sevenperforce

A reasonably well-designed solid-core NTR is absolutely less radioactive (in terms of its exhaust) than a coal plant. No question. Orders-of-magnitude difference.

But it read like "Mary Had A Little Lamb" in comparison to just about any end-of-season Who episode.

Meltdowns are not as easy to achieve as people often think. There are numerous ways to scram a reactor.

Arianespace has a better launch inclination for GEO sats.

Well, yes, but I was speaking in comparison to the SL-expanded engine. Sorry for not making that clear. Nozzle area ratio is a function of pressure drop ratio. Given an engine that comes in a SL-expanded variant and a vacuum-expanded variant, the difference in nozzle area ratios (and, correspondingly, the sizes of the nozzles) is going to decrease as the pressure increases. An engine with relatively low chamber pressure has a moderate pressure drop at SL and a significantly greater pressure drop in vacuum, so its vacuum variant is going to need to be much much larger than its SL variant in order to take advantage. An engine with a much higher chamber pressure has less of a pressure drop between SL and vacuum, and so the nozzles will be closer in size. All that to say: if the BE-3 was a low-chamber-pressure engine, then we would expect to see a very large vacuum nozzle in comparison to the SL nozzle. But since the BE-3 is deeply throttleable, which means it is high-chamber-pressure, then the vacuum nozzle will not be quite so gigantic, since most of the expansion has already happened in a SL-sized nozzle. Since when is Vulcan not going to have a Centaur upper stage? ULA needs Vulcan to remain competitive at all.

At least there will be a gorram LES this time.

Depends more than anything else on the chamber pressure of a SL BE-3. BE-3 is the first combustion-tapoff turbopump engine to ever fly, so there is no good way to estimate its chamber pressure. But given its high SL throttleability, I'm guessing they have a pretty high chamber pressure, which means a somewhat smaller engine bell than a lower-chamber-pressure solution.

And done:

CRS-14 Dragon 1 Rendezvous and Docking When we last left off, Dragon had managed to make it into orbit and was setting up for a rendezvous with the ISS.

Bigger Falcon Rocket: "Oh, I didn't see you there."

Photons have momentum, so momentum is conserved, and the energy-momentum-4-vector of the photon takes the place of reaction mass in the rocket equation. This is one of the reasons why I just laugh whenever anyone claims to have a reactionless drive. You can't have one. It would break all of relativity. Mass would no longer be coupled to gravity. All the stars would explode. "Do you see? Do you see what you do? Trying to build a reactionless drive and you destroyed the universe. Well, I hope you're happy."

A photon has energy and momentum, which couple to gravity. Particles with rest mass (or bound relativistic mass, which functions as rest mass for all intents and purposes) couple directly to gravity, but massless particles couple to gravity through their energy-momentum-4-vector. Interestingly enough, you could make a "dark Kugelblitz" by firing a bunch of massive particles at the same point at relativistic speeds from a bunch of different directions, and thus produce a black hole using relativistic kinetic mass-energy. But in that instance, the additional mass-energy would be the rest of timelight interactions between the particles in SR, which would produce bound states (transient or no) which will couple to the GR tensor.

Well, thanks to conservation of momentum, that kinetic energy has to have come from somewhere, so the effect on the gravitational tensor is conserved by virtue of whatever reaction mass was acted upon to produce the dV. The relativistic mass gained by an object with an increase in kinetic energy must be defined with respect to an outside reference frame, and SR spacetime curvature effects will be present for an observer in that outside reference frame even though there is no actual change to the GR tensor.

Rotational energy is non-inertial, so yes. Most mass is, in fact, relativistic kinetic mass-energy, rather than rest mass. But it is trapped inside chemical and nuclear bonds so it affects the tensor. Rotating black holes appear smaller if you are orbiting them prograde than if you are orbiting them retrograde. For a rotating black hole, there exists orbital radii for which the orbital velocity is lower than c when orbiting prograde but greater than c when orbiting retrograde.

Right. The answer is no; it does not. As long as you are in an inertial reference frame, your kinetic energy does not curve spacetime. If you are in a non-inertial reference frame, all bets are off. A rotating object, for example, has centripetal acceleration and thus produces frame-dragging effects.

It's impossible even in theory. Simply put: there is no universal reference frame, so spacetime has no idea how fast you're going, so it can't make a black hole around you.

Good question, but no. You can't accelerate yourself into a black hole because the fabric of space has no universal coordinate system. Space does not notice how fast you are going; your gravitational interactions with space are the same no matter how much you accelerate. If you think about it, it HAS to be this way, because all motion is relative. Your kinetic energy has to be measured with respect to some outside object; you can be moving at 200 km/s relative to Earth but only 2 km/s relative to some other object way out in space. If relativistic kinetic energy caused the formation of an event horizon, then you could be a black hole w/r/t Earth but not w/r/t another object, and that's obviously very problematic.

Dragon capture successful! Now to fly this part of the mission...

That was just the number of restarts in the mission.

Love the way you seamlessly integrated the radial intake into the engine modules.

A Hohmann transfer is a transfer between two orbits, not between two objects.

Well, it IS the transfer stage. No. Deep Space 1's single ion thruster ran for 10,000 hours but only restarted 34 times.

An SSME's thrust grew by 23% from SL to vacuum, and you're going to see a much bigger jump with a vacuum-optimized bell. 600 kN should be no problem.

Ions give much lower thrust IRL than in KSP. You'd be looking at hundreds or even thousands of periapsis kicks.

Not enough thrust. High-thrust chemical engines can get from LEO to LLO with a little over 4 km/s, but an ion-based low-thrust system must spiral out slowly and much less efficiently. Costs about twice as much dV. What they could do, I suppose, is use Block 2 to send the whole station into a lunar slingshot at just the right trajectory that the moon's gravity raises the perigee very high; that would enable a low-thrust solution to insert into a NRHO relatively fast.