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For a Senior Design Project (Which has been completed and submitted) I (and another teammate) made a numerical simulation for a single stage launch vehicle from mars that included a lot of various effects like gravity varying as a function of height, atmospheric drag (with the $C_d $ varying as a function of mach number) and the gravity turn.

We never really figure out how to calculate the orbit at each and every time step which would have been very useful. We did have the altitude and velocity at every time point. I was wondering how one would go about calculating the orbital parameters at every point.

I was thinking that because the launch location was known and the altitude, velocity and down range distance were all calculated at each time step, could I find the position and velocity vectors for each time step and work from the vectors to find the orbital parameters? or would some other method need to be be used?

In addition what is the radial vector aligned to? As in if the launch site is Eberswadale Crater (24°S, 33°W) with a altitude of 0 m (just using this as an example altitude), what would the starting radial vector and velocity vector be?

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marked as duplicate by Russell Borogove, Nathan Tuggy, Hohmannfan, ForgeMonkey, Brian Tompsett - 汤莱恩 Jun 1 '16 at 7:31

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  • $\begingroup$ Not at all as this question is very different. I am asking if my idea would work and if I even could calculate the radius and velocity vectors given the information I have. $\endgroup$ – B787 Jun 1 '16 at 4:23
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    $\begingroup$ Yes: launch position + altitude + downrange direction + downrange distance determines position at each time step; that's just Cartesian coordinates and maybe a little trig. Difference in position from one time step to the next determines velocity. The only tricky part is deciding what coordinate system to use and remembering that the planet is rotating. Or maybe I'm misunderstanding; I don't see how you could develop the launch simulation in the first place without this knowledge. $\endgroup$ – Russell Borogove Jun 1 '16 at 4:30
  • $\begingroup$ Right which is why I clarified my original post just now to include that question about the coordinate frame, as I am sure that there is some definition that I have not found that is commonly used. $\endgroup$ – B787 Jun 1 '16 at 4:33
  • $\begingroup$ The current sim assumes a perfect sphere and no planetary rotation so that there is no starting velocity and launch location does not matter (essentially assuming equatorial launch in any direction) . I want to make it more accurate for personal use and to make sure that the answer does not change by much (other than now we have extra fuel to use or get rid of). the Target final orbit is 24 deg by 267 km circular. $\endgroup$ – B787 Jun 1 '16 at 4:39
  • $\begingroup$ Rotational speed at latitude is equatorial speed x cos(latitude). To turn scalar rotational speed into a vector, rotate it to the longitude of the launch site. So depending on your choice of coordinate system (I don't know if there's a standard one), that might be [K cos long cos lat, K sin long cos lat, 0]. where K is the equatorial rotational speed -- this means that the rotational velocity vector at lat 0 long 0 alt 0 is defined as [K, 0, 0] $\endgroup$ – Russell Borogove Jun 1 '16 at 5:18

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