the force is the partial derivative with respect to coordinate, so in this trivial case, time-dependence won't matter. F = mg Sin(θ). So the cart is still going to experience acceleration g Sin(θ), and starting with velocity -v, so long as v = u, it will come to rest at the bottom of the hill, which is what we are looking for. However, if you try to use energy conservation here without taking into account recoil to the ramp, you are going to make a mistake. So with all of this in mind, we are going back to the case of orbital motion. You considered motion of the ship subject to potential -μ/r. Here, the gravitational parameter μ takes reduced mass into account, and r is distance from barycenter. Lets do this in 2D. Potential energy can then be written U(x, y) = - mμ/Sqrt(x² + y²). If your ship is at location x, y traveling at velocity v, it has total energy mv²/2 + U(x, y). Lets figure out how its energy is changing at that instance. dE/dt = m/2 d(vx² + vy²)dt + dU(x, y)/dt I re-wrote v² in terms of components so that I can chain-rule it. 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 So this is a very nice result. Energy of the ship is conserved. We don't need to consider the whole system. We can forget that the planet/star is there at all, and just work from conservation of energy of the ship. Now lets consider a moving coordinate system. It will be co-moving with the ship, but let me name that velocity u, as before, so as not to confuse it with ship's velocity that goes into kinetic energy. So again U(x, y, t) = U(x - uxt, y - uyt). And again, we are interested in energy of the ship and how it changes with time. dE/dt = m/2 d(vx² + vy²)dt + dU(x, y, t)/dt The first term is zero, because v = 0. Ship is at rest in this frame. So we only need to see if dU(x, y, t)/dt is zero or non-zero. dU(x, y, t)/dt = ∂U/∂x dx/dt + ∂U/∂y dy/dt + ∂U/∂t Here, dx/dt = vx = 0 = vy = dy/dt. All that's left is verifying that ∂U/∂t ≠0, which I'm sure you can handle. So in coordinate system where the ship is at rest, dE/dt of the ship is non-zero. Energy of the ship is not conserved. Yes, energy of the ship-planet/star system is still conserved, but the lot of good that does you. Yes, you've optimized the rate at which ship's energy is increased. But that is not a conserved quantity, so whether or not you achieved some sort of optimal in transfer, you don't know. You have to solve for the entire trajectory and see what's going on. In contrast, if you start with coordinate system in which barycenter is not moving, you know that if you maximized E that the ship gets from the burn, you've maximized ship's energy throughout the trajectory. Because it is conserved. Hopefully, it's now clear why choice of the coordinate system matters, and why there is a "right" and "wrong" choice here. But let me know if you have any more questions.

(-v)² = v²

It's an article in a popular publication. These are almost always worse than useless.

I know that these are the 2nd and 3rd brightest objects in the sky, but I still find it pretty amazing that we've gotten to a point where you can just snap that picture with a phone camera. As I recall, you had to go to some trouble to do this with film.

Her foot wasn't caught in it. You might be remembering wrong. The line was looped around her foot, but it was sliding pretty much freely.

I figured there might be a case of bi-stability, but I wasn't sure, so I didn't want to say anything. The problem, though, is that even if you eliminate that point-first stability, you can still have a dynamic instability. Id est, if you start with a bit of a tumble, it's likely that it would grow only worse. Offsetting CM wouldn't really change that. But I guess you can try to eliminate or reduce this effect by careful choice of the heat shield curvature and the aspect ratio of the capsule. Is the other reason lift, by the way? Seems like a capsule with an offset CM is going to glide a little better, which should allow for a slightly gentler deceleration profile.

No, that last bit is actually important. It shows that such a wormhole follows the no-communication theorem of the entanglement in Quantum Mechanics. I mean, yeah, it's obvious, but it's an important feature, so it's worth mentioning.

Here is a problem for you to consider. A cart is parked on a hill of some height h and has potential energy mgh. It then rolls off the hill onto a flat surface, and rolls with velocity v. So its energy is now mv²/2 = mgh. Good? Now picture this as you ride by in a train. The train just happens to travel at the same velocity v. You initially observe the cart sitting at the top of the hill having energy mgh + mv²/2. Then the cart rolls off the hill, matches velocity of the train, and now has energy 0. Where did the energy go? Yes, energy of the system is conserved in a closed system an inertial coordinate system. But we aren't dealing with a system. Did you take motion of the planet/star into account in your problem? No. You threw that away completely and substituted an effective central potential. Your sub-system is now your ship and the gravitational potential. And if that potential is not moving in your chosen coordinate system, that's fine. You still have a conservation law, and you can say that ship's energy is conserved. If that potential is moving relative to your chosen coordinate system, you do not have energy conservation in your chosen sub-system. Coordinate system that's moving relative to barycenter of a 2-body problem has energy exchange between the ship and parent body. If you consider the whole system, that's fine. If you just talk about the ship, and maximizing energy of the ship, ship's energy is not conserved. As for why you optimize energy, try to prove that Hohmann Transfer is an optimal transfer between two co-planar circular orbits without making use of energy and angular momentum conservation laws. P.S. Just as a point of interest, energy conservation follows from time-translation invariance of the Lagrangian and Noether's Theorem. In Classical Mechanics, this requires time-independent interactions. And if you follow this rabbit hole as far as it goes, you'll find that in general, a time-dependent interaction can only arise from something that's equivalent to an accelerated frame of reference. But we're talking advanced field theory by this point.

Hold it against what? She wasn't attached to the rope in any way. It was freely sliding against her foot. This will provide a tiny amount of friction as you are moving away, slowing you down, but if there is a consistent force pulling you the other way, what is that going to do? The only reason she didn't slide away as well is a bit of tension built up in the rope due to the fact that they were both sliding away from the station, which allowed for a bit to recoil. And since her spacesuit was much, much lighter than Kowalski's, that was enough. And like I said, this is an absolute minimum of the force generated. lajoswinkler goes into more detail based on footage, which raises the lower bound dramatically.

That's a factor, but it's not quite that simple. If you naively put center of drag at the center of the heat shield, then no matter how low center of mass is, center of drag is always in front of it, so the system will be unstable. What really happens is that center of drag shifts around as the capsule tilts, providing stability. Reasons why it works are way beyond the scope of this topic. But a simple analogy is the way a boat floats. If you try to tilt the boat to the right, center of buoyancy also shifts to the right, providing stability. Up to a point, of course. If you rock the boat too much, it will capsize. Same thing can happen to re-entering capsule.

You should be seeking to optimize orbital energy in a frame of reference in which energy is actually conserved. That's the coordinate system in which barycenter is stationary. If you got optimal exhaust velocity at zero, then you probably chose rocket as your frame of reference. A frame of reference which moves relative to barycenter does not conserve energy. So it's useless optimizing energy in that coordinate system.

As usual, horrible reporting. What scientists found is another gravity-QM duality. There are actually a few known ones already. And it's not really surprising, since most of the QM is built on the same symmetry principles as General Relativity, so it's likely that you'll find systems in both that behave in dual ways.

It doesn't take "severe" spinning. It takes very little, in fact. Length of the rope is about 15m. (Estimated from picture on previous page.) They weren't moving too fast before getting caught in the ropes, but fast enough to make it difficult to grab things. Lets go with a low-end estimate of 1m/s. Now, how close did they pass to the point where ropes are attached? Lets be generous and say no more than 1m. The specific angular momentum, then, is 1m²/s. Like I said earlier, ropes would not kill angular momentum. In that case, at 15m from origin, they would retain about 1/15th or about 0.06(6)m/s of transverse velocity. That's not much. In fact, only about 0.05 RPM. This is just a little more than two and a half times faster than a minute hand on the clock. Lets do forces, shall we? Clooney's character is wearing an EMU. That's about 125kg. Lets go on the skinny side for the astronaut for a nice, round total of 200kg. At the above transverse velocity, this generates 0.06N of force. This is about a weight of a marble here on Earth. Really not a lot. But given that there are absolutely not other forces save for friction in the practically free-floating ropes? It is going to be enough to produce something very similar to what we see in the movie. And keep in mind, this is the absolutely lowest possible estimate.

Because you minimized energy for given amount of momentum. What you are neglecting is the fact that this also increases amount of propellant used.

A tether cancels only radial motion. Any angular momentum they'd have flying by the station, they'd still have. And it would take a very small amount of angular motion to generate enough centrifugal force for there to be a problem. They didn't show it all that great, but that's what would have happened, yes.