I just started researching this topic and therefore I do not know much about it. Nonetheless, I am still curious about what determines the direction an object takes after passing by a planet. I think the new direction is most likely determined by the gravity of set planet and the new direction can be calculated using vectors. But I also think that the speed of the planet can have an influence in the new direction of the object. In both cases I believe that the new direction is found by adding vectors or by using Pythagora's. Can someone correct me?

enter image description here

  • $\begingroup$ The planet's speed definitely has an effect, but you can't get an answer just by "adding vectors". You simply have to integrate $F=ma$ while there's a non-negligible force, calculating the rocket's new speed and position at each instant, using $F=GmM/r^2$, $F$ the force acting on the rocket, $m$ its mass, $M$ the Earth's mass, and $r$ the distance between them. That can easily (even trivially) be done numerically, but I'm not sure whether or not there's a closed-form solution, especially since the total $F$ on the rocket would have to include the always-non-negligible Sun in addition to Earth. $\endgroup$ – John Forkosh Jan 11 at 20:30
  • 1
    $\begingroup$ @JohnForkosh I think there is a closed form if we assume the flyby happens fast enough that the no other bodies exert a meaningful impulse to the spacecraft during it. I don't have time to post an answer right now though. $\endgroup$ – jacob1729 Jan 11 at 20:36
  • $\begingroup$ Actually, in retrospect, I think there probably is a closed-form solution, but definitely not by simply adding vectors. It's a gravitational three-body problem, which has no general closed-form solution. But in this case, two of the bodies are essentially infinitely massive (and hence negligibly affected) with respect to the third. And I believe there's a closed-form solution for that special case (among various other special cases). $\endgroup$ – John Forkosh Jan 11 at 20:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.