I've played a lot of Kerbal Space Program and always wondered: When in orbit around a body without an atmosphere, which trajectory does a lander take to approach and land using minimum delta-v, without wasting too much time, and landing at the exact right place that has been chosen during mission planning?

  • $\begingroup$ "using minimum delta-v" That would be to reduce the perihelion (not sure of the term as it applies to random masses, that is more about stars) to a point where the orbiting object is not coming back out of the atmosphere due to drag. The vector to fire the engines would be in the direction the craft is moving (to slow it down and reduce the perihelion of the orbit). The place to do that burn is pretty much the exact opposite side of the larger object that is being orbited, to where you intend to reach the perihelion. But really I think this question is 'too broad' for a single answer. $\endgroup$ Jul 26 '15 at 20:32
  • $\begingroup$ @AndrewThompson "periapsis" is the general term, "perihelion" refers specifically to the low point on orbits around our sun. $\endgroup$
    – 1337joe
    Jul 26 '15 at 20:36
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    $\begingroup$ Are you talking about a landing on a body without an atmosphere or an atmospheric re-entry on Earth? $\endgroup$
    – Philipp
    Jul 26 '15 at 21:46
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    $\begingroup$ Very, very carefully. $\endgroup$
    – Mark Adler
    Jul 26 '15 at 22:33
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    $\begingroup$ You can easily exercise "suicide-burn" landing in KSP with KER mod, it calculates (approximate) suicide-burn duration.(And then your mission fails as your lander rolls down some crater ;p) $\endgroup$
    – PTwr
    Jul 27 '15 at 10:33

If the body on which you want to land has no atmosphere, you need to reduce the spacecraft velocity using the onboard propulsion system.

If you perform an Hohmann-like manoeuvre you end up having a touch down which is not vertical and the resulting $\Delta$v is actually an underestimation of the real value.

Hohmann-like manoeuvre

Instead, if you try to perfom a landing like the one shown in the second figure, you will have a vertical terminal trajectory but the $\Delta$v required for the break makes this manoeuvre unfeasible.

second type of manoeuvre

A better estimation of the actual $\Delta$v needed can be found applying an optimal control theory, which will lead to a trajectory more like the one in the third picture.

enter image description here

EDIT: The optimal control theory approach

Like any optimization problem, you need to set a goal (what you want to optimize, in this case could be the amount of fuel used for the manoeuvre), then choose a model for the process and parameterize it (to have a finite number of parameters to be tuned). Finally tou have to define a cost function and some constraints. The third case I have presented represents a 2-phases landing:

  1. coasting phase, which is ballistic
  2. powered descent phase

For this case, you need also to make some assumptions about the panet gravity field (which is assumed to be perfectly spherical) and also you have to assume the the manoeuvre is 2D and that during the power descent the magnitude of the thrust is constant.

When you have set all these things you are ready for the numerical optimization which will give you the resulting trajectory which will minimize the amount of fuel needed.

All what I have told you was part of a lecture I had last year. The lecturer had developed all this during his master thesis and you can download it here: http://hdl.handle.net/10589/79781


The minimum $\Delta V$ is effectively a Hohmann transfer. You would de-orbit just enough to barely touch the surface at periapsis, and right at the surface you would do an instantaneous impulsive burn (also known as a "suicide burn") to exactly cancel your velocity relative to the surface. Done! A perfect landing.

In the real world however you don't have instantaneous burns. So you need to start sometime before you get to the surface and do a gravity turn to minimize the gravity loss during the finite burn. The higher your thrust-to-weight ratio, the shorter the burn and the less gravity loss that you'll have.

Then the time it takes will be the time it takes. If your objective is to minimize $\Delta V$, then you can neither "waste" time nor be more efficient with the time.

As for the accuracy, you want to have the landing site in the plane of the orbit. Then its just a matter of timing your de-orbit properly. The residual error can be taken out with a smidge more $\Delta V$ and some means of sensing your location relative to the intended landing point, ideally before you start your landing burn.

  • $\begingroup$ Is there a way to calculate the landing burn time depending on the TWR ? $\endgroup$
    – Magix
    Jul 27 '15 at 5:41
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    $\begingroup$ Well, you can start with the instantaneous impulse required and divide by the thrust. That will get you in the ballpark. Then you'll need to numerically integrate the trajectory to get the actual burn time. $\endgroup$
    – Mark Adler
    Jul 27 '15 at 6:27
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    $\begingroup$ @MagixKiller: Also note that a minimum-ΔV landing trajectory is commonly called (particularly in KSP jargon, although the term itself is far older) a "suicide burn". The reason being that minimizing ΔV means delaying the burn to the last second possible and then burning at maximum thrust, so that you'll have no safety margin, very little time to react, and no easy way to abort the landing (since your thrust is already maxed) if you find that you've miscalculated. There's a reason why real-world landing trajectories don't usually go for absolute minimum ΔV. $\endgroup$ Jul 27 '15 at 9:04
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    $\begingroup$ I like the term "suicide burn". The way it's really done is to bring the lander to some height above the landing location, measured in meters, and to go down from there with a constant velocity to the surface. That allows for some uncertainty in that height. $\endgroup$
    – Mark Adler
    Jul 27 '15 at 17:17

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