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About GoSlash27

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    Sliderule Jockey

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  1. K^2, No, it sounds completely wrong Your total energy at the bottom of the well prior to the burn is zero (or very close to nil). You had 5GJ of kinetic energy and -5GJ of potential energy, which cancel out. You then added .5MJ of kinetic energy by increasing your velocity 1,000 m/sec, which you keep as you leave. The kinetic energy is transferred back into potential energy to pay off the debt you incurred by falling into the hole, leaving your 1,000 m/sec DV. Best, -Slashy Lislias, Again you are confusing the mathematics of Oberth with the mechanics of gr
  2. Lisias, Not at all. K^2's example was given in the ship's frame. The example neutron star had no defined motion, thus no kinetic energy to steal. I'm afraid you're confusing Oberth with gravity assist, which are 2 different subjects. Best, -Slashy
  3. Lisias, This is correct when the body is in motion. You're stealing kinetic energy from the moon. However, this doesn't apply to K^2's example. Best, -Slashy
  4. K^2, I think this is where you went off the rails: This is not correct. The conservation of energy demands that the total energy remains the same in a closed system. You converted chemical energy into kinetic energy, adding 500 kJ. That's all you did. You should've algebraically added the kinetic and potential, which in this case is subtraction rather than addition. Whatever potential energy you have converted to kinetic energy on the way in will be converted back into potential energy on the way back out. Your total energy at the end of this maneuver isn't 100.5MJ, it's 0.5 M
  5. I get where you're coming from. I used to think the same way. But (yet again) that's not how it actually works. The goal isn't to inject energy as cheaply as possible, it's to get the vehicle to the destination as cheaply as possible. If you model the situation you're talking about, you'll see the pitfall in that line of reasoning. Best, -Slashy
  6. HMV, No, at least not in our solar system. There's only a couple targets that have a gate orbit anywhere near the moon, and you don't need to get close to the lunar surface to hit it. Farther targets are more efficiently reached by a LEO burn and simple slingshot. K^2, Again, that's not how it works. You don't actually pick up "extra" DV, even if you do your burn scraping an event horizon. The depth of the gravity well itself doesn't help or hurt you, it merely determines how far you can bend your trajectory at a high inbound velocity. This is the illusion of the Oberth
  7. HMV, "Yabut" That's what I'm trying to get across... Oberth isn't what most people think it is. It's not like "free parking", where you automatically get bonus DV for going faster. It's simply math behaving correctly, but in a counter-intuitive way. For every Vinf, there is an ideal R for the burn. Above it, and you waste kinetic energy. Below it, you waste potential energy. Going too low is *much* worse than going too high. The idea of "the lower and faster, the better" is a gross over- simplification, and leads to a false conclusion. Best, -Slashy
  8. HMV, see below... K^2, Actually, it doesn't work that way. Any velocity you pick up from the Mun's gravity well is lost escaping it. If you come in (say) at 1 km/sec with respect to the moon, you will leave with that same 1 km/sec with respect to the moon.... even if it *temporarily* accelerates you to relativistic speeds. Think of it as coasting down a hill and up the other side, except with no drag. It doesn't give you anything for free. You leave with the same speed you had when you came in, no matter how fast you're going at the bottom. Whatever you gain on the way in, you lose o
  9. HMV, It's always cheaper to burn direct from LEO unless you can refuel in lunar orbit using reaction mass collected from the moon itself. Since the moon is a black hole, that's not an option. AFA the Oberth advantage, it's offset by the gravity well. Every destination has a gate orbit; a radius where the Oberth and gravity well balance at a DV minimum. The closer destinations have a gate orbit above the current moon's surface. The more distant ones would have a gate orbit below the moon's surface, and thus would save some DV. But again, you can't acquire fuel from a black hole. TL/
  10. Yes, that's the advantage. It's not all that much of an advantage though, since we can get nearly any direction we want now. My point is that the most DV you can gain from an encounter is 2Vorb, regardless of whether it's a moon or a black hole. Best, -Slashy
  11. I don't think there'd be much advantage there, since you leave with the same relative velocity you came with. Gravity assists harness the body's orbital velocity, which wouldn't change. Best, -Slashy
  12. About the size of a gnat's head, for perspective. A gnat's head orbiting Earth with the moon's mass at the moon's SMA. But in reality, it wouldn't actually *be* that size. The gnat's head would be just the event horizon. The actual moon would continue to compress beyond that point into a singularity; beyond our ability to image, even with an electron microscope. Best, -Slashy
  13. Ah. So we've got some time, then
  14. Nothing, really. It might confuse the hell out of some insects (sudden loss of their navigation beacon) and cause the extinction of nocturnal hunters... Otherwise, the difference in gravity to other planets and Earth is zero. But in the long run, the moon would lose its mass through Hawking radiation. This would cause the extinction of life on Earth, since our core would cool, the magnetosphere would die out, and we would no longer have a tidal force to keep our rotational axis stable. Best, -Slashy
  15. My second stages are usually around that t/w. First stage is 1.4 t/w and 1700 m/sec DV. It burns out at 45° pitch, 28km altitude, and 750 m/sec velocity. Second stage is 0.7 t/w, 1600 m/sec DV. It burns out right about LKO. Best, -Slashy
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