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Is it possible for a metal (or dirt/ice) asteroid of 1km average diameter to enter the Earth-Moon system in such a way that the asteroid behaves as if it were dropped from a height instead of as a high-speed impactor?

I'm imagining a non-catastrophic asteroid "landing", and wondering if anything other than probability prevents it.

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    $\begingroup$ It'd probably have to be slowed by the moon first (or be slow to begin with) just perfectly to enter an aerocapture with earth that slowly reduces the apoapsis after multiple passes without dipping too low. Then again this is just conjecture. I have no idea if asteroids could survive an aerocapture, and it would depend on the mass, shape and various other factors such as how much heat resistance the asteroid has. Regardless... the landing will not be pretty, even over water. Imagine a boulder dropped from a plane at best case scenario. $\endgroup$ – Magic Octopus Urn May 11 '18 at 14:02
  • $\begingroup$ @MagicOctopusUrn Right, I wouldn't expect a gentle landing. A giant rock falling from several km would not be something to observe up close. $\endgroup$ – Eric Hauenstein May 11 '18 at 14:17
  • $\begingroup$ sciencealert.com/… - there's a few examples... Haven't been able to find much on the logistics of the entry into the atmosphere though. I'd assume with how old those instances are, any posts about the trajectory would be conjecture. $\endgroup$ – Magic Octopus Urn May 11 '18 at 14:28
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    $\begingroup$ My first thought is "unlikely" because the chances of a low velocity object near the Earth accelerating and escaping to heliocentric orbit are not likely either. For some types of problems you can run a problem backwards in time since gravity is a conserving force. In a strict circular 3-body (CR3BP) scenario there is a conserved quantity called the Jacobi integral so whatever the answer turns out to be, the proof will likely come from that. $\endgroup$ – uhoh May 11 '18 at 15:35
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The slowest you can have the asteroid enter Earth's atmosphere without active propulsion is LEO orbit velocity. It is indeed possible to have such an object enter the Earth-moon system, do a very close gravity assist flyby of the moon with an approach V∞ equal to the moon's orbit velocity around Earth (~1.02 km/s), and come to a dead stop with respect to Earth, thus falling straight down to Earth. [This is a variant on the "stop'n'drop" airless-body landing technique I developed while at JPL] That gravity assist orbit has a hyperbolic bending angle of ~94°, so it's nothing out of the ordinary. Small variants on approach angle and V∞ result in non-zero but small residual velocities that don't affect the ultimate outcome much at all: the object drops from ~375,000 km to Earth, impacting at ~11 km/s. That velocity gives it a specific energy of ~60 MJ/kg, and 4 MJ is the equivalent of ~1 kg (~4 sticks) of dynamite—Whack! You get a result something like the Barringer meteor crater, just much larger and more damaging. The probability of such a trajectory is exceedingly small.

It is also possible to have the lunar gravity assist place the object on a trajectory that just grazes Earth's atmosphere. The object then does a long series of aerobraking (note: not aerocapture!)(The Wikipedia article is more informative) passes through the atmosphere, slowing its periapsis velocity until you get a near-circular orbit with the periapse still aerobraking, and ultimately the orbit decays and the object enters—at nearly 8 km/s, 32 MJ/kg specific energy. The probability of this kind of entry is so small it makes the stop'n'drop entry look commonplace!

But assuming that grazing entry scenario from LEO with the 1 km object, iron with a little nickel (see below), its density will be roughly 8 gm/cm^3, or 8000 kg/m^3, giving it a ballistic coefficient of roughly 5.3 million kg per square meter of projected surface area. For comparison, the Apollo CM had an entry ballistic coefficient of ~490 kg/m^2. The atmosphere will remove less than half this object's speed before it impacts (the back-of-envelope calculation I did suggested ~25%, but an accurate estimate needs a full-up integration), despite the grazing entry, so 10-20 MJ/kg of impact energy, still several sticks of dynamite per kg. The object gets blasted into little pieces. At these speeds there's no "rolling contact" as someone mentioned in responses to the similar question.

Metal asteroids are thought to be the broken-up remnants of the core of a once-planet (possibly more than one) that was competely disrupted by one or more collisions. The planet(s) were large enough that they fully differentiated, so the really heavy stuff like metals, mostly iron and nickel, sank to the core. Heavier silicates (rock-forming minerals) formed a mantle, and lighter silicates formed a crust. The meteorites (and the asteroids they came from) with mixtures of silicates and metals are from the core-mantle boundary region. Meteorites and asteroids that are purely metal are from farther down in the parent body. At some time during the planet formation process those metals were liquified—melted—by the heat of formation plus subsequent radiogenic heating. If that core solidified before the disruption event, a slow process reminiscent of annealing, the shards remaining might be relatively free of cracks. If not, and big blobs of liquid metal were exposed to space, they would cool very quickly and probably develop significant cracks.

The uncracked objects would probably survive to Earth's surface relatively intact, just some stuff spalled and ablated off the surface. I calculated the stagnation pressure at the leading face of the 1-km object at LEO orbital velocity (i.e., assuming no atmospheric deceleration before reaching sea-level atmospheric densities) and it is ~35-40 MPa, where the compression yield strength of iron is listed as 8-17 GPa, ~2.5 orders of magnitude greater. Pre-cracked objects would probably break up.

The orbital dynamics for non-metallic objects would be the pretty much the same as for the metallic objects, until aerodynamic forces became large. Aerobraking would go a bit faster due to a smaller ballistic coefficient. But with much lower strengths, they would break up in the atmosphere, like the Tunguska or Chelyabinsk events. The solar system seems to have it in for Russia! :)

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  • $\begingroup$ This is a really intriguing answer! I'm trying to reconcile two sentences in the first paragraph; "The slowest you can have the asteroid enter Earth's atmosphere without active propulsion is LEO orbit velocity." together with "... and come to a dead stop with respect to Earth, thus falling straight down to Earth." I'm stuck here because they appear to be contradictory which means I'm missing something. Can you help me sort this out? $\endgroup$ – uhoh May 12 '18 at 2:17
  • $\begingroup$ Oh, "...the object drops from ~375,000 km to Earth." I see, never mind. $\endgroup$ – uhoh May 12 '18 at 2:30
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    $\begingroup$ @uhoh , if an asteroid comes to a stop wrt earth at about a lunar distance away, it would be moving nearly escape velocity by the time it reached earth's atmosphere. An asteroid could enter the atmosphere at LEO orbit velocity if it makes numerous perigee drag passes through the upper atmosphere. $\endgroup$ – HopDavid May 12 '18 at 2:31
  • $\begingroup$ I once discussed this with Nathan Strange, a good friend & JPL orbital dynamics expert (as of ~2 years ago, Dr. Strange! He went back and got his Ph.D!), who told me that the restricted 3-body (and 4-body, etc.) problems don't get around deep potential wells, they just merge them in interesting ways. The way you get an insignificant-mass 3rd body, at a large distance from a massive body, to wind up near to & at a very low speed with respect to the massive body, is to have an equally (or nearly so) massive 2nd body approach closely to the 3rd body, while near the 1st. (continued...) $\endgroup$ – Tom Spilker May 13 '18 at 0:35
  • $\begingroup$ The moon & sun never get near Earth so the topology of the total potential surface (the potential surface that includes all solar system objects) in the vicinity of Earth always looks pretty much the same. $\endgroup$ – Tom Spilker May 13 '18 at 1:06

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