# Fastest orbit transfer constrained by maximum g-force magnitude

Let's consider an imaginary constant 1g rocket from Earth to Mars. What is the fastest possible trajectory?

Intuitively it seems that the fastest travel time is achieved with the "radial" trajectory (the blue line below). However, as both Earth and Mars already travel at their orbital speeds (30 and 26 km/s if I remember correctly), the "straightest" (shortest by distance) trajectory is S-shaped (the green line).

Is it almost-optimal trajectory, or we can do better?

(Yes, I know that it's impractical compared to Hohmann transfer, "short burns" transfers, or "low thrust" multi-orbit transfer. The question is about limits on transfer time if we have extremely good rockets - we are still limited by low-g survivability limits of humans)

The question is "How fast we can travel to Mars if the only constraint is human survivability". The simplest allowed such survivability constraint is maximum allowed acceleration magnitude.

The picture shows a sketch of Mars and Earth orbits, which for the purpose of the queston can be treated as cicular and coplanar.

• Really interesting question! Fastest constant-acceleration transfer between circular orbits. – uhoh Mar 14 '19 at 22:49
• Truly constant acceleration, or are you allowed to shut the engine off during turnover? – Mark Mar 15 '19 at 1:54
• You are allowed any lower acceleration. The idea, basically, is fastest survivable earth-to-mars trip. – nponeccop Mar 15 '19 at 2:09
• @nponeccop I think what's still missing is that acceleration is a vector and I think that you would allow that changing the direction of the acceleration without changing the magnitude counts as "constant" acceleration. So I wonder if you want to add to your question that you mean "constant magnitude" acceleration. Also, double checking, the initial and final states are traveling in the circular orbits at orbital velocity, in the same direction, right? – uhoh Mar 15 '19 at 8:52
• Scott Manley explains this concept in a video: youtube.com/watch?v=toMnjO8aJDI – Elad Stern Mar 20 '19 at 12:46

One-g is large enough to make this easier: Times are short.

Boosting straight out from Earth (taking $$g = 10 m/s^2$$ and ignoring the $$0.006 m/s^2$$ acceleration toward the Sun) takes

$$d = (230 \times 10^6 \times 10^3 - 150 \times 10^6 \times 10^3)/2 = 1/2 g (t/2)^2$$

$$t = 2 \times \sqrt{ {2 \times 40 \times 10^ 6 \times 10^3} \over {g}}$$

For a 1 g acceleration, that's $$180 \times 10^3 s$$ or about 50 hours.

Turning that acceleration to also change the speed along the orbit by $$6 km/s$$ only requires a $$6,000 / 180,000 = 0.033 m/s^2$$ acceleration along the orbit, or changing the angle of the thrust by just about 3 mrad. And that, in turn, only reduces the radial acceleration by $$g (1 - cos(0.0033)) = .05 mm/s^2$$, much less than the acceleration due to the Sun.

So I propose that the fastest way to get to Mars at 1g is to thrust straight toward the rendezvous point and orbital velocity you want, and turn over when half way there.

When you do that at 1 g, you build up outward speed so fast that the initial along-orbit speed is swamped. It takes less than an hour to build up an outward velocity larger than the initial orbital velocity! You really don't see the curves at the end of the sketch above:

(small custom python simulation; mars is red, earth is blue) When you draw the rocket path with one-hour dots, you can see it's really moving at turnover:

Finally, to answer the question about the fastest possible trip given g tolerance: That scales (in hours and g's) as $$50/ \sqrt{g_{max}}$$. If you can handle 4g for an extended time, you can be at Mars in about a day!

(Edit to include some comments)

Is this the fastest strategy?

In the simplified case (1) transfer between two straight-line moving objects (2) with no overall gravitational effects from the Sun and (3) no local gravitational effects from the starting and ending at planets, perhaps in orbit, then it's clear that a "go straight, turnover at the appropriate moment to make speed" is fastest. Not a single Gal of acceleration wasted!

How important are those assumptions?

For accelerations large enough that times are short (10 days?), curvature (1) and the Sun's 0.006g acceleration (2) don't change the strategy. You're still going as fast as you can to get to the place you need to be. For exact arrival, you've have to include it in calculations of direction and time, but this is a little bit like the difference between celestial navigation for an airplane and a sailboat: The boat moves slowly enough that navigation to a mile or two accuracy matters; much less accuracy is needed for the plane because it can get within 10's of miles and just home on the destination without major inconvenience.

Questions about departure and arrival details are more complicated; this Answer has ignored them. Leaving when your orbit is pointed at the destination is faster than leaving when it's pointed away. The initial and final orbits might be inclined, the planet orbits are askew, etc. But at large accelerations (i.e. anything near "survivability limits of humans") these are just computational inconveniences: A constant 1g flight starts off slower than many transfer orbit insertions (through you have to start in orbit; you can't leave earth's surface at 1g) but then you keep going all day, ending that day going 10X as fast as LEO in whatever direction you want.

• Since python is so ubiquitous (we even have a python tag) I usually post the python script itself or add a link to a copy in pastebin. Options for posting incidental python scripts or other code? It's hard to verify your answer without it. How do we know you've found the "Fastest orbit transfer constrained by maximum g-force magnitude"? – uhoh Mar 20 '19 at 0:18
• Well, you don’t really know it’s the fastest, but it’s hard to see how there could be anything faster than a straight line acceleration and deceleration (the algebraic part at the top). Not a single Gal wasted! The python is just to visualize that. – Bob Jacobsen Mar 20 '19 at 0:56
• I think the choice of the large 1.0g tends to hide the complexity and intrigue of the problem in a very small range of times about the straight-line time. For a moment suppose the question were about 0.1g or 0.01g; those shortest-time trajectories then would be dramatically different than a straight line. Yes, in the limit of large acceleration they would become close to a straight-line trajectory, but they would always have some interesting curvature, and I think this is the kernel of the question. – uhoh Mar 20 '19 at 2:30
• @uhoh Agreed. We know the solutions for very low thrust (ion drive trajectories) and for impulsive transfers (Lambert's problem) but it would be interesting to see the the continuous boost higer acceleration trajectories in general, and to see exactly what the small corrections do in this case -- for instance do you take off 25 hours before closest approach, or a few minutes earlier or later? – Steve Linton Mar 20 '19 at 13:13
• @SteveLinton 25 hours is just a single-digit approximation. I added some more to the end of the answer to cover some of the comments. As to the small-acceleration case: Not sure that's what's being asked, but in any case the ion-drive answer could be used to study this by raising the acceleration; no approximations would intervene. A nice paper on this for LEO to GEO transfer is issfd.org/ISSFD_1999/pdf/OC1_1.pdf but unfortunately I don't know of a similar one for Earth to Mars transfer; would be great to hear of one. – Bob Jacobsen Mar 20 '19 at 14:05