Could a trajectory exist that gave an object kinetic energy relative to the Earth/Moon that somehow passes zero as the Earth moved across the other's path, tugging on it? And having it happen at a potential energy minimum (e.g., a flat place on the surface of the Earth or Moon)?

I had been wondering about this as a way to soft-land on the moon with minimal $\Delta\ v$. A recent question mentioned something similar as a way to soft-land on Earth, but didn't appear set up to cancel the relative kinetic energy. My question also doesn't specify a keplerian orbit for the "lander", just whether a trajectory could exist and what features might describe a trajectory satisfying this scenario.

A compelling answer would show quantitatively the potential energies being considered and the range of relative kinetic energies for which a solution, if any, exists. I presume the lander's relevant kinetic energies come from the E/M system moving in orbit around the sun, the moon around the Earth, and their rotations ranging from effectively zero at the poles to an absolute maximum at their equators. I'm also presuming that inclinations, tilts, topography, and (Earth's) atmosphere don't drastically effect the answer's validity unless of course "making it" actually relies on any of those. Since some responses so far suggest it might be possible only elsewhere in the solar system, an answer that lets other planet and maybe moon masses, radii and velocities be substituted would be extra useful, though not necessary.

  • $\begingroup$ Most of the energy required to go from moon orbit to landing on the moon is in removing the velocity the object required to be in orbit. So I'm thinking 'no - no such magical place exists'. $\endgroup$ Jul 8, 2014 at 1:57
  • $\begingroup$ @AndrewThompson Even as I initially wrote the question, I didn't want to restrict the "lander" to being in an orbit, per se, just "what, if any, trajectory would yield the desired result, or if none, why?" Thinking perhaps a retrograde orbit might be able to yield such a trajectory or come close, I might have muddled the question to the point that it makes you think "orbit" when I mean "trajectory". Whether some orbital arrangement could easily be converted into such a trajectory would be a much different question. Maybe I'd better remove some of the offending text. $\endgroup$
    – lionel
    Jul 8, 2014 at 5:17
  • $\begingroup$ On fast spinners perhaps, but they would have to rotate close to their surface orbital speed. And these would be horizontal landings, so their surface also has to be flat without any mountains intersecting your strictly ballistic low altitude insertion trajectory. I.e. a shallow attack angle w.r.t. the surface, that rotates close to your approach velocity. $\endgroup$
    – TildalWave
    Jul 8, 2014 at 11:30
  • $\begingroup$ What distinguishes a "trajectory" from a "Keplerian orbit"? $\endgroup$
    – DJohnM
    Jul 8, 2014 at 18:33

4 Answers 4


Proof By Contradiction:

Assume such a "Soft" Landing Trajectory (SLT) exists.

Then anything that happens to be sitting at the SLT landing spot could be "launched" into the reverse SLT with the same tiny residual energy input. Just run the equations backwards in time...

Do we see chunks of the moon just drifting off?

  • 2
    $\begingroup$ This is a very elegant answer. $\endgroup$ Jul 8, 2014 at 21:52
  • 1
    $\begingroup$ This is T-symmetry that is applied here? I'm not clear on when in orbital mechanics it can be applied and when (transfer of angular momentum maybe or something else?) it can't be applied. I agree that the approach is elegant, interesting, and promising, just not quite enough specifics written here yet to call proof. $\endgroup$
    – lionel
    Jul 10, 2014 at 22:37
  • $\begingroup$ @lionel I tend to agree with your take, T-symmetry cannot be applied here. For one, soft landing still assumes some impact force being absorbed by lithobraking. There's no T-symmetry analog to this, gravity as we know it doesn't work as a repulsive force (unless, possibly, if we include antimatter into the mix). Or in a different example, applying time reversal to horizontal landing trajectory would require a surface propulsive force as an inverse to friction. And on bodies with atmosphere, buoyancy without drag... $\endgroup$
    – TildalWave
    Jul 12, 2014 at 2:09

I don't think so, at least for the Earth/Moon system. It's tempting to imagine, say, an Earth-centered elliptical orbit that just touches the Moons surface, but that won't work out well.

Take a look at the Wikipedia pages for "Sphere of Influence (astrodynamics)" and "Hill sphere." Once you get within about 60,000km of the Moon, the Moon's gravity dominates. From there on, to a first approximation, you're falling towards the Moon from whatever initial trajectory you entered on.

So, think about it as a single body problem: is there any trajectory you could fall from space on that has you hitting the ground at an acceptable speed? Start with your spaceship at the edge of the moon's Hill sphere. At the very least, it has to be approaching the moon with velocity of zero or greater--otherwise, it would move away from the moon back into an Earth orbit. So, the lowest speed it can reach the Moon's surface with is what it would have falling from 60,000km.

As TildalWave pointed out, you might be able to pull something off if you're landing on a fast spinning object. Not sure if that's true, the thing might have to be spinning faster than its orbital velocity for that to work, but there's probably some narrow window of barely stable fast spinning and lithobraking that would work.

I think your idea will only work for bodies that are smaller larger than their own Hill sphere--in that case, you can treat the spaceship as in orbit around something else until it reaches the surface.

  • $\begingroup$ "I think your idea will only work for bodies that are larger than their own Hill sphere" My understanding of the Hill sphere suggests that if a body is so rarefied as to be larger than its own Hill sphere, it would be 'torn apart' by whatever is depressing the sphere below the 'outer surface' of the object. Or to put that another way, I don't think the definition of the term allows for it to be smaller than the physical size, at least not in any stable object. $\endgroup$ Jul 8, 2014 at 14:34
  • $\begingroup$ I agree for planet sized bodies, but I think it's possible if things are small enough that mechanical forces are relevant, e.g., the Space Shuttle has a Hill sphere of 120cm, so its own gravity doesn't bump it up against things. Thinking about it, I bet you're right, Hill sphere < radius can only be true for objects so small their escape velocity is negligible anyway. $\endgroup$
    – user42551
    Jul 8, 2014 at 15:08
  • $\begingroup$ There is a Kerbal Space Program mod that adds a planet called 'Inaccessible' which, at most latitudes, rotates not only faster than orbital velocity, but faster than escape velocity for the Inaccessible system.(it's a devil to land on. You have to hit the poles, and spot-on) Such a planet could never form, natch, but soft-landing-trajectories are only possible on rapidly-rotating(or very light) planet-like things. $\endgroup$ Jul 9, 2014 at 18:49

A recent question mentioned something similar as a way to soft-land on Earth, but didn't appear set up to cancel the relative kinetic energy. My question also doesn't specify a keplerian orbit for the "lander", just whether a trajectory could exist and what features might describe a trajectory satisfying this scenario.

The specific gravitational potential follows a simple $1/r$ function. But of course, neighboring bodies exist with their own gravity well. So you can be at the same potential in two different places. This is my best thinking of what these questions are trying to communicate.

Let me use the xkcd gravity wells infographic to illustrate those thoughts. (license is non-commercial CC-SA-2.5, so this is actually legal use)

weird idea

I can't say that you can't do something like this, but just that there's no obvious way with rocket technology. In quantum physics, things jump straight through potential barriers like this all the time. But our ships are (extremely) sufficiently non-quantum.

However, there's a much more simple manifestation of such a "jump", through space elevators/tethers. Consider a space elevator from the moon's surface to EML-2. You could make the space elevator fully self-powered, if delivering material from the moon's surface to high Earth orbit. Just have material on the "up" side of the elevator get pulled by material going down the "down" side of the elevator. Since the "down" side isn't limited by the moon's surface, you can get energy out of the process. You can also deliver things from a high Earth orbit (really a pseudo-orbit, I'm glossing over some details) to the surface of the moon in a zero-energy process. Main drawback is that you need a space elevator in the first place.

  • $\begingroup$ I love the xkcd illustration, but don't see how it captures, for instance, the various motions as described at the start. It's possible to land a ball on a plate with zero kinetic energy relative to the plate (notwithstanding that here, the moving plate exerts gravitational force). Now it may be that some math shows it impossible given the actual kinetic and potential energies of an object approaching the Earth/Moon regardless of relative trajectory. But that has yet to be demonstrated here. $\endgroup$
    – lionel
    Jul 8, 2014 at 16:34

The answer is yes if you only land part of the craft. You use a "wheel" of long lines or whips (one is enough) that rotate so that as they pass the landing point or orbit, the cable end is like a huge wheel running on the surface and a payload can be set down with zero velocity. The complete dynamics of inter-solar system travel with whips was described by the late Robert Forward. His company after he left Hughes Research, Tethers Unlimited, has made some devices for NASA and works on various tether ideas.

It has potential for a Moon landing x-prize - a no-motor no-fuel soft landing.


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