When spacecraft land on Earth, they use the atmosphere to slow down.
On the Moon there's no atmosphere, so spacecraft must instead use rockets to slow down from orbital velocity.

Would it be possible to use a large pool of fluffy material or low density foam to slow down instead?

Lunar orbital velocity is "only" 1,700 m/s, compared to 7,800 m/s for Earth, so presumably the forces involved are much smaller. I understand the pool would have to be large, as even at 10 Gs, several kilometres are required to come to a halt.

But would hitting a pool at 1,700 m/s smash a spacecraft apart, regardless of how thin and fluffy the contents are? Do we have good materials for this purpose?

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    $\begingroup$ I wanted to argue this is off-topic but after thinking about it I couldn't. It's really intriguing, and if it were possible it would be useful especially if it were reusable. 98 m/s^2 for 17 seconds is almost 15 kilometers, yikes! $\endgroup$
    – uhoh
    Commented Oct 24, 2020 at 16:29
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    $\begingroup$ I'd model this as hitting a non-Hooke spring which exerts an even 10Gs of force until completely compressed. $\endgroup$ Commented Oct 24, 2020 at 17:32
  • $\begingroup$ I have heard of theories of moon dust runway landings from orbit. Sounds scary - a pebble would hit like an air piercing slug. $\endgroup$
    – ikrase
    Commented Oct 25, 2020 at 1:59
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    $\begingroup$ All this is interesting but I'm wondering, and based on what has already been mentioned here, if we wanted to go down this route, why bother manufacturing "cotton candy"? If "cotton candy" needs a containment cylinder some 15 km high, why not have and artificial atmospheric column, where the containment cylinder is filled with a non oxidizing gas & possibly have the landing craft act like a piston as a means of losing kinetic energy? Just something else to think about ... $\endgroup$
    – Fred
    Commented Oct 25, 2020 at 14:41
  • $\begingroup$ Lunar bouncy castle? Stay-pufft marshmallow man? Massive pile of shaving cream? $\endgroup$
    – DrSheldon
    Commented Oct 27, 2020 at 3:19

2 Answers 2


Decelerating from 1700 meters a second at even quite high accelerations like 10G (98 meters a second) involves a fair bit of distance, nearly 15 km per Uhoh in comments.

For the first segment of deceleration while traveling above the speed of sound in the materials this will look more like aerobraking where the deceleration force will be produced by punching a hole in the cotton candy. The mass of the candy being accelerated will produce a force balanced by craft deceleration.

So for the first second of deceleration from 1700 m/s at 10G each kg of craft will need 98 newtons of force, which would come from accelerating 57 grams of material to 1700 m/s.

This material would be spread along a 1700 meter path and if we model our payload as 1kg of water in a 10cm cube that gives a front face of 0.01 square meters for a total volume of 0.01*1700 = 17 cubic meters.

So the first section of our cotton candy brake pad needs to have a density of 57g/17 = 3.3 g per cubic meter. Which is pretty low, air at sea level is over one kg per cubic meter. Some random googling got 1200 g per cubic meter for cotton candy so far to high.

It is not of course required to make the cotton candy at human edible density, so this structure starts to look like a tunnel or channel threaded with a carefully engineered web where the first 10 kilometers or so are only a couple of strands per meter, each strands as fine as possible to reduce the ablative forces on craft itself, which would be suffering something akin to a vigorous sandblasting during this process, though 1700 m/s is not totally impossible to armor against.

As the density of the craft increases the needed number of threads increases as well, closer to actual cotton candy but so do the ablative forces on the shielding surface of the craft.

Unsure how to model the moment to moment interaction of the craft shield surface with the incoming threads - the average energies are lower than for earth re-entry so presumably achievable but each thread strike will involve substantial energy transfer.

So the final product is 15-20 km of pathway/tunnel of carefully engineered threads that will need to be fully replaced after each use, and the craft itself will need to be engineered to withstand this process.

  • $\begingroup$ It would probably be easier to use a capture wire to snag and decelerate the craft similar to how aircraft carriers catch jets. $\endgroup$
    – Dragongeek
    Commented Oct 25, 2020 at 12:42
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    $\begingroup$ @dragonGeek, Speed of sound in steel is 5km/s so not outright impossible coming in at 1.7kms but whole bunch of exciting physics happen when you hit that wire. For starters if it weighs much just bringing it up to orbital speed in the hook will impose massively more than 10G. Would suggest a magnetic method might be more sensible than anything involving physical objects contacting at 1700ms. $\endgroup$ Commented Oct 25, 2020 at 13:01

I think you're coming at this the wrong way. Remember, escape velocity is a scalar quantity--so long as it doesn't intersect the planet it doesn't matter what direction you're going. That means you can also approach in any direction.

A 15km landing pit would be a mammoth feat of engineering, but since the approach can be in any direction we don't need a pit at all--do the same thing with a 15km track instead.

(Note that if you're already in orbit this is even easier, you just lower your periapsis to the altitude of the track.)

You still have a problem with getting the mass density right but since it's spread out horizontally that will be a lot easier.

(Note, however, that instead of any resistance medium I would build a maglev track.)

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    $\begingroup$ I mean, obviously it's horizontal. Orbital velocity is not vertical. $\endgroup$ Commented Oct 25, 2020 at 7:55
  • $\begingroup$ @SE-stopfiringthegoodguys disagree strongly and recommend we don't assume everyone is a mind-reader and refrain from condescending "I mean, obviously it's..." type comments. Depending on the history an orbit could intersect the surface at oblique, non-tangential angles. Answers are posted for all readers and supplemental answers may benefit others even if they aren't what you were hoping for. $\endgroup$
    – uhoh
    Commented Oct 27, 2020 at 1:30
  • $\begingroup$ +1 because I did actually assume an oblique descent rather than a nearly parallel trajectory and visualized a tall, inclined chute. Your answer caused me to face-palm and exclaim "doh!" which is a good thing! $\endgroup$
    – uhoh
    Commented Oct 27, 2020 at 1:31

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