Given two arbitrary orbits around a point mass, there exists some optimal transfer between them in terms of delta-v.
What's the highest number of impulses such a transfer could require? (That is, I'm asking for a specific quantity of solutions to a variant of Lambert's problem.)

"Optimal" in a mathematical sense. Burns not being perfectly impulsive, transfers taking unreasonable amounts of time being undesirable, perturbations, three-body effects and so on can all be ignored.

The number is obviously larger than 1, since not all orbits share a common point.

If all optimal planar transfers are bi-tangential orbits, the answer is 2 for planar orbits.

The number is larger than 2, since solutions with 3 impulses are better for some types of transfers.

An infinite apoapsis generalised bi-elliptic transfer, which is sometimes optimal, has two non-zero impulses and two zero-impulse manoeuvers. Whether this counts as 2 or 4 impulses is less important since: 1) There can at most be 2 zero-impulse manoeuvers in any optimal transfer, and 2) Any optimal transfer containing a zero-impulse manoeuver can at most have 2 non-zero impulses.

Does an optimal transfer requiring 4 or more non-zero impulses exist?

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    $\begingroup$ I was thinking a bit about adding a bounty to your question (because I have a hunch this will turn out to be really interesting) but I don't know if it interferes with your timing or intent. How would you feel about that? $\endgroup$
    – uhoh
    Nov 16, 2020 at 0:15
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    $\begingroup$ @uhoh Sure, go ahead. I too would like to know the answer to this. (Even though bounties haven't proved very effective for this kind of question in the past). $\endgroup$ Nov 16, 2020 at 1:03
  • $\begingroup$ This is quite interesting. For ill-conditioned transfers (practically everything with 30+ degrees inclination change; generally vast majority), the answer is "4" (1. raise apoapsis to near-infinity, 2. circularize+plane change to enter a slow orbit to a point co-axial with target apoapsis, 3. drop periapsis + plane change for target orbit, 4. drop apoapsis.) Considering the "mathematical perfection" cost of 2. and 3. is infinitesimally small, (and transfer time between them is infinite), but the whole thing costs $\sqrt{2} ( v_{Pe1} + v_{Pe2} ) + \epsilon $ $\endgroup$
    – SF.
    Nov 16, 2020 at 8:49
  • $\begingroup$ This is quite a bit, and there will be a margin of degenerate transfers that are not-quite-as-ill-conditioned, that can be done on less delta-V but on more burns. $\endgroup$
    – SF.
    Nov 16, 2020 at 8:53

3 Answers 3


Whether you realize it or not, this is a very fundamental and challenging question in astrodynamics. It's personally one of my favorite topics in the field, and has been very rigorously studied by some of the greatest minds in astrodynamics/trajectory optimization for decades. The question was first formally posed by T. N. Edelbaum in his paper "How many impulses?" (url: https://arc.aiaa.org/doi/pdf/10.2514/6.1966-7).

From what I gather in your question, it seems you are interested in transfers between orbits (not between two fixed points fixed in space on two different orbits) and also transfers in which the time-of-transfer is not fixed and also unbounded. There is no way to mathematically guarantee a solution to this problem is globally optimal. It becomes very challenging finding solutions with this much freedom since the time-of-transfer and terminal states are coupled. On top of this, there's also an infinite number of revolutions around the central point mass that can potentially be considered. Anyone who can give a more mathematically rigorous and/or better explanation of why this is so challenging please feel free to jump in.

If you care to consider fixed-time-of-flight minimum-delta-v transfers between two points fixed on two different arbitrary orbits around a point mass for a moment, a recent open access paper by Ehsan Taheri and John Junkins answers Edelbaum's question "How many impulses?" very well and in a very interesting way (url: https://link.springer.com/article/10.1007/s40295-019-00203-1). To summarize what they do, they first solve the transfer problem using continuous-acceleration instead of impulsive-acceleration, i.e., acceleration (I’m using “acceleration” here just because a massless spacecraft is considered in the paper) is applied continuously in time with a maximum allowed level of acceleration enforced. Then, using this solution as an initial guess, the same problem is solved with an increased level of maximum acceleration allowed. Eventually, the maximum acceleration allowed is increased until the solution looks very similar to an impulsive solution, i.e., the continuous-time acceleration arcs are so short in duration that they look instantenuous. This provides a very good initial guess for when, where, and how many impulses should be considered for an impulsive trajectory optimization algorithm.

There are of course many many more details considered, but the optimal minimum-delta-v solution is able to be confidently recovered using rigorous optimal control theory. Examples are shown for problems in which the optimal number of impulses is 4 or more and for different numbers of revolutions around the central mass. The paper also gives lots of interesting discussion on all the many nuances surrounding Edelbaum's question, just as Edelbaum gives in his own paper.

One line of discussion you’ll find that will shed some light on your original question is that while this recent paper technically solves fixed time-of-transfer problems with fixed initial and final states, it’s shown that some solutions can reveal what is known as “early arrival” and “late departure” in which no impulses are performed at the initial and final times, revealing that a different time-of-transfer and boundary conditions (within the same orbit) will give the same optimality. I.e., this method can in fact reveal possible solutions to your original question in which time and boundary conditions are free, but there is still no guarantee of global optimality.

While there is no definitive answer to your original question, hopefully this still sheds some light on it. The papers I referenced can be good starting points for finding almost all the relevant work done to answer yours/Edelbaum’s question as there are many works referenced in them on top of the insight they already give if you want to take a deep dive.


For coplanar orbits, a bi-elliptical transfer is more efficient than an Hohmann transfer when the ratio of the initial and final radii is greater than 15.58. When the ratio is less than 11.94, an Hohmann transfer is more efficient. (Thanks to notovny for correcting me.)

A bielliptic transfer is effectively two subsequent Hohmann transfers. Section 6.3.2 of "Fundamentals of Astrodynamics" by Vallado (p. 328 in the 4th edition) compares Hohmann transfers to the bi-elliptical transfer. In a bi-elliptical transfer, you will need three burns: one to depart the initial orbit onto an elliptical orbit (you must depart when your flight path angle is zero), then perform an apogee burn on the elliptical orbit, and finally perform a final maneuver on the destination orbit, also where you should get a flight path angle of zero.

For any other transfer, it really depends on the problem you are trying to solve, and the variables of the problem (e.g. how many times can you reignite the engine, what will be the errors in the thruster performance, where are the ground stations placed for navigation, etc.).

For example, for interplanetary or lunar missions, one would set up the problem to assume 4 to 8 control points, i.e. positions in the trajectory where you should place a maneuver. One would rarely place more than 8 control points. Each control point is assumed to be a point in the trajectory where a maneuver will be executed, and those require some operational overhead. As such, we ensure there is some time between each potential maneuver. For example, before a maneuver, it is important to have very good knowledge of the position and velocity of the spacecraft before the maneuver (i.e. a good navigation solution), and be able to continue tracking the spacecraft soon after the maneuver. In short, the fewer the maneuvers, the easier it is to fly the spacecraft. So there's a trade off between the fuel savings and the overhead needed for each maneuver.

Moreover, optimizers (like SNOPT) would be used to optimize the placement of these control nodes and the optimizer will try to minimize the delta-V at each node. This approach is called "multiple shooting" and is used for Ballistic Lunar Transfers to libration point orbits. The optimizer may show that some of the control nodes have extremely small delta-Vs (e.g. less than a few millimeters per second), and in which case, you can omit that maneuver, and rerun the optimization problem.

A similar approach would be done for Earth orbits on different planes. As you also correctly stated, one would generally start with a Lambert solution for a first level approximation. Then, you would place the control points at different positions and let the optimizer find the best solution.

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    $\begingroup$ This is a really interesting answer and I'll bet there's a question about three-body orbits that would suit it better. The current question post begins: "Given two arbitrary orbits around a point mass..." so I think at least the intent is to ask about transfers between Keplerian orbits. It doesn't specify anything else about those orbits, so while "planar" is mentioned in an example of the counting procedure, my guess is that the maximum number will be between non-coplanar elliptical orbits that are either awkwardly asynchronous or "maliciously synchronized" to make it hard. $\endgroup$
    – uhoh
    Nov 15, 2020 at 1:35
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    $\begingroup$ While interesting about trajectory optimisation in general, this doesn't answer my question. $\endgroup$ Nov 15, 2020 at 12:52
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    $\begingroup$ "It really depends on the problem you are trying to solve". Optimal delta-v transfers between two orbits around a point mass. $\endgroup$ Nov 15, 2020 at 12:54
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    $\begingroup$ @ChrisR: No mission profile ever involves maneuvers that would lean on delta-v equivalent of turning inclination of LEO by 180 degrees ("""optimally"""). There's no point bringing real-life missions to this discussion because these almost-worst-case scenarios are already so bad nobody even takes them into consideration. $\endgroup$
    – SF.
    Nov 16, 2020 at 21:36
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    $\begingroup$ @ChrisR I see you're confused because you take a lot of engineer's approach, not mathematician's, considering factors which are non-factor in asker's conditions (like delay between maneuvers). For example, " in which case, you can omit that maneuver," - if you see my comment about all the "worst scenario" transfers, the solution consists of two large burns and two minuscule ones, and the minuscule ones are critically important. It also has completely impractical duration, which again is non-factor. Or consider a delta-v optimal bielliptic transfer - transfer orbit being infinite is OK. $\endgroup$
    – SF.
    Nov 17, 2020 at 10:29

In searching the title of my paper, I accidentally came across this post! In addition to the excellent responses above, you can check this recent paper published in the AIAA Journal of Guidance, Controls and Dynamics (https://arc.aiaa.org/doi/10.2514/1.G007409) in which I have provided answers to your questions and provided analytic bounds on the lower (required) and upper (allowable) number of impulses for three classes of maneuvers: fixed-terminal-time rendezvous, free-terminal-time rendezvous, and phase-free transfer (directly related to Edelbaum's question). I encourage you to first read the "How Many Impulses Redux" paper (https://link.springer.com/article/10.1007/s40295-019-00203-1), since it is a good starting point (in particular for the results regarding the Earth-Dionysus problem), and then continue to read the recent JGCD paper. In case you don't have subscription to JGCD, I have uploaded a version of the JGCD paper before being formatted by JGCD on ResearchGate (https://www.researchgate.net/publication/372525067_Existence_of_Infinitely_Many_Optimal_Iso-Impulse_Trajectories_in_Two-Body_Dynamics)

In summary and considering the simplest example possible, the minimum number of impulses is one. Consider GTO and GEO, where the apogee of the GTO is tangent to the GEO. You can also assume that GTO has a non-zero inclination value. If the starting point of the maneuver is perigee of the GTO, at precisely half of the orbital period of the GTO, the spacecraft reaches the apogee point and one single maneuver is needed to optimally inject the spacecraft onto the GEO. This is the case in which there is one intersecting point between the two orbits. I coined the term "impulse anchor position" since there are cases that there are two impulse anchor positions (check Earth-Dionysus problem in the JGCD paper).

How about the maximum number of impulses, which appears to be a far more complex problem? However, it is possible to divide the single delta-v vector at the intersection point (impulse anchor position) into many small-magnitude, same-direction impulses such that the total sum of the smaller impulses is still equivalent to the magnitude of the original impulse. The only inevitable change to the problem is that we need extra time. In principle, any minimum-deltav (respecting Lawden's primer vector theory) maneuver can be broken down into 4 main phases: 1) motion on the original orbit, 2) motion on an unknown number of phasing orbits, 3) motion on a connecting arc (associated with a phase-free transfer between impulse anchor positions) and 4) motion on the target orbit. Depending on the available time, the spacecraft can make multiple revolutions (around the central body) on these 4 main phases. Regarding the GTO-GEO problem, each small impulse, places the spacecraft on an intermediate elliptical phasing orbit. Thus, if we are interested in a transfer maneuver (and if the time has no bound), you can have infinitely many small-magnitude impulses that achieve the transfer (i.e., move the spacecraft from one orbit into another). So, the answer to your question is infinity! Note that if the time of flight is finite/bounded, then, the maximum number of impulses is finite.

For all infinity of solutions, the total delta-v is equal to the single-impulse optimal one, and we have essentially stretched the maneuver time without sacrificing optimality with respect to the delta-v. Why do we want to stretch the time? Well, if the propulsion system of the spacecraft can only produce a maximum value of impulse, with this flexibility, we can throw as many phasing orbits as needed to ensure that the largest impulse is less than the maximum value (threshold). So, this flexibility is significant from an operational point of view at the cost of increasing the mission time, but without incurring any additional cost on delta-v, namely, delta-v optimality is retained.

On the other hand, if the problem is a rendezvous-type maneuver and the time of flight has an upper bound, the highest number of impulses is always finite. How can we determine the maximum number of impulses? We can use the available time of flight and use the lowest orbital period (among the 4 phases of motion) and analytically obtain the maximum number of phasing orbits that can be added to the motion and also determine the maximum number of impulses (again, check the Earth-Dionysus problem in the link to the JGCD paper on (ResearhGate)). More examples are given in the paper (with entire solution envelops for possible ranges of time spent on each phase of motion), but the entire idea is based on the key observations that I outlined for the GTO-GEO problem.

Updates to the response regarding the "required" versus "could" number of impulses. My response is quite lengthy to be included as a comment and I had to update my response.

In orbital mechanics/astrodynamics, the class of transfer maneuvers is characterized in a specific manner, i.e., it corresponds to time-free, phase-free maneuvers that make it possible for a spacecraft to move from one orbit and to be inserted into another orbit. Since according to the definition both time and phase are free, orbit transfer maneuvers are the least constrained type of trajectories that are typically used for gaining additional insights into the theoretically optimal (with respect to delta-v) solutions. With this definition in mind, and as I explained earlier, infinitely many impulses are the answer to the highest number of impulses.

Allow me to provide more explanations to clarify my point above. The original question posed by Edelbaum in his 1966 paper is raised within the context of inverse-square restricted two-body dynamics. In principle, we are seeking solutions to a nonlinear dynamical system. Potential solutions to nonlinear dynamical systems may (and frequently do) exhibit bifurcations with respect to both temporal component (i.e., time) and other parameters of the problem. It is important and crucial to take both of these components into consideration when we are interested in analyzing complex nonlinear dynamical systems.

For example, consider the problem of maneuvering a spacecraft from a planer circular LEO to another large-raidus circular orbit. It is a classic result that if the ratio of radii of the larger orbit and LEO is less than 11.94, then, the Hohmann maneuver is the most efficient transfer maneuver. But, in this result, it is implied that time is unbounded since by definition we are dealing with transfer maneuvers. Question: If the ratio of radii is less than 11.94, how many impulses are required to perform an optimal (in the sense of delta-v) orbital transfer? The answer depends on the value that we choose/allow for the time of flight. If time of flight is constrained to be less than half of the orbital period of the Hohmann ellipse (T_Hoh), then, we have no option, but to solve a time-constrained Lambert problem, which results in a bi-impulsive maneuver. We can find the best (least delta-v) bi-impulsive maneuver, but the solution will never be optimal from a delta-v perspective. However, if the time of flight is exactly equal to T_Hoh, then, the optimal number of impulses is two and we can guarantee optimality of the solution (w.r.t. delta-v). If the time of flight is constrained to be greater than T_Hoh, then, we have to split the impulses at the anchor positions to match the time of flight without sacrificing optimality in delta-v. So, we may need more than two impulses to retain delta-v optimality. From these three cases, it is clear that we have to relax any constraint on the time of flight to attain the optimal solution. Thus, for the range of the ratio parameter of the radii, the minimum number of impulses is two (corresponding to the classic Hohmann transfer), whereas the maximum number of impulses is infinity and all of these are optimal with respect to delta-v. But there is more to these solutions than meets the eye!

Another example will shed more light on what I explained above. What if in the above example the ratio of the radii is between 11.94 and 15.58? Again, it is theoretically proven that we can have Hohman and Bielliptic maneuvers that have the same total delta-v value, i.e., delta-v_Hoh = delta-v_BE. In other words, there is a separatrix between Hohmann and Bielliptic maneuvers when the ratio is between 11.94 and 15.58. The key point is that time plays a significant role in the (required) number of optimal impulses. If we relax/ignore time, again, there is an infinity of two- and three-impulse solutions (corresponding to the separatrix curve) that are optimal with respect to delta-v. If we fix the ratio of the radii (between 11.94 and 15.58), as I explained, we can have unique two-impulse (Hohmann) and three-impulse (Bielliptic) maneuvers that have exactly the same delta-v. So, the number of impulses is not unique. The difference between the two solutions is the time of flight. Check chapter six of either third or fourth editions of the Book by Howard D. Curtis (Fig 6.8 in the Third/Fourth editions) for a plot that summarizes the results and also depicts the separatrix (corresponding to the equal-delta-v curve).

As is evident, time and other components (e.g., ratio of the radii) can result in different classes of solutions. Characteristics of the solutions differ according to these values. Note also that in dynamical systems (while often ignored), feasibility is a more fundamental criterion than pure optimality. In the circular-to-circular example with a ratio of radii less than 11.94, if the imposed time of flight is less than T_Hoh, we cannot expect any optimality since time being less than T_Hoh restricts the set of feasible solutions to the class of two-impulse Lambert solutions. We can find the best (with respect to delta-v) two-impulse solution, but that solution is sub-optimal compared to the delta-v of the Hohmann maneuver. That is why we have to relax the time of flight to enlarge the class of feasible solutions, which is the idea in studying transfer maneuvers.

Without going into the details, the same analysis is true for other problems. In the JGCD paper, I have shown that for the Earth-Dionysus problem, if we limit the time of flight to be fixed at 9.7 years, 139 different classes of equal-delta-v solutions exist. The lowest number of required impulses is three and the largest number of required impulses is 8. All of these solutions require the same amount of delta-v. In addition, I have shown that if the time of flight is constrained to be less than 9.7 years, a natural consequence is reduction in the set of feasible equal-delta-v solutions. However, there is a specific time of flight below which we have to abandon delta-v optimality. To address the time-feasibility problem of the Earth-Dionysus problem, I have mentioned that if the time of flight is less than 1840.31 days, it is impossible to obtain any optimal solution (with respect to delta-v), and as such, a two-impulse solution is the only option; however, the delta-v corresponding to the bi-impulsive maneuver is never going to be less than its value for the three-impulse maneuver. In other words, we have proposed an algorithm that allows us to determine time-feasibility of a minimum-delta-v maneuver.

In summary, when we are interested in determining the maximum number of required impulses, time-feasibility has to be considered and a constraint on the time of flight has to be imposed. Below a certan time, it is impossible/infeasible to obtain any theoretically optimal solution. There are ranges of the time parameter for which the number of impulses can change to retain optimality. For transfer maneuvers, the maximum number of impulses is infinity since time has no bound by definition. You can check the details of the Earth-Dionysus paper in the JGCD paper.

Best, Ehsan Taheri

  • $\begingroup$ This is great! Thank you so much for continuing to perfect your answer, and wow! for uploading a version of the JGCD to researchgate! $\endgroup$
    – uhoh
    Jul 26, 2023 at 20:14
  • $\begingroup$ I'm not sure this actually answers the question though. The argument for "infinitely many" seems to be how many impulses could be used, not how many are required. It's fairly trivial that an optimal transfer requiring, say 5, impulses could be modified in the way you describe to use 6 impulses. The crux of the question, however, is that you can't always do away with fewer impulses, without sacrificing delta-v $\endgroup$ Jul 26, 2023 at 23:04

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