• 9 Posts
  • 350 Comments
Joined 2 years ago
cake
Cake day: June 29th, 2023

help-circle




  • That happened to me a few years back, a friend just had his clutch fail in his car. I started thinking how to make a better way to transfer power from motor to wheels.

    Turns out I just reinvented the fluid coupling used in automatic transmissions…needless to say my idea wasn’t that impressive after that.












  • 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/d2; 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.



  • 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.

    • 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.

    • 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.


  • absGeekNZ@lemmy.nztoScience Memes@mander.xyzLittle Pea Shooters
    link
    fedilink
    English
    arrow-up
    12
    arrow-down
    1
    ·
    3 months ago

    No, it’s hard to explain without diagrams.

    But as you fall towards a planet (any gravity well); you pick up speed, if the planet is moving away from you, you fall for longer before you catch up. As you climb back up, you don’t spend all of the energy you gained on the way down. That difference is the Slingshot effect.

    It also works in reverse, if the planet is moving towards you. You catch up quicker, thus gain less speed. And spend overall more energy than you gained when you climb back out. Slowing down in the process.


  • You are gaining (or losing) energy based on if you are traveling in the same direction at the planet or not.

    If you are coming from behind (travelling in the same direction) you an falling into the gravity well for longer. Thus gaining more energy. The extra energy is based on the speed of the planet through space.

    Conversely if you an coming from the front, you fall for a shorter period. You lose energy at you climb up the gravity well.