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An asteroid approaches and the Moon 'catches' it in the same way that a sports player catches a ball - that is to say by matching the velocity of the hand to that of the ball.

Could a lucky slingshot approach cause this to happen?

Assuming that the Moon has no atmosphere, my intuition tells me that there must be a direction and velocity such that an asteroid can do this. Does mathematics say otherwise?

Gravity assist

In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by is the use of the relative movement (e.g. orbit around the Sun) and gravity of a planet or other astronomical object to alter the path and speed of a spacecraft.


NOTE

I mean that the asteroid approaches the Moon in a direction opposite to the Moon's Earth orbit. It then starts to swing around behind the Moon and just happens to contact the Moon's surface at the time the exact matching of speed occurs. Thus landing relative to the surface at zero velocity in any direction.

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    $\begingroup$ Think about it in reversed time. There's no way to launch from zero velocity without actively altering your momentum. The same goes when trying to land. $\endgroup$ – MackTuesday Nov 30 '18 at 2:41
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In order to land on the Moon, you must, at some point, be moving towards the Moon (decreasing your distance from it, to be more precise, you may also be moving sideways) and close enough that the Moon's gravity dominates that of the Earth and the Sun. From that point on, your kinetic energy (relative to the Moon's centre of mass) can only increase as you get closer to it (you are converting potential energy to kinetic), so gravity cannot slow you down. So you will always crash at roughly the Moon's escape velocity (2.3 km/s) or more. You might crash straight in, or graze the surface while moving almost horizontally, but there will be a substantial relative velocity.

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    $\begingroup$ To get a genuinely smooth contact, you would need the equator of the Moon to be spinning exactly at escape velocity, so that the asteroid, also coming in at escape velocity could "dock" with it. The problem with this is that the rocks at the equator wouid escape, ie fly off into space. If the Moon was small enough that the strength of the rock held it together then this could work, but not if it's meant to be held in by gravity. $\endgroup$ – Steve Linton Nov 29 '18 at 11:55
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    $\begingroup$ The asteroid would also have entered into the gravity wells of earth and sun, and taken on kinetc energy from that, so the moon would have to spin significantly faster than escape velocity. $\endgroup$ – bukwyrm Nov 29 '18 at 12:01
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    $\begingroup$ And then SevenEves happens. $\endgroup$ – Carl Witthoft Nov 29 '18 at 16:16
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    $\begingroup$ @CarlWitthoft only if the asteroid is going at relativistic speeds, or made of antimatter. To actually gravitationally disrupt the Moon takes an insane amount of energy. $\endgroup$ – Steve Linton Nov 29 '18 at 17:21
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    $\begingroup$ @MSalters Asteroids usually come from farther out than earth orbit, that makes them automatically come from 'higher' up in the suns gravity well - any movement towards earth will take them deeper into the suns well. $\endgroup$ – bukwyrm Nov 30 '18 at 16:02
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@SteveLinton's answer is right, no matter how gently you try, by the time you get to the surface the Moon's gravity will have accelerated you to something like 2,400 m/s. There are ways to use the gravity of the Earth and Sun to make a tiny reduction in this, but it's a very small effect.

The simplest way to argue this is that rocks on the Moon don't suddenly, spontaneously jump up and fly into deep space. Classical mechanics works basically the same forwards and backwards in time (in lossless systems as @Mołot importalty points out). So if something can not happen in one direction in time, it can't happen in the other either.

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    $\begingroup$ Hmm, yes. The argument about rocks jumping off does make absolute sense and gives me an intuitive feel for @SteveLinton's answer. $\endgroup$ – chasly from UK Nov 29 '18 at 12:43
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    $\begingroup$ I think you're getting the idea, but to help you along, think like Einstein: Imagine the Moon as a bowling ball rolling along a large trampoline surface. At any instant the surface of the trampoline is deformed into a deep well at the ball. No matter how you approach it, in order to land on the Moon, you have to fall into that gravity well. Whether you or it is moving to begin with doesn't matter - the Moon is always below you. $\endgroup$ – Oscar Bravo Nov 29 '18 at 13:35
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    $\begingroup$ @OscarBravo that's a really good way to look at it; if you have a few minutes, consider posting an answer. Maybe add one of those plots of a gravity well, or an image or video of those displays in science museums where the balls orbits in a concave surface representing the well. $\endgroup$ – uhoh Nov 29 '18 at 13:46
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    $\begingroup$ It always amazes me that these gravity-well models on a rubber sheer require gravity to make them work. It seems a bit of a cop-out. $\endgroup$ – chasly from UK Nov 29 '18 at 16:03
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    $\begingroup$ "Classical mechanics works basically the same forwards and backwards in time." — only in vacuum and without fiction. Happily, this condition is fulfilled in this case. $\endgroup$ – Mołot Nov 29 '18 at 23:52
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In theory a soft landing is possible, although not with our moon.

As others have said, the rock inherently comes in at a minimum of escape velocity. You can't actually spin the moon that fast or it would fly apart--but what if your object is a solid piece of rock spinning at a high rate? It barely grazes the moon, it's spin makes up for the lack of enough spin of the moon. It bumps lightly and eventually rolls to a stop. (Note that it will have to be spinning far above it's escape velocity and thus must be held together by chemical bonds, not gravity.)

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    $\begingroup$ Sorry? What makes it stick to the Moon? It comes in on a tangent at ~2 km/s; at the moment of contact it is roughly stationary with respect to the surface. Then it keeps going. At 2 km/s. The moon's surface, now peels away as the Moon rotates under it. To stay put, w.r.t the surface, it needs to start rotating with the Moon. So it needs to to hang on rather tightly to overcome the centrifugal force. It's like running along and jumping onto a merry-go-round; you need to grab onto a pole pretty firmly as soon as you arrive. $\endgroup$ – Oscar Bravo Nov 30 '18 at 7:23
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    $\begingroup$ Nice idea. Such a material would need a tensile strength on the order of tens of gigapascals. Monocrystalline silicon should do the trick. $\endgroup$ – MooseBoys Nov 30 '18 at 7:34
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    $\begingroup$ @OscarBravo if the object had just slightly higher energy than escape, the slight bump on the surface (with low relative velocity) could be enough to cause capture. It would need to do a bunch of highly eccentric orbits but could eventually slow to surface orbit speed and roll to a stop. $\endgroup$ – MooseBoys Nov 30 '18 at 7:37
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    $\begingroup$ If the rotation speed was similar to (or greater than) escape velocity the asteroid would simply brush the surface and fly away. There would be no way to transfer the energy in order to slow down the asteroid. $\endgroup$ – Ister Nov 30 '18 at 8:49
  • $\begingroup$ @Ister Every brush would lower the apoapsis until it slowed to orbital velocity. $\endgroup$ – Loren Pechtel Nov 30 '18 at 23:27
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I don't know the details, but in the right configuration, if two bodies that are orbiting each other approach a third body, one of the bodies can be captured into orbit around the third body and the other body is ejected from the system. A considerable part of the kinetic energy acquired by one body from gravity is eventually carried away by the other.

Perhaps an extreme case can be constructed where the captured body is actually left at zero orbital velocity on the surface of the capturing body.

As another answer notes, classical mechanics is reversible, so the reverse sequence is that an incoming body passes near the Moon, gravitationally picking up and carrying away a mass sitting on the surface of the Moon.

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    $\begingroup$ Yes, I think this will work as a three-body problem. The most simple case would be two billiard balls colliding just above the surface, leaving one of them with zero velocity while the other bounces off into space. I think you're right that the interaction could be via gravity rather than a collision, but that scenario is much harder to visualise or calculate. $\endgroup$ – craq Dec 2 '18 at 22:01
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I'm going to take a different approach here. Let's suppose that such a rock does in fact land on the surface of the moon. The reason I do this is to show a more substantial reason why it cannot happen.

We know to begin with that the only forces that could be acting upon the meteor are gravity and electromagnetic forces. We know this for a fact because it is in the vacuum of space and those are the only two forces that could feasibly interact without a medium. We know that it must be moving towards the surface of the moon or have a component of motion in that direction. The reason for this is that the meteor moves towards the moon and will slow down as it reaches it. Since we are only considering the moon and the meteor we know that the moon has to be pushing on the object so as to decelerate it. We will assume that motion tangential to the ground on the moon is somehow matched by trajectory. This is possible since the object could have a tangential motion speed equal to that of the moons rotation and that speed wouldn't change. However it would get faster as it approached the surface in the sense that it has to travel a greater distance to keep up when on the surface vs in orbit. So we mean rotation speed of the moon at the surface and not in orbit. This also means that the exact speed is determined by the precise landing location's elevation. Since there are many craters on the moon and it is not perfectly flat this does make the situation less likely.

However, now you have to stop and think about this for a moment. The moon is pushing the object away in such a way that it decelerates it. Now unless this force causes the meteor to fracture at the surface, this means that the moon can push the object away and there is nothing that will stop it from immediately doing so. Meaning that the meteor will hit the moons surface very lightly (can't be exactly at 0 velocity or it would stop midair) and then start being repelled by the moon and be launched into space.

However, this situation will not occur because the moon is not magnetically charged as far as I am aware, and furthermore such a planet with a charge greater in magnitude then the force of gravity would likely be unstable and start repelling itself. The force of gravity is what we primarily see as the strongest non-contact force in the universe specifically because (barring theoretical dark matter) it does not cancel.

The only other alternative is that something external to the moon pulls on the meteor causing it to decelerate and that said object slowly stops pulling on the meteor such that the acceleration is almost 0 at the surface of the moon. However your scenario does not include such parameters so I cannot justify it as being a valid case.

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