Is there any way to land a rover on the moon without using any thruster, with the help of Thermocol, Cotton, Bubble wrap or any other packaging material, like we can receive from online shop?
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10$\begingroup$ In theory, sure. You just need a big enough crumple zone. But since weight is the limiting factor, I doubt that it would be feasible. $\endgroup$– PolygnomeSep 11, 2019 at 15:19
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5$\begingroup$ NO thrusters for landing or deorbiting? You need to slow down from an orbital velocity to reduce altitude and land, so would you edit your question to make it more clear? If it can use deorbiting retrothrusters, but not to land it can follow some of the ways described here pages.citebite.com/s3e3l8g4j0poy $\endgroup$– Rajath PaiSep 11, 2019 at 16:22
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8$\begingroup$ @Rajath Pai: You don't actually need to orbit the moon to "land" there. For instance, the early US Ranger probes were launched on a direct course. You just need to build a probe that can withstand abrupt deceleration from Lunar escape velocity :-) $\endgroup$– jamesqfSep 12, 2019 at 2:59
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7$\begingroup$ Would bubble wrap explode in a vacuum? Hmm. $\endgroup$– SchwernSep 12, 2019 at 5:13
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8$\begingroup$ It's called Lithobraking: en.wikipedia.org/wiki/Lithobraking $\endgroup$– InfrisiosSep 12, 2019 at 11:35
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8 Answers
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It is not practical to use this approach from orbital (1.6km/s) or escape velocity (2.4km/s), for two major reasons. The first is the acceleration reason.
The kinematic for bringing objects to a stop under constant acceleration is $$d = \frac{v^2}{2a}$$ from which we can quickly solve for the acceleration to be $$a = \frac{v^2}{2d}$$ Even with 10 meters of padding, as @zeta-band used, the acceleration would be $283000\frac{m}{s^2} = 29000g$, which is roughly the acceleration experienced by electronics in artillery shells. So in theory we could build the electronics to survive the impact.
However, this is where the second issue arises. Even if a crumple zone of arbitrary size were to be used, there would be mechanical and material issues associated with the speed of sound in the material. The problem, in a nutshell, is that in ultra-high velocity impacts, even strong materials like steel splash and crack rather than deform neatly. In order to be effective, a crumple zone must effectively reduce the speed of sound in the material, spreading the shock of impact. However the velocities involved are too high. Our crumple zone would not act like the crumple zone in a car. In fact, the pressure wave indicating the start of the impact with the ground may only reach the payload ~50% faster than the ground itself, and that is if we use a sturdy crumple zone out of a material like aluminum or steel (former better for weight).
Items like bubble wrap, cardboard, foam cushions, things that we consider soft and suitable for packing do not have a high speed of sound. They would not even transmit the ground impact force until the spacecraft itself hit the ground.
Finally, I want to debunk the idea that one could "land" (come in on a strong tangent, rather than straight down) on a long runway on the moon. In theory, this is possible. Using ultra-strong alloys, one could (barely) make wheels that could spin up to the required velocity of 2.4 km/s. However the gyroscopic issues will be severe, not to mention the wheel balancing (and what happens when a wheel breaks?). One could even go simple and just slide it out on the longest slip n' slide ever built on or off this world. But hypervelocity issues strike again. Any rubbing surface at these speeds won't just heat up, they will plasmify. Atoms in the materials will impact so hard they simply get dislodged off the material completely. Even diamonds will degrade. Its not that any of these methods are impossible in principle, they are just impractical and the engineering challenges to make them work are monumentous.
So the upshot is that there must be some form of propulsion to slow the spacecraft down to land on the moon.
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8$\begingroup$ @JamesJenkins I reference the speed of sound in the material. $\endgroup$ Sep 11, 2019 at 17:35
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5$\begingroup$ Ah, I see it is not the speed of sound on the moon, it is the speed of sound in the material, which is not dependent on air (or water) pressure. $\endgroup$ Sep 11, 2019 at 17:41
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4$\begingroup$ can't we do litho braking like we do aero braking, bouncing off the surface, scratching and loosing a bit of velocity at each turn? $\endgroup$– njzk2Sep 12, 2019 at 3:43
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5$\begingroup$ Magnetic levitation was used in the Red Moon novel. No clue if it's realistic, but I don't see any immediate reason it would be impossible $\endgroup$ Sep 12, 2019 at 8:24
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4$\begingroup$ @JollyJoker Better hope that your insurance is paid up for that maglev catcher, if there are no emergency thrusters to correct or ditch a payload that's coming in just slightly out of tolerance. $\endgroup$– notovnySep 12, 2019 at 15:00
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Well, lunar escape velocity is 2.38 km/s. So this is about the speed that a rover dropped in from orbit (with no sideways velocity) will hit at. So let's take a guess at how many g's deceleration will be. Assume it has 10 meters of crumple to stop in. It will take it about 10/2380 seconds to stop. Which is .0042 seconds. Deceleration will be 2380 / .00042 = 566,666 m/s squared. Which is about 57,823 g's.
That's pretty tough to survive and still function.
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7$\begingroup$ This is a good estimate for the order of magnitude of the acceleration, however I would go a step further and use the kinematic equation. $a=\frac{v^2}{2d} = 283000 m/s^2 \approx 29000g$. $\endgroup$ Sep 11, 2019 at 17:08
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1$\begingroup$ I Googled 2.38 km/s > MPH = 5,323.908 ; KPH = 8,568 I think we can rule out a wheels down landing as well. $\endgroup$ Sep 11, 2019 at 17:26
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1$\begingroup$ You can land sideways and bounce a while, reducing velocity on each bounce. Furthermore you only need to shed about 1.6km/s, assuming you start in orbit. $\endgroup$ Sep 11, 2019 at 17:43
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$\begingroup$ @Polygnome The bounce will either put you back on the route to Earth (if you're not in orbit around the Moon), or have the next bounce far away from the first one. Additionally, you're trying to design a spacecraft that can repeatedly bounce at huge relative velocities. I don't see how this could ever be practical. $\endgroup$– LuaanSep 13, 2019 at 10:24
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1$\begingroup$ @Polygnome By the time the MER got to the bounce, it was already slowed down considerably through aerobraking (it was designed for an impact speed of about 100 km/h, which are speeds we can handle safely quite easily). The same approach wouldn't work with the Moon, where you'd still be moving at orbital velocities on impact. $\endgroup$– LuaanSep 13, 2019 at 10:32
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Yes, this can be done by using a circular magnetic track, like for a maglev train. A spacecraft should approach with low vertical and high horizontal velocity, landing on the short straight section like an aircraft on the runway.
For horizontal adjustment, the landing part could be shaped like a funnel horizontally and V shaped vertically, allowing some imperfections of the landing approach. If say 20 km length, we have 10 seconds for precise displacement. We can move the spacecraft hundreds of meters in this time under 10 g acceleration.
The "magnetic runway" should then guide the spacecraft into closed circuit track, like a toy railway. There it could run around as a proton in a particle accelerator, gradually slowing down. The possible deceleration force does not matter much as the spacecraft can then complete many loops of the circular track before it comes to halt.
The magnetic track must be strongly angled and actually more on the outer side of the track than above the ground ("wall of death"). It needs to provide enough force for keeping the spacecraft in a circular path. The necessary acceleration is $$a = \frac{v^2}{r}$$
Assuming we have a 2 km/s = 2000 m/s velocity and accept a heavy but human survivable 10 g acceleration, we only need a circular maglev railway with radius of $$\frac{{2000}^2}{100} = 40000$$
So 40 kilometers. It may not be easy to build but definitely not something that only supercivilization could do. If we opt for 400 km radius, we can simply use the existing todays technologies to land a train, not some tiny lunar lander, under just 1 g acceleration.
For a non zero descent angle, the landing funnel can be built on the side of a mountain, in the form of an arc-shaped ramp. For instance, the Mons Huygens is over 5 km high, so can embed on the side a ramp supporting the descent angle as high as 7 degrees.
The same system can be used also to launch the spacecraft.
This looks almost real: a landing strip of maglev track enters the turn inside a tunnel (source).
I do no think that any very special magnets or anything the like would be required very much above that today's maglev already possess. A simple model railway where a train levitates never touching any part of the track and staying stable when tilted can be built with permanent magnets without any electricity.
It is out of scope of the question how did the spacecraft arrived to the destination but there are ways of space propulsion others than a rocket.
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1$\begingroup$ Cool idea, but I'd argue it's still using thrusters. The thrusters are electromagnetic rather than chemical, and on the ground rather than on the spacecraft, but we're still using an apparatus that generates an external force to slow the spacecraft down. $\endgroup$ Sep 12, 2019 at 19:28
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4$\begingroup$ @NuclearWang but the lander itself doesn't have a thruster. The lander only needs a chunk of magnetic material and/or some passive inductor, so I think it meets the criteria of the question as asked. $\endgroup$– uhohSep 12, 2019 at 22:52
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5$\begingroup$ My primary research deals with magnets of the size that would be needed for this approach. I can tell you that this approach doesn't work due to issues related to controlling the magnetic fields involved. Unlike in a particle accelerator, this machine would be handling substantial magnetic dipoles with substantial mass, and the field fluctuations for control would be enormous. Coupled with the inductance of the magnets, one could not achieve the speed of control necessary. This is an engineering issue unrelated to space travel, just thought I would enlighten based on my knowledge of magnets. $\endgroup$ Sep 13, 2019 at 0:45
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1$\begingroup$ @h22 Those models also have other correction mechanisms - most often, the model levitates, but it's also held in place through wires or strings. Static magnets alone aren't enough - they'll bounce you back off into space. It's basically the same problem as balancing a pencil on its tip. $\endgroup$– LuaanSep 13, 2019 at 10:29
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1$\begingroup$ @uhoh True, but by that criterion, it's acceptable to have a second probe with thrusters, have it clamp onto the first and perform a retrograde burn, and then detach from the first before it lands. The first lander itself has no thrusters, and needs no special apparatus to decelerate, but it seems like cheating to just move the thrusters to another object that's not technically part of the lander. This apparatus in this answer isn't materially different from the one I just described. $\endgroup$ Sep 13, 2019 at 13:03
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I'm going to take a different approach here, which only makes sense in the context of repeated landings after we already have people or robot-like things up there.
Create a large lattice of materials on the Moon, then fire the lander at the lattice.
This is effectively like giving the Moon a localized atmosphere and using drag to slow the lander down. The advantage of doing this would be greatly reduced mass on the lander, since we don't need to launch a rocket with the lander, or the fuel for said rocket.
I've never done this, nor heard of it being experimented with, and can't speak to the feasibility of it. However, there's nothing physically preventing it.
One issue would be aiming the lander that well from far away. Normally, we get in the ballpark on the way over, then do a few course corrections as we get closer. And that requires thrust of some kind. But the Moon is really close compared to other planet-like bodies, so maybe we could pull it off.
But it would also require an ability to generate a lattice on the Moon with materials available there, which would likely require mining, smelting, and refining operations. So I doubt it's anything we'd be doing short-term, if ever.
If would also require a "heat shield"-like surface on the lander that could survive impact at those speeds, which might well require more mass than the rocket we're replacing. Some kind of carbon nanotube mesh for the lattice would spread the forces more evenly, requiring less shielding, but there's no way we could create that scale of mesh on Earth right now, let alone the Moon. Steel beams would be easier (not easy) to create on the Moon, but would require substantially more shielding since the impact forces will be very localized.
Note that iron and aluminum are both present in relatively large quantities at the Moon's surface, but carbon isn't, so steel might be out of the question, requiring a structure made of aluminum. See Wikipedia's article on Geology of the Moon for large-scale deposits. An Artemis Project article, Carbon on the Moon, says carbon is found at 82 PPM in the upper 1 to 2 meters of the ground. One ton of carbon would require smelting 12000 tons of surface material, but mild steel is only 0.05% carbon, so a ton of mild steel would only require smelting 6 tons of surface materal (just for the carbon -- you'd also have to get the iron somewhere). Doable, but maybe not feasible (especially if you can't get 100% of the carbon through smelting).
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$\begingroup$ I might even add that you don't need to ship the impact material to the moon, just the building materials to hold it. I surmise that lunar regolith is a nice soft material, and given enough of it packed lightly (to keep density and cohesion down), we could just impact into the box of regolith. Still substantial acceleration, but can be handled by specialized electronics. Accuracy would likely not be an issue. We are pretty good at rocket science these days, even with inter-planetary transfers. We land inside smallish craters on Mars. $\endgroup$ Sep 12, 2019 at 13:30
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$\begingroup$ Set up a snowblower at one of the poles and make a big snowbank. No, really! (Okay, not really.) $\endgroup$– RogerSep 12, 2019 at 14:15
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I think you could do it with a rotating skyhook
The skyhook would be in a low lunar orbit, and the payload would catch the upper end at about escape velocity, or twice orbital speed. It's then swung down and backwards until it's almost stationary and released.
The tallest mountain on the moon is about 5km, so potentially it would have to withstand a 5km freefall - although I guess you could pick an orbit that avoids tall mountains and time it to land on a high plateau and 'only' freefall a km or two and crash at 50-100m/s, which still causes a few hundred g's of deceleration with a 10m crumple zone.
Or maybe the skyhook would unreel a tether of a carefully calculated length, detach the rover with millisecond accuracy and reel the tether back in before the next rotation...
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$\begingroup$ If the approaching spacecraft has no thrusters to finesse its approach to the skyhook, how big a target window can the skyhook move within? How big is its "catcher's mitt?" $\endgroup$ Sep 12, 2019 at 20:20
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1$\begingroup$ @Camille - good question. A skyhook wouldn't need to be as precise as an ISS docking because it's not trying to make an air-tight seal between two hatches, but there's an obvious trade-off between accuracy and the weight of the link. I guess it could fire off some sort of harpoon from some distance but realistically a few small thrusters would be the lightest and most reliable way to do it. They'd only need a delta-V of a few m/s, unlike the 1-2000m/s required to land from orbit. Similarly, a bigger, more expensive skyhook would give more time there is to catch it and a gentler turn. $\endgroup$ Sep 13, 2019 at 7:55
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One option might be a penetrator -- a hard dart-shaped vehicle designed to hit the ground point first at 2.4 km/s and come to a stop in a few meters of regolith. The deceleration would still be severe but specialist electronics could be expected to survive it. The speed of sound issues mentioned above would be alleviated because the penetrator is made of something hard which will have a high speed of sound.
answered Sep 12, 2019 at 11:34
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2$\begingroup$ Even if the electronics survived, the rest of the rover wouldn't. $\endgroup$ Sep 12, 2019 at 19:38
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3$\begingroup$ I missed the "rover" aspect. I think you can do a useful fixed scientific station this way, but it would be challenging to build something that could land this way and still "rove", I admit. $\endgroup$ Sep 12, 2019 at 20:11
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$\begingroup$ If the electronics survive the impact but the solar cells and cameras not, the rover ise useless. But what is the use of a rover burried under a few meters of regolith anyway? No light to the solar cells, no radio link. If the penetrator hits a point without a cover of a few meters of fine grain regolith it would be destroyed. $\endgroup$– UweSep 12, 2019 at 21:20
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$\begingroup$ But how do you control the attitude of your hard dart-shaped vehicle without using any thrusters? $\endgroup$– UweSep 12, 2019 at 21:38
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$\begingroup$ @Uwe: If the device doesn't pick up too much rotational momentum, reaction wheels should be able to provide attitude control without having to eject any material. $\endgroup$– supercatSep 13, 2019 at 17:15
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Some of the Ranger probes (including Ranger 3) to the moon had balsa-wood landers. Unfortunately, none of the ones that were working when they reached the moon carried them.
So we believe it's possible. It just comes down to how much crumple-zone you can afford to carry and how much impact you can stand.
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$\begingroup$ related: Puzzler - which spacecraft(s) incorporated real wood structural elements? and also What was the first piece of wood to reach the far side of the Moon? Or the first spacecraft? $\endgroup$– uhohSep 12, 2019 at 2:38
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10$\begingroup$ Note that Ranger 3 did use a rocket to drop the speed from km/s to something under 44ms and question is asking about rocket free descent. Does provide an interesting data point on volume of padding needed even at that low speed. $\endgroup$ Sep 12, 2019 at 8:01
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There is a way to land on the moon using less energy than it requires to reach Lunar orbit. It requires a megastructure, but one that we have the materials to build today.
Just dangle a rope down from the moon to (roughly) geosynchronous orbit distance from the Earth.
A ship would have to use thrusters to leave the Earth, but orbit is fast not far away. The "rope" (skyhook) would be moving much slower than geosynchronous orbit; so a non-orbital trajectory -- a ballistic trajectory -- from Earth would work.
In theory you could even do a high-altitude balloon launch (to get past the atmosphere), then fire the package up to the moonhook. Given perfect aim you could reach the moonhook with near zero relative velocity (the moonhook isn't in orbit, it is dangling; you aren't in orbit, you would be about to fall), grab the moonhook, then simply crawl your way up to the moon.
In practice you'd probably use thrusters to leave Earth and to refine your aim as you approach the moonhook, but nothing like actually landing on the moon.
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$\begingroup$ I don't think this will work the way that you planned. Even if a rope was dangled from the moon down to roughly GEO, once the object crossed the moon-earth L1 point it would start accelerating toward the moon up to nearly lunar escape velocity. It would need a braking system of some kind, which includes a system to manage the angular velocity as well, since this is in a rotating reference frame (not the same as free fall straight down). This speaks to thrusters of some sort. $\endgroup$ Sep 13, 2019 at 13:58
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1$\begingroup$ @Quietghost Yes, it is a really tall tower up to L1, then a rope counter-weight on that tower. $\endgroup$– YakkSep 13, 2019 at 14:01
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$\begingroup$ If you have the tech and materials to dangle a rope from the moon, it would seem to be a better idea to use it to build a pair of proper space elevators -- one from earth to GEO-and-a-bit, another from the moon to L1-and-a-bit. $\endgroup$ Sep 13, 2019 at 18:00
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$\begingroup$ @HenningMakholm See the link -- a moonhook is trivial compared to an earth-to-space beanstalk. Beanstalks do not like being built on planets with as much gravity as Earth, and the fact that the moon orbits the Earth makes a beanstalk (aka, moonhook) there even easier, as it means you can counterweight using Earth gravity instead of Centripetal force. A Centripetal moon beanstalk has a problem because of how slowly the moon orbits; a moonhook down past L1 fixes that. Saying "you might as well build a beanstalk" is like "you can add, you might as well break RSA". $\endgroup$– YakkSep 13, 2019 at 18:48