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My goal is to get a graph "spaceport latitude"<>"delta-v to reach mars".

Using "Trajectory Optimization Tool" (Matlab one) I have successfully generated porkchop graph for Earth->Mars transfer. It does not take into account orbital parameters around Earth i.e. it is the trajectory we are going to use regardless of space port location. Is that correct to assume that spaceport latitude would not affect optimal Earth->Mars trajectory?

To get to this specific trajectory, I do departure burn calculation in the same tool for initial circular 200km LEO with different inclinations of the orbit.

I found that lowest delta-v is needed to depart from 21° orbit inclination, although I assume achieving this orbit will require more energy than 0° one from equator.

My further thinking is that I need to record delta-v requirements for all orbital inclinations, calculate all possible delta-v's for all spaceport latitudes to reach initial orbit with all inclinations, try all orbital plane changes on LEO and by combining all this (i.e. finding optimal combinations of initial inclination and plane change for each spaceport location) with delta-v I got for departure burn to Mars for all inclinations I might get the graph I want.

Is there a more straightforward way of doing so? Maybe there was already some research done on this topic which I am overlooking?

enter image description here

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  • $\begingroup$ Wow! Does the optimization include low-energy transfers and ballistic capture? $\endgroup$
    – uhoh
    May 9, 2016 at 3:28
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    $\begingroup$ @uhoh As far as I understand it calculates only for 1 burn departure. With regards to delta-v/C3 1 burn is more efficient than low-energy transfer. So different program would be needed for that. With regards to capture, program allows you to set a specific maximum velocity on arrival. $\endgroup$
    – BarsMonster
    May 9, 2016 at 3:32
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    $\begingroup$ Looks like the 2018 opportunity. $\endgroup$
    – Mark Adler
    May 9, 2016 at 5:55
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    $\begingroup$ @uhoh If this is the same tool, then it is just a Lambert solver using patched conics. $\endgroup$
    – fibonatic
    May 9, 2016 at 11:55
  • $\begingroup$ @fibonatic Yes, this is the one. $\endgroup$
    – BarsMonster
    May 9, 2016 at 16:35

2 Answers 2

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In addition to the C3 for any given pair of departure and arrival dates, you will also need to know the departure declination for that pair. (The Lambert solver you are using must be generating that as well.)

You then want to optimize mass delivered to the interplanetary trajectory. If the departure declination is less than the latitude of the site, then the inclination of the parking orbit is the latitude of the site. There is no penalty due to the declination, so you can pick the minimum C3. If the departure declination is greater than the latitude, then there is a mass penalty since you now have to go to an orbit inclination equal to the departure declination. You will need some way of estimating the mass penalty on the launch vehicle. Given that, you can optimize for mass. You may find that a higher C3 with a lower declination delivers more mass to the interplanetary trajectory.

For the selected the departure conditions, the parking orbit velocity and radius can be used to easily compute the $\Delta V$ from that orbit to the departure C3.

This handbook provides more information on the departure geometry.

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I don't know how that Matlab tool works, but if it requires you to start from some specific orbit about Earth, it is doing it wrong. You are addressing the first phase of trajectory planning. The best thing to do in this phase is to ignore that the spacecraft starts from some LEO orbit. That's a detail, something that should be ignored in up-front planning. If the subsequent detailed plan involves launching into LEO, and from there to Mars, the optimal trajectory plan should instead inform you of the inclination of the optimal LEO orbit.

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  • $\begingroup$ It is exactly as you say - "porkchop" is not related to orbit at all. It is used to find 1 optimal transfer trajectory form Earth. The interesting part starts when I try to figure out how to get to that trajectory from Earth from different latitudes/different LEO inclinations. $\endgroup$
    – BarsMonster
    May 10, 2016 at 9:47

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