I'm looking for some realistic data about the size (in meters) and mass (in kg) of a typical small asteroids that impact the Earth, without disintegrating in the atmosphere.

Any suggestion ?

  • $\begingroup$ By "impact" do you mean "impact at high velocities capable of forming a crater in rock" or "have some rocks reach the Earth as meteorites"? By "disintegrate" Is it required that the asteroid must reach the Earth in one piece, or will you allow for several pieces to reach the Earth. Is the type of asteroid important, as the answer will be different for iron, rocky and rubble-pile asteroids $\endgroup$ – James K May 26 '17 at 14:29
  • $\begingroup$ The type of asteroid isn't important. I just need the size and mass of a "typical" asteroid. I need to calculate the kinetic energy of such a rock impacting the ground. The asteroid must reach the Earth in one piece, yes. $\endgroup$ – Cham May 26 '17 at 14:39
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    $\begingroup$ There is no typical size, only a size distribution, which roughly follows a power law. $\endgroup$ – AtmosphericPrisonEscape May 26 '17 at 14:50
  • $\begingroup$ Then I only need the size and mass of a medium/small or large asteroid, that could do the Meteor crater in Arizona. Something relatively impressive, but certainly not a planet killer ! $\endgroup$ – Cham May 26 '17 at 14:59
  • $\begingroup$ 1 second of googling reveals the most probable impactor size for this crater to be 50 m. Now take a typical rock density and calculate the mass. $\endgroup$ – AtmosphericPrisonEscape May 26 '17 at 15:05

For the asteroid not to break up it needs to either be big, or tough. You can experiment with the impact effects calculator but you should notice that a rocky asteroid with a diameter of less than about 1km will partially break up on impact with the atmosphere, and there may be multiple craters formed.

If the asteroid is less than about 100m in diameter than the breakup will be complete enough that the resulting meteorites will fall at terminal velocity, and won't form craters in rock.

Iron meteorites are tougher, but even these are likely to break up if they are less than 500m in diameter.

Needless to say, these are not small events, and we have never actually witnessed any object hitting the Earth that is large enough to make it through our atmosphere in one piece.

The direct answer to the question is that there are no typical small impacts of bodies that don't break up. The Barringer "meteor crator" impactor was an iron asteroid, with a diameter of about 40 or 50m, that broke up on impact with the atmosphere, and fell in a circle of about 150m in diameter, forming the 1km diameter crater we see now. The mass of the impactor was about 1 to 2 million tonnes. It hit the atmosphere at about 15 to 17 km/s but by the time it reached Earth, it had slowed to less than 12 km/s. It exploded with the energy of about 10 Megatonnes of TNT. The dispersal of the fragments was not enough to from several craters. See a barringer type impact Or read about the impact

  • $\begingroup$ Ok, then what was the estimated size and mass of the asteroid that created the meteor crater, before it broke into pieces ? $\endgroup$ – Cham May 26 '17 at 15:15
  • $\begingroup$ The numbers from the link now included in the body. $\endgroup$ – James K May 26 '17 at 15:22

It is best summed up by this graph of meteorite diameter versus frequency of impact, divided by the probability of total atmospheric penetration.

Typical meteorite sizes, are similar to typical road lengths and typical piece of string length.

Micrometeorites (particles normally less than 1 mm in size) constitute the main part of the flux of extraterrestrial matter accreting on Earth (1–3). (from arctic samples)

Over the whole surface area of Earth, there's 18,000 to 84,000 meteorites bigger than 10 grams per year. (study by P. A. Bland and was published in Monthly Notices of the Royal Astronomical Society.)

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  • $\begingroup$ Thank you for your answer. Could you please give more details of the reference "study by P. A. Bland"? You have taken the picture from there I guess. Thanks in advance. $\endgroup$ – Dr. Wolfgang Hintze Oct 2 at 13:56

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