3 ms·

I see it this way. You have to accelerate the mass to the escape velocity. You also have to achieve a net effect of being in orbit (some distance x above surfac

• ars 12y ago
  > being in orbit (some distance x above surface of earth). The most direct vector to that distance is perpendicular. That's not what being in orbit is. In fact that's the opposite of being in orbit. To be in orbit you need to move parallel (tangent) to the surface of the earth, not perpendicular. The distance about the surface is entirely for air resistance, and has nothing to do with being in orbit.
• SamReidHughes 12y ago
  i'll spare everybody else the snark here
  • has2k1 12y ago
    Zero feelings involved, just analysis before numerical computation. The concepts need to make sense first.
    • SamReidHughes 12y ago
      Okay. In this scenario where you accelerate upward then turn 90 degrees and accelerate horizontally, are you talking about this being powered by a rocket or by a track?
    • has2k1 12y ago
      @SamReidHughes > ... accelerate upward then turn 90 degrees and accelerate horizontally I am not for such a scenario at all. The point of the upward (upward only and not considering the atmosphere) acceleration to the desired location is to give the lower bound of the energy requirements. This is the baseline (baseline-1) and the rocket equation is as simple as possible. In reality with an atmosphere and to put the object in orbit, the aerodynamics change and the energy requirement increases beyond the above baseline-1. If you "accelerate upward then turn 90 degrees and accelerate horizontally", you can calculate an energy requirement for that and it is easy. Only two vectors involved. That should give some limit (call it baseline-2). We should expect to do better than baseline-2, how better? A calculation using the diagonal of the vectors involved in baseline-2 should give us baseline-3. We shouldn't do better than baseline-3. Our launch designs and ingenuity should have an energy requirement between baseline-2 (this is bad, we are not thinking) and baseline-3 (this is maybe closer to ideal). The rockets and shuttles do "pitch-over manoeuvres" to turn the straight upward acceleration into an elliptical acceleration. * Note, I have not addressed the complications of the variations in atmospheric drag, but if it varies close to linearly along the vertical cross-sectional then how I think about it above does not change unless there is some other oversight.
      • SamReidHughes 12y ago
        I don't get what idea you're communicating here. You seem to be saying that the best option for rockets is to accelerate upwards, out of the atmosphere, gradually transitioning to accelerating horizontally (if this is what you mean by baseline-3). Well okay, that's what we do, when using rockets. Doing that with a track would be expensive because the track would have to be built hundreds of miles high over all of its length. It would be cheaper to build most of it lower, and maybe accept that we'll have to handle the air resistance somehow. If we build a track that doesn't go out of the atmosphere, we could still use it to build up a lot of speed and then turn the rocket upwards before the thing is self-powered. If we do build a track that goes out of the atmosphere, we'd still want to get as much ground-level acceleration as we can. Maybe it's more practical to build the track on the Moon, where there's no atmosphere.
      • has2k1 12y ago
        The scenario for baseline-3 is just a conceptual tool(pythogras theorem) to establish a bound. It still doesn't get you into orbit. What I was grappling with, is. I presumed @ars (parent) knew he was talking and in expressing my contention it would get addressed with a little bit more information in what I was missing. I now see some emphasis in his explanation and additional links. The key is was that tangential acceleration opens up none fuel based acceleration mechanisms i.e change to the type of energy and the quantity (you accelerate less fuel to burn up the fuel).
• lmm 12y ago
  It's escape velocity, not escape speed - and it's orbital velocity that matters here, we're not escaping entirely. If you're in the same place, travelling at the same speed, but pointed down, you're not in orbit, right? It's the same if you're pointed straight up. You need to be at an orbital altitude and travelling at the right speed in the right direction. (well, any combination of speed, direction, and position will be an 'orbit' in some sense, in that there's a conic section you're on that you would follow if you were in freefall. But if you're sitting still over a planet then it's the degenerate ellipse that's a straight line down to the planet's core). So in the absence of atmosphere the ideal ascent trajectory would look pretty much like a Hohmann transfer orbit: you'd accelerate horizontally until you were in orbit at surface level, and then you'd do the minimum energy transfer from that orbit to a higher orbit. In reality it's worth getting to altitude where the air is thinner before turning horizontal, but even so, the vast majority of a rocket's acceleration is horizontal, not vertical. You can do the maths, but if you want to really understand these things, play Kerbal Space Program. Seriously.
  • baq 12y ago
    escape speed is technically correct. you need to go fast enough, direction doesn't really matter.
    • ubernostrum 12y ago
      Some directions may require more speed than others. Or perhaps you are having a bad problem and will not go to space today.