1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Computing parabolic orbital trajectory

  1. Jan 12, 2012 #1
    I wrote a game with a rocket that orbits a planet, and you can change the orbit with thrusters. As part of it I draw the predicted orbit for the rocket. To do this I compute the eccentricity vector, then the eccentricity e, then the major and minor axis of the ellipse or hyperbola as appropriate. With these values computing the equation of the orbit's elliptical or hyperbolic path is straightforward. This all works great in my app. However it's entirely unclear to me how to compute the parabolic path when e=1. I know its the limit of the elliptical orbit as e approaches 1, but I don't know how to compute the parameters when e=1 exactly. Specifically, how do I compute the distance from the focus to the vertex of the parabola?
  2. jcsd
  3. Jan 14, 2012 #2
    It turns out this can be solved geometrically.

    Solution 1: The eccentricity vector points at the periapsis, so the directrix of the parabola is perpendicular to the eccentricity vector. Since the distance from the planet to the rocket is equal to the distance from the rocket to the directrix (by definition) you can find the location of the directrix by moving away from the rocket in the direction of the eccentricity vector for a distance equal to the orbital distance.

    Solution 2: The goal is to draw the parabola, and I'm using a quadratic bezier to draw it which requires 3 control points. One of the control points is the rocket itself. The opposite control point is the mirror position on the opposite side of the parabola. The central control point is the intersection of the lines tangent to the parabola at the two opposoite control points. The tangents are defined exactly by the rocket's velocity vector, and by symmetry the intersection is also the intersection with the eccentricity vector, so that gives us the middle control point without knowing the opposite point. I can then compute the opposite control point easily.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook