2
$\begingroup$

So I am modelling an Earth - Mars low thrust trajectory in a simple manner. I'm telling my model to not model any trajectories where the spacecraft is moving any more than 200 m/s quicker or slower than the planet upon rendezvous. This is just for the conic section. So when the spacecraft arrives at Mars, I calculate its speed relative to the planet at that point, and I am treating that as the hyperbolic excess speed (V_infinity), which is constrained to be less than 200m/s. Now, I am getting trajectories that fit the criteria, and they happen to burn very little fuel throughout the journey - my question is, is this physically possible, is it real? Is it even desirable to have low arrival C3's? All the literature I read have arrival C3's/V_infinities in the order of km's/s (or km's^2/s^2) but I have it constrained to mere m's/s. I would have thought, intuitively, the closer the s/c can get to matching the planets velocity, the less of a burn needed for orbital capture, ergo fuel saved?

Best,

EDIT

My orbit looks like a spiral thats made 2 revolutions of the sun. Time of flight, approximately 1400 days. Delta - V is much larger than a chemically propelled ship, but that's OK I guess as the fuel usage is lower. Plotting Mars as an elliptical orbit, not just circular. Hope that helps.

$\endgroup$
  • 1
    $\begingroup$ It could be wrong or right, but there is too much you haven't yet explained. What do you mean a "Earth - Mars low thrust trajectory"? If you've really slowly spiraled out over decades, then your final orbit could potentially roughly match that of Mars, especially if your unmentioned software or algorithm allows for elliptical orbits. Answers to Why would a slow spiral from a C3 of zero take about 2.4 times as much ΔV as an impulsive maneuver? will explain that you've used much more ΔV than an impulsive transfer would have used... $\endgroup$ – uhoh Aug 18 '18 at 17:25
  • 1
    $\begingroup$ ...but if you're using fuel-efficient electric propulsion, and don't mind the time, that's okay! Why not add a few more details about what you've done and what your transfer orbit looks like? $\endgroup$ – uhoh Aug 18 '18 at 17:26
  • 2
    $\begingroup$ Thanks for the edit! You may not have included gravitational attraction from Mars yet. You can match the velocity of mars easily if you haven't. However, if Mars' gravity is active, then the spacecraft will begin to accelerate it which means you'll slingshot away, or burn up in the atmosphere, depending on the details of the approach. If you want to stay there in orbit, or safely land, you'll still need that burn! Alternatively, since you have a lot of time, you could slowly aerobrake. Either way, you'll need to turn Mars' gravity on. $\endgroup$ – uhoh Aug 18 '18 at 17:38
  • 1
    $\begingroup$ Massive help, that was precisely the problem! $\endgroup$ – Harvey Rael Aug 18 '18 at 18:24
1
$\begingroup$

I found the issue, thanks to @uhoh. I was trying to model a low-thrust equivalent to a Lambert's solver, with all the juiciness that comes with that. The trajectories were resulting in artificially optimal C3/Fuel burnt figures, and the reason was because I was neglecting to model Mars' gravity as I approach it "from infinity", which, in "real life" would cause the spacecraft to accelerate toward it. Normally, burns would be needed to counteract this in order to prevent smashing into Mars' atmosphere or sling shotting away from it. I simply had not modelled that. Arrival C3's while ideally would be 0, are usually higher because of this gravitational effect.

$\endgroup$

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.

Not the answer you're looking for? Browse other questions tagged or ask your own question.