Does anyone have quantitative information on the delta-v overhead involved with low thrust interplanetary transfers? Looking at accelerations of roughly 0.00032 m/s/s from LEO to the Earth-Mars region. Paper references or handbooks would be useful.

  • $\begingroup$ This might be useful: cds.caltech.edu/~koon/book/KoLoMaRo_DMissionBk.pdf $\endgroup$ – Jerard Puckett Jan 4 '15 at 6:11
  • $\begingroup$ Thank you that has given me some ideas about how to implement a simulation. However, I am under time pressure so I am trying to get some rough results like those available for chemical systems. I'm just not sure how much of an impact the lack of the Oberth effect has. Several sources say it leads to an increase in delta-v, but no-one is putting up quantitative information. $\endgroup$ – Lochie Ferrier Jan 4 '15 at 6:44
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    $\begingroup$ a Google search for "low thrust delta v" turned up this: lr.tudelft.nl/en/organisation/departments/space-engineering/… Does include some figures for low-thrust LEO-Mars. $\endgroup$ – Jerard Puckett Jan 4 '15 at 7:00
  • $\begingroup$ Thanks, I saw that document but didn't see the values. I need to get your Google skills. $\endgroup$ – Lochie Ferrier Jan 4 '15 at 7:10
  • $\begingroup$ For people trying to solve this problem in the future, JPL recently wrote a fantastic paper with a clear method. This free version has the solution on p14: erps.spacegrant.org/uploads/images/images/… This AIAA Journal of Spacecraft and Rockets version has nicer formatting but the same content on p615: arc.aiaa.org/doi/abs/10.2514/1.A32326 . Read about the limitations of the predictions however, as they are important. $\endgroup$ – Lochie Ferrier Jan 5 '15 at 4:42

To get the delta V of a slow, low thrust spiral from one circular orbit to another coplanar circular orbit, use the difference between the speeds of the two circular orbits. See this Wikipedia article.

Speed of earth orbit is about 30 km/s. Mars is about 24 km/s. Delta V would be (30-24) km/s, about 6 km/s.

LEO to escape would be a 7.7 km/s orbit to a 0 km/s orbit at infinity. Though you wouldn't have to go infinity for escape, more like just the edge of earth's sphere of influence about a million kilometers out. So I'd say more like 7 km/s.

Low Mars orbit is about 3.4 km/s. Spiraling in from the edge of Mars sphere of influence to low Mars orbit would take around 3 km/s, I believe.

7+6+3 is 16. 16 km/s is my guestimate for a very low thrust trip from Low Earth Orbit to Low Mars Orbit.

  • $\begingroup$ This works well for co-planar orbits only. $\endgroup$ – ChrisR Aug 20 '16 at 14:28
  • $\begingroup$ @ChrisR Good point. Answer amended. $\endgroup$ – HopDavid Aug 22 '16 at 3:13

Good books on this topic include "Interplanetary Mission Analysis and Design" by Stephen Kemble (Springer, 2006) and "Orbit & Constellation Design & Management" by James Wertz et al. (Microcosm, 2001).

The approximate delta-v for a slowly spiralling transfer over many orbits is, as pointed out already by HopDavid, the difference in orbital speeds for a transfer between circular coplanar orbits. This approximation does not apply to a Mars transfer as your 0.00032 m/s^2 acceleration can reach the required delta-v in much less than one orbit. You may however use it as an upper bound and the Hohmann transfer delta-v as a lower bound.

In order to get to Mars with a low-thrust electrical propulsion system, you will need to (1) escape Earth orbit (2) transfer to Mars and (3) capture into Mars orbit.

The spacecraft will start in whatever orbit the launch vehicle initially injected it, then use either continuous thrust or thrust-coast arcs (coasting near apogee to increase efficiency) to get into a translunar orbit from where escape is possible by gravity assist (cutting about 800m/s off the delta-v requirement). It is possible that the launch vehicle may put you directly in a trans-lunar orbit but something similar to geosynchronous transfer orbit is more likely. Delta-v to trans-lunar orbit in that case should be around 3km/s for continous thrust (3.2 km/s in the slightly lower acceleration example given by Kemble; much more than the 750m/s for high thrust). To solve the problem with your own numbers you would use Lagrange's orbital equations. (Starting on p. 226 of Kemble or p. 63 of Wertz. This post is already getting too long...)

After lunar flyby, you will still not have enough hyperbolic excess to reach Mars. That would be about 3km/s for a typical launch window with high-thrust propulsion minus the small hyperbolic excess you have after the lunar flyby. Since the Oberth effect is no longer working in your favour, you will need roughly this amount of delta-v (with high thrust). That amounts to >3 months of thrust in your example but still much less than the semi-period of the transfer orbit. Expect somewhere in the lower half of 3km/s < delta-v < 6km/s to get you from Earth escape to Mars-crossing orbit.

Hyperbolic velocities at Mars insertion will be similar to Earth escape but there is no major moon for gravity assist. (It is possible to get a free injection by making use of the Lagrange points. I will neglect this as taking too long.) How much delta-v this requires will depend on your requirements and I don't think it is part of the question.

You can further reduce the delta-v requirement with only slight increase in transfer time if you assume that the spacecraft performs an Earth flyby a short while after Earth escape (with a deep-space manoeuvre in-between). Mars flyby before insertion is also useful but less so.


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