Comment on Little Pea Shooters

• absGeekNZ@lemmy.nz ⁨3⁩ ⁨days⁩ ago

  It is difficult to conceptualise.

  But you also have to choose the most convenient observer to help you get it.

  I would say the easiest way to “get it” would be to consider it from the Suns observation point of view. Choosing the planet or spacecraft just means that you have to consider a lot more relative motion.

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  • Situation A: Your space craft is catching up to the planet;
  • Situation B: You have gained “29” speed as you fall toward it.
  • Situation C: You spend 20 speed climbing back out of the gravity well; for a total speed gain of 9.

  Image

  • Situation D: Same setup, you catch the planet quicker because it is now traveling toward you.
  • Situation E: You have only gained “11” speed rather than the 29 when the planet was moving away from you.
  • Situation F: You still spend 20 speed to climb out of the gravity well; for a total speed loss of 9.

  This is obviously simplified and the numbers are meaningless. But the concept stands.

  Depending of the incoming and outgoing angles; the energy changes are more or less…

  Hope this illustrates it a little better.

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  • kuberoot@discuss.tchncs.de ⁨3⁩ ⁨days⁩ ago

    Well, relative motions are more intuitive to me - they make sense, and I can use calculations for them.

    In the first example, you presented 101 speed - this means only 1 speed relative to the planet, and that’s all that’s getting redirected (in the planet’s frame of reference your enter and exit velocity should be the same, since that’s how orbits work). The number is just too small, but your velocity would be planet velocity + 1 on a different vector, which will be less than 101 total.

    If we estimate the angle on the picture is about 50 degrees from the velocity vector, and the speed to be 100+v1, the speed from the planet’s frame of reference is v1 - so, the exit velocity will have components of (100+v1cos(60°)) and (v1sin(60°)), so the final speed relative to the sun should be

    sqrt((100+xcos(60°))^2+(xsin(60°))^2)

    Wolfram alpha suggests this simplifies to sqrt(x(x+100)+10000), and comparing the equation by appending <x+100 gives the solution of x>0

    This means, if my math is correct, with an entry angle of 0° and exit angle of 60°, you always lose speed.

    I could try replacing the angle with a variable and setting a constraint of x>0 and see if the free version of wolfram alpha would spit out something, but just replacing the 60 with y is spitting out some convincing solutions, since in those x is never greater than 0.

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    • absGeekNZ@lemmy.nz ⁨2⁩ ⁨days⁩ ago

      Taking the planet as the reference point. Complicates the situation a lot, but here we go.

      If you contrast “A” and “D”. The initial velocity in “A” is 1, whereas “D” is 201. The acceleration due to gravity in “A” SEEMS LOWER (this is why external observer is way easier) on the way in and in “D” it seems higher. In “A” you are literally falling for much longer (gaining much more speed); than in “D”.

      In “C” and “F” the situations are also different, I over simplified a bit too much. In “C” you would spend more energy than in “F”; since the acceleration due to gravity would seem higher, but not that much more. I should have made the exit angle 90°, to make them exactly equivalent…

      The calculations are significantly more complex from the point of view of either the planet or space craft.

      Thinking about trying to solve a real set of equations is a bit much; there are other concerns; like the fact that gravity drops off at 1/d^2^; so distance between the objects matters, the integration over distance of the equations is beyond me (I haven’t had to do that since uni, 20yrs ago). But the concepts are not too complicated; and for me at least the external observer makes it so much less complicated.

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      • kuberoot@discuss.tchncs.de ⁨2⁩ ⁨days⁩ ago

        So, the issue is, as far as I know the calculations are dead simple - you “enter” and “exit” the planet’s influence at the same distance from the planet, which means your potential gravitational energy didn’t change, so from the orbital mechanics point of view, from the planet’s frame of reference, your velocity should stay the same.

        As you “fall” in the orbit around the planet, you’re converting potential gravitational energy to kinetic energy, but as you “climb” you convert it back into potential gravitational energy, ending with the same amount of each kind of energy. The only change is that the velocity is redirected.

        With that in mind, it’s why, from my knowledge, the equations are really simple, with the only complications being trigonometry (to resolve the angles) and pythagoras (squaring, adding and getting the square root make the result unintuitive).

        Going back to your graph, if I were to do the math, according to my theory:

        • In A, let’s say you go in with 200 speed, and 0° angle (for simplicity). That means relative to the planet you have 100 speed.
        • In B, you gain some speed by converting potential energy to kinetic. We can’t say how much you gained, because we’re missing any real measure of distance and mass, but the neat thing is - it doesn’t matter, because:
        • In C, you turn that kinetic energy back into potential energy, and end up with the same speed you entered at, at the same distance. This means you now again have 100 speed relative to the planet, but aimed at a 60° angle. We can now add the velocity vectors of the planet and the velocity relative to the planet to get the velocity relative to the sun, using the planet’s velocity as one axis, getting a vector of [100+100*cos(60°); 100*sin(60°)], or [150; 86.6025], with magnitude of 173.2051, which is less than the 200 we went in with.

        If you want an intuitive example of what I’m referring to, consider a planet approaching you as you are stationary relative to the sun. If we assume ideal, presumably impossible, entry and exit angles of 0° and 180°, leaving the planet’s gravity field moving in the exact opposite direction than what you entered, you’ll note you’ll be gaining speed on exit either way, despite not moving towards the planet on the approach and “catching up”.

        The graph doesn’t really show anything other than illustrate your thoughts - but there’s absolutely nothing backing that as being true :/

        Either way, it does feel like we’re going around in circles, and I don’t want to be taking up your time unnecessarily. If you have something to disprove my math (maybe my understanding of orbital dynamics is wrong, and it’s not that simple), that’d be a starting point to try to figure out what’s wrong; if you’re interested, I could try to make diagrams, though I feel like they might kind of look the same, just with different numbers based on calculations.

        I guess one last thing I can offer is a video somebody replied to me with elsewhere in the thread, explaining this idea: youtube.com/shorts/kD8PFhj_a8s

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