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From what I understand, an object entering the atmosphere will start to burn up from the tremendous resistance of the atmosphere. Presumably, for asteroids under a certain size, they will burn up completely and never impact the surface of the earth.

Do we have a way of determining the minimum size needed for actual impact?

If so, roughly what is the size and how does it compare to the average size of asteroids that pass by us regularly?

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    $\begingroup$ The answer will depend strongly on the composition of the object... $\endgroup$ – dmckee Dec 28 '12 at 1:28
  • $\begingroup$ I understand this, but neglected to mention it to keep the question short. I'm assuming that there are a few ''typical'' compositions of asteroids, and its these most common compositions that I'm interested in. $\endgroup$ – Todd R Dec 28 '12 at 1:34
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    $\begingroup$ Original poster should also clarify what does “impact” mean. Obviously, a millimetre-sized grain falling at terminal velocity doesn’t make an impact, but what does, how quick it must be to match OP’s undeclared criterion? $\endgroup$ – Incnis Mrsi Aug 23 '14 at 11:21
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As mentioned in NotAstronaut's answer, objects smaller than 25 meters will typically burn up in the atmosphere. One can very easily see why this should be the case using Newton's impact depth formula. This is based on approximating the problem by assuming that the matter in the path of the object is being pushed at the same velocity of the object, so as soon as the object has swiped out path containing the same mass as its own mass, it will have lost all of its initial momentum. All its kinetic energy will then have dissipated there, so if this happens in the atmosphere it will have burned up before reaching the ground.

This is, of course, a gross oversimplification, but it will yield correct order of magnitude estimates. We can then calculate the critical diameter as follows. The mass of the atmosphere per unit area equals the atmospheric pressure at sea level divided by the gravitational acceleration, so this is about $10^4\text{ kg/m}^2$. If an asteroid of diameter $D$ and density $\rho$ is to penetrate the atmosphere, its mass of $1/6\pi \rho D^3$ should be larger than the mass of the atmosphere it will encounter on its way to the ground, which is $5/2 \pi 10^3 D^2\text{ kg/m}^2$. Therefore:

$$ D > \frac{1.5\times 10^4}{\rho} \text{ kg/m}^2$$

If we take the density $\rho$ to be that of a typical rock of $3\times 10^3 \text{ kg}/\text{m}^3$, then we see that $D>5\text{ m}$, which is reasonably close order of magnitude estimate to the correct answer.

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  • $\begingroup$ I wouldn't have known where to begin this calculation. Awesome! $\endgroup$ – WetSavannaAnimal Apr 20 '17 at 1:29
  • $\begingroup$ "as soon as the object has swiped out path containing the same mass as its own mass, it will have lost all of its initial momentum. All its kinetic energy will then have dissipated there" I get that part. "if this happens in the atmosphere it will have burned up before reaching the ground" I don't get that part. $\endgroup$ – PM 2Ring Apr 6 at 9:29
  • $\begingroup$ @PM2Ring If we focus on large objects, then the kinetic energy of the object that will dissipate will for some part heat the object and cause it to explode. This can then happen before the object reaches the ground or after. In the later case we'll get a crater, in the former case we'll get an airburst. $\endgroup$ – Count Iblis Apr 6 at 9:46
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It needs to be more than 25m or otherwise it will burn up in the atmosphere according to this Nasa article "Space rocks smaller than about 25 meters (about 82 feet) will most likely burn up as they enter the Earth's atmosphere and cause little or no damage." https://www.nasa.gov/mission_pages/asteroids/overview/fastfacts.html

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Consider a specific example. The leonids arrive at the top of the atmosphere at $72$ km/s with a maximal mass of around $0.5$ g. According to the article these particles are $0.01$ m across. They reach the ground. If such a particle strikes the $30$ km high atmosphere at $45$ degrees it must travel around $4 \times 10^4$ m before it hits the ground, mostly burning up. If it started at rest at the top of the atmosphere it would only accelerate to about $1$ km/s. As you can see it very much depends on azimuth, latitude, meteor composition, and speed.

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    $\begingroup$ Are you sure they "impact" ? The article on Wikipedia states merely that the showers deposit several tons of material on Earth, but that does not mean that the particles reach the ground intact, just that the material eventually falls to earth (e.g. as tiny dust particles). $\endgroup$ – StephenG Apr 19 '17 at 19:44
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    $\begingroup$ "If it started at rest at the top of the atmosphere it would only accelerate to about 1 km/s" what-if.xkcd.com/28 PS: I, like @StephenG, am a little skeptical of the Leonids claim. Could you give a reference, even if it is a museum geology collections with confirmed Leonid objects in it. $\endgroup$ – WetSavannaAnimal Apr 20 '17 at 1:34
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Below is quoted from "How big does a meteor have to be to make it to the ground?" 10 October 2000. HowStuffWorks.com. http://science.howstuffworks.com/question486.htm 27 December 2012.

"So how big does a meteoroid have to be to make it to the surface of the Earth? Surprisingly, most of the meteoroids that reach the ground are especially small -- from microscopic debris to dust-particle-size pieces. ... Typically, though, a meteoroid would have to be about the size of a marble for a portion of it to reach the Earth's surface. Smaller particles burn up in the atmosphere about 50 to 75 miles (80 to 120 kilometers) above the Earth."

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