5 ms·

Given that earth is a ball, aren't you always pointed heavenward if you are between 0-180 degrees. So you might as well start of by pointing directly heavenward

• ars 12y ago
  You don't want to turn. To orbit you need to have velocity tangent to the earth (i.e. along the ground), not directly up. You only turn ever so slightly heavenward, mainly to avoid air resistance.
  • has2k1 12y ago
    I'm still lost. Isn't being tangential along the equator less advantageous than being perpendicular at the poles. When you start tangential, as soon as you cover a distance equivalent to the radius of the earth, you are then perpendicular (maybe not directly so but there is no practical difference as far as the gravity well is concerned).
    • SamReidHughes 12y ago
      When going tangentially, you'll maintain tangentiality until you're going so fast that gravity isn't pulling you down fast enough to keep you pressed against the track you're on (because the Earth's surface curves downwards like the widdle spheroid it is). At that point, you're in low earth orbit (though you might want to fire the rockets a little bit to give yourself an orbit that doesn't intersect the planet). The whole point is that you gain your velocity while you're pushing off the earth, instead of your own propellant. That way, you don't have to carry so much propellant. > When you start tangential, as soon as you cover a distance equivalent to the radius of the earth, you are then perpendicular (maybe not directly so but there is no practical difference as far as the gravity well is concerned). If you were assuming the tangential take-off would continue going a straight line instead of curving in an orbit, I still don't get what you mean -- after covering one Earth radius, you'd be traveling at a 45-degree inclination relative to the surface of the planet, not perpendicularly.
      • has2k1 12y ago
        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 surface of earth). The most direct vector to that distance is perpendicular. The two combined should give you the minimal energy requirement. Any engineering (and aerodynamics) creativity cannot give you anything better.
        • 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
    > You only turn ever so slightly heavenward, mainly to avoid air resistance. And then you turn Earthward, to avoid crashing into the Earth.
• claudius 12y ago
  You are always pointed "up", but as long as you don’t accelerate too much, earth’s gravity keeps you on a circular trajectory. This is especially true if you fix the vehicle to the ground using e.g. rails. It would then be conceivable to accelerate to a high speed on ground-level rails (which is considered “free” as you can get energy delivered through the rails) until your speed is such that gravity alone does not hold you down any more and then unlock from the rails and fly off. I did not do the maths.
  • has2k1 12y ago
    You would have to build up a lot more speed if all you are going to do is decelerate once you "take-off". Also noteworthy, once you get to the escape velocity you would require an additional force just to keep you circular.
    • avoid3d 12y ago
      Staitght up is also wrong if you want a circular orbit.
    • claudius 12y ago
      You can still have a rocket on your vehicle to accelerate further after take-off, but you certainly need to accelerate less if you already at a certain velocity than if you are stationary. Since the “certain velocity” is free by assumption (i.e. supplied by ground-based electricty generators or something similar), I don’t see how you could be off worse. But I still didn’t do the maths fully.
      • nitrogen 12y ago
        Air friction, perhaps? There is more air in a tangent line.
• eterm 12y ago
  You don't get to orbit by going up far enough, if you did you'd just end up falling back to earth. You get to orbit by going fast enough, so that acceleration due to gravity acts perpendicular to your velocity and so acts to change your direction and pulls you around the earth.
• phreeza 12y ago
  What really helped me understand why you do not want to go up is the following thought experiment: Imagine you are standing on Mt Everest (and the atmosphere has magically disappeared), and you would like to shoot a bullet once all the way around the earth, how would you have to fire it? It seems pretty intuitive that shooting it up into the air won't do the trick. If you want to shoot around the earth, you need to fire in parallel to the earths surface, with a sufficiently high muzzle velocity. And really an orbit is nothing else than a shot around the earth.
  • ericd 12y ago
    Yep, you just need the ground to fall away from the bullet as quickly as the bullet is falling towards the ground.
• ajuc 12y ago
  Turning requires acceleration, but you can convert velocity direction with reasonable efficiency by pushing on something (like rails or magnetic field in a tube or just road).
• peterfirefly 12y ago
  https://what-if.xkcd.com/58/ https://what-if.xkcd.com/58/
• wlievens 12y ago
  If you go straight up, you can never attain orbit.