9
$\begingroup$

Consider using solar electric propulsion for, e.g. Earth-Mars trip, or for asteroid retrieval mission. Is there a benefit to first entering an eccentric orbit so that part of the time is spent relatively close to the Sun, to maximize the available solar power and the effect of thrust, similar to the Oberth effect?

Does circular -> eccentric -> circular have potential benefits compared to simply spiraling from one circular orbit to another?

Also, is there any software that can calculate optimal solar electric trajectories and taking effects like this into account?

Assume the solar cells can withstand being closer to the Sun, and operate at the same % efficiency at any light level; also assume the engine can effectively use all available power.

$\endgroup$
1
  • $\begingroup$ What you're proposing boils down to using bi-elliptic transfer instead of Hohmann or spiral, and for Earth-Mars transfer the answer is "no", their orbits are too close to benefit from bi-elliptic. If the transfer was to Trans-Neptunian objects, the answer could be quite different, $\endgroup$
    – SF.
    Nov 30, 2015 at 7:01

3 Answers 3

10
$\begingroup$

Optimal low-thrust trajectories for interplanetary missions are complicated to calculate. It's a little too broad of a topic for a Q&A. In fact, I spent an entire internship at NASA Johnson Space Center working on this problem (building of the work of many others years before and hence), and wrote my first (non-conference) academic paper on exactly this topic.

A Genetic Algorithm and Calculus of Variations-Based Trajectory Optimization Technique

A genetic algorithm is used cooperatively with the Davidon–Fletcher–Powell penalty function method and the calculus of variations to optimize low-thrust, Mars-to-Earth trajectories for the Mars Sample Return Mission. The return trajectory is chosen thrust-coast-thrust a priori, has a fixed time of flight, and is subject to initial and final position and velocity equality constraints. The global search properties of the genetic algorithm combine with the local search capabilities of the calculus of variations to produce solutions that are superior to those generated with the calculus of variations alone, and these solutions are obtained more quickly and require less user interaction than previously possible. The genetic algorithm is not hampered by ill-behaved gradients and is relatively insensitive to problems with a small radius of convergence, allowing it to optimize trajectories for which solutions had not yet been obtained. The use of the calculus of variations within the genetic algorithm optimization routine increased the precision of the final solutions to levels uncommon for a genetic algorithm.

TLDR: As first-year orbital mechanics would suggest, it was marginally more efficient to perform the plane change during the Mars escape. The general solution was using electrical propulsion to spiral to escape velocity, coast to Earth, and spiral back in.

$\endgroup$
8
$\begingroup$

Solar Electric Propulsion (SEP) is unlikely to benefit on many missions from periapsis burns since, albeit it can achieve high specific impulse (high exhaust exit velocity), is a relatively small thrust (small exhaust mass flow rate) propulsion method. We have yet to see actual SEP implementations for human missions, and development of those is aiming for much higher thrust than what we're seeing currently, but it will still be considered low thrust and constant acceleration systems compared to impulse burn chemical rockets. E.g. NASA's NEXT (Evolutionary Xenon Thruster) is currently at about one quarter Newton thrust. So SEP will operate pretty much all the of the time, meaning there's no net gain due to Oberth effect with highly eccentric orbits during periapsis burn, since it would only linger longer at apoapsis then (where no net gain of periapsis burns at constant thrust is a direct consequence of orbital-energy-invariance law).

In that sense, from thrust efficiency perspective, there is therefore no incentive to start the climb from a highly eccentric orbit. From solar power's perspective, any eccentricity will result in more or less same insolation since it will still be close to the primary with orbit's semi-major axis within the primary's Hill sphere, otherwise it's already on its way somewhere else anyway. It will be more important to pay attention to other orbital elements such as inclination and node and make sure that the spacecraft stays for as short period as possible in the primary's shadow (both umbra and penumbra), or that it avoids other potentially damaging regions (such as, e.g., Earth's Van Allen radiation belts). Equally, it's important to realize that constant thrust will naturally reduce initial eccentricity anyway for the same reason I already mention in the first paragraph (efficiency of shorter periapsis burn will be offset by inefficiency of longer apoapsis burn - actually, come to think of it, this can be visualized pretty well with Kepler's second law of planetary motion, and why $\Delta A/\Delta t$ remains constant - but i.e. difference between apoapsis and periapsis will remain more or less the same while semi-major axis constantly increases, thus decrease in eccentricity).

So using SEP in deep gravity wells will be more akin to slowly climbing out of orbit in many spiraling out orbits up to reaching characteristic energy needed to escape the gravity well ($C_3$) and be on its way to some interplanetary destination. At that point any additional trust translates to hyperbolic excess velocity ($v_\infty$) with respect to the primary body and shortens transfer time, but it doesn't orbit it any more. I.e. it's the point where the spiraling geometry of climbing out of the gravity well changes to either a heliocentric Hohmann transfer orbit's hyperbole without any additional thrust for the interplanetary transfer leg, or the first accelerating leg of a brachistochrone curve with continuous propulsion.

As for if there are any tools that can calculate optimal solar electric trajectories, yes, some previous versions of NASA's General Mission Analysis Tool (GMAT) already had third party Electric Propulsion plugins available, but that's now built into the latter releases, starting with R2015a:

GMAT now supports modelling of electric propulsion systems. Below is an example showing GMAT modelling a cube-sat with electric propulsion in a lunar weak-stability orbit. You can model electric tanks, thrusters, and power systems (both Solar and nuclear).

GMAT Electric Propulsion

See the Electric Propulsion tutorial for more information.

$\endgroup$
5
  • $\begingroup$ SMART-1 would do it's burns while in the neighborhood of perigee. See en.wikipedia.org/wiki/SMART-1#Spacecraft_design $\endgroup$
    – HopDavid
    Nov 29, 2015 at 21:56
  • $\begingroup$ @HopDavid Sure, but it needed nearly 14 months to reach EML1. The question asks about manned missions, so slowly rising apogee with perigee only burns won't be an option, it would take too long. It's still an interesting source for that decreasing eccentricity tho, even if it only did 1/3 the orbit perigee burns initially. Here's its trajectory (slide 10). I find it quite interesting and might add it to the answer, time permitting. Cheers! $\endgroup$
    – TildalWave
    Nov 29, 2015 at 22:32
  • $\begingroup$ People often ask about manned Earth to Mars missions in this Stack Exchange but I don't see Alex specify if the mission he has in mind is manned or robotic. And usually asteroid retrieval is assumed to be robotic. $\endgroup$
    – HopDavid
    Nov 29, 2015 at 22:51
  • $\begingroup$ @HopDavid Well I read the question as asking about human missions, that's why I also briefly mention current development of SEP for it in my answer. I might as well be wrong, but Earth-Mars trip sounds like manned to me. And I read asteroid retrieval as e.g. the retrieval part of the ARM mission (i.e. manned rendezvous in distant retrograde orbit, not the redirect part into DRO, for which I'd assume to be meant as robotic, too). $\endgroup$
    – TildalWave
    Nov 29, 2015 at 22:58
  • $\begingroup$ @TildalWave: "slowly rising apogee with perigee only burns" - that's exactly what I was thinking of. Aha, so it does work! Btw for the most part I was thinking robotic missions (esp since a big part of a manned mission can be robotic to pre-position stuff anyway). Sorry if that was confusing. Also thanks for pointing me to GMAT :) $\endgroup$
    – Alex I
    Nov 30, 2015 at 2:38
4
$\begingroup$

You do not mention what acceleration your SEP is capable of. Presently power sources have to be rather massive. When, for example, your thrust is 1 newton but spacecraft mass is 10 tonnes, acceleration is .0001 meters/sec^2. See The Need For a Better Alpha.

If acceleration from the ion engines is a tiny fraction of the acceleration the sun exerts, the trajectory is a gradual spiral. The delta V from one circular orbit to another is well approximated by |speed of departure orbit - speed of destination orbit|. See the Stack Exchange question General guidelines for modeling a low thrust ion spiral?.

For example earth's average speed is about 29.79 km/s. Mars average speed is about 24.13 km/s. 29.79 - 24.13 = 5.66. A slow ion spiral from a 1 A.U. circular orbit to a 1.52 circular orbit takes about 5.66 km/s. (Here I'm assuming circular coplanar orbits and disregarding the planetary gravity wells).

Now let's look at a Hohmann orbit for the same scenario, a Hohmann from a 1 A.U. circular orbit to a coplanar circular 1.52 A.U. circular orbit.

Departure Vinf is 2.94 km/s. Arrival Vinf is 2.65 km. These total 5.59 km/s.

5.59 km/s vs 5.66 km/s? That's a difference of only .07 km/s.

Here is a graphic comparing sum of Hohmann Vinfs to the delta V needed for a gradual ion spiral:

enter image description here

The blue and red portion is like a sand chart. The red part laid on top of the blue gives the sum of the two Vinfs. The grey part in the back is |destination orbit speed minus earth orbit speed|.

You can see for neighboring planets like Venus and Mars, they're pretty close. But a Hohmann to Jupiter beats slow ion spirals by a good margin. A Hohmann to Uranus wins by a larger margin.

But as you move from the sun, burns can be much more leisurely. Something in low earth orbit is moving about 4 degrees per minute. If you wanted your burn within 8 degrees of perigee, you'd have less than two minutes.

In contrast earth moves around the sun at about a degree per day. At 1 A.U., a burn within 8 degrees of perihelion would be about 8 days.

Jupiter moves around the sun at about a degree every 12 days. So if perihelion is 5.2 A.U. from the sun, you'd have more than three months to execute a burn within 8 degrees of perihelion.

So as you move outward from the sun, even ion burns start looking like impulsive burns.

$\endgroup$
4
  • $\begingroup$ "You do not mention what acceleration your SEP is capable of." - For the purposes of this, assume the thruster is flexible - any propellant flow rate and any exhaust velocity is possible - but of course it does need power and propellant :) I think this is a reasonable simplifying assumption to look at trajectories as separate from the specifics of the thruster. $\endgroup$
    – Alex I
    Nov 30, 2015 at 1:48
  • $\begingroup$ @AlexI It's a trade off. Rockets with higher exhaust velocity have lower thrust. Rockets with lower exhaust velocity tend to have more thrust. Are you thinking actual rockets or a device to use in a science fiction story? As it is I can imagine a mutlitude of scenarios. The question is too vague to give a specific answer. $\endgroup$
    – HopDavid
    Nov 30, 2015 at 2:17
  • $\begingroup$ Actual device - could be VASIMR, a throttleable ion thruster, or some more exotic options with variable Isp (including potentially extremely high values). $\endgroup$
    – Alex I
    Nov 30, 2015 at 2:32
  • $\begingroup$ (Googling VASIMR...) Well maybe up to 50 km/s exhaust velocity. Newtons? Mass of spacecraft? And throttled down to ? Km/s with ? newstons and ? kilograms for spacecraft engine and power source. If the numbers you're providing are (mumble, mumble), (waving hands), I don't see how your question can be answered. $\endgroup$
    – HopDavid
    Nov 30, 2015 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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