Lunar landing in cotton candy?

Question

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?

Answer 1

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.

Answer 2

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.)