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Hello and thanks for helping. I am exploring orbital equations given position, velocity, mass, and gravitational constant ($r, v, G, M$). There likely are answers somewhere here that explain these equations, but I have implemented them in Desmos. I would appreciate if someone took a look at my equations and told me where I am wrong. The reason why I think I am wrong is because the path of the orbit does not intersect with the initial position (which, of course, should lie on the orbital path).

Here are the equations:

$$ h = r \times v $$ $$ E=0.5\cdot|v|^2-\frac{GM}{|r|} $$ $$ a=-\frac{GM}{2E} $$ $$ e=\sqrt{1+ \frac{2Eh^2}{G^2M^2} } $$ $$ b=a\cdot\sqrt{1-e^2} $$

where h is angular momentum, E is specific mechanical energy, a is the semi-major axis, e is eccentricity, and b is the semi-minor axis. I then drew the ellipse with the following equation:

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$

My Desmos implementation can be found here: https://www.desmos.com/calculator/kcc3afkuj0

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  • $\begingroup$ Thank you PM 2Ring. The mass of the rocket is neglible, and is disregarded. Thus the ellipse is centered on the large mass around which the rocket orbits. I used this page for the equation on specific energy: en.wikipedia.org/wiki/…. $\endgroup$
    – kakben
    Jan 24 at 12:52

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I then drew the ellipse with the following equation: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$

This is your error. That's the equation for an ellipse with the center at the origin. You need the origin to be at a focus of the ellipse. The distance $c$ between the center and one of the foci is given by $c = \sqrt{a^2 - b^2} = ae$. Offset the ellipse by this amount (in this case, to the right) and all will be fine.

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  • $\begingroup$ Thank you! This solves half the problem, because now my starting point intersects the path if, and only if, the starting point is on the x axis. My thought is that I need to offset the ellipse in the other direction as well? How do I divide the offset distance c between the x and y axis? $\endgroup$
    – kakben
    Jan 24 at 13:08
  • $\begingroup$ In different words: how do you know that I need to offset the ellipse to the right in this case? What tells you that this direction is the correct one? $\endgroup$
    – kakben
    Jan 24 at 13:21
  • $\begingroup$ @kakben I knew to offset to the right because of your drawing. In general, an orbit lies on a plane in three dimensional space. There are four angles of interest: inclination, right ascension of ascending node, argument of periapsis, and true anomaly. I suggest you Google those terms. $\endgroup$
    – David Hammen
    Jan 24 at 13:23
  • $\begingroup$ Right, I realize I come off as a complete newbie here. I am no astrophysicist but merely enjoy these things as a hobby, and I am trying to draw the trajectories of a rocket in a small "game" if you will, but without using simulation. The game is in 2D, hence the simplifications. $\endgroup$
    – kakben
    Jan 24 at 13:26
  • $\begingroup$ @kakben Since you're using 2D for simplifications, I suggest you read the wikipedia article in the PQW reference frame. The orbit lies on the PQ plane. You'll only need two angles of interest in the 2D simplification: The angle between the +x axis and periapsis (closest approach to the central body) (this is the "argument of periapsis"), and either the angle between the +x axis and the spacecraft ("argument of latitude") or the angle between periapsis and the spacecraft ("true anomaly"). $\endgroup$
    – David Hammen
    Jan 24 at 13:44

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