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Obviously the asteroid belt is pretty sparsly populated. But how sparsly exactly?

What is the state of knowledge regarding how many particles are there in a given size bracket and a volume, what would be typical distances between asteroids in a given size bracket?

Edit to add:
The answers so far talk about asterois sized around a km. I was thinking also about smaller asteroids or even dust, down to the scale of a mm. I understand that we may know little about such small asteroids.
Anyway, an answer that gives me the mean free paths for several different sizes, all to within an order of magnitude, would be great.

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I just want to add that a lot of work goes into predictions of abundance, for objects including those we have not detected so far. There is some similarity to exoplanets - where we know that a method has a detection bias. If you can quantify the detection bias perfectly, then you can get the total abundance for different sizes.

One source gives a pretty good idea. Many references employ the D^-2.3 relationship, but it can occur as D^-1.3 as a cumulative metric. It's interesting that smaller sizes have been found to follow a different pattern. I strongly doubt that this pattern would continue to extremely small sizes, below D=0.1 km. As a mathematical statement, integrating to zero diverges.

abundance with size

I built some numbers for coefficients of the above relations with simple Excel calculations. To give you an idea of how the count changes with size, I took two regions of the above pdf. The region from 6 km to 1000 km contains on the order of 90,000 objects. But you're also interested in small bodies, so I also integrated the D^-4 relationship from 0.1 km to 6 km. That gives about 15,000,000,000 objects (15 billion).

If you decreased the lower bound from 0.1 km to something smaller, the count would be even more (likely by orders of magnitude). But we have no idea what relationship should hold there. Notice that the error bars grow as you go smaller? We don't have a good idea what the prevalence of small and microscopic bodies should be. Go small enough and you're at molecular sizes, so you could get a measure of the particle density in the solar system. But at that point, completely different factors are driving it (like atmospheric escape and solar wind), as opposed to the gravitational collection and breakup of bodies, which is what we think of as asteroids. We can put some bounds on the numbers - because space probes apparently didn't get smashed by micrometeorites. Regolith studies could also give some indication.

To get particle density, take the number you like for number of objects and divide by some rough metric of the volume that defines the asteroid belt. The uncertainty from that will probably be less than the count itself so I wouldn't worry too much about accuracy.

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  • $\begingroup$ I think this is what I wanted, but I don't understand the graph - D is mean distance?, what's n and n/n_10? $\endgroup$ – mart Jul 22 '13 at 13:43
  • $\begingroup$ Ah, maybe I should edit. For now, D is diameter of the object. The variable n has a shifting definition. In the graph, uppercase N is the cumulative number of objects, and lowercase n is the differential number of objects. This is CDF versus PDF. n10 is a reference for an object of 10 km diameter. The log is used to scale the graph. $\endgroup$ – AlanSE Jul 22 '13 at 13:46
  • $\begingroup$ If you can explain what the axis mean, it'd be great. The number of objects is total for the belt, or for a specific volume? $\endgroup$ – mart Jul 22 '13 at 13:59
  • $\begingroup$ @mart Do you mean the axis of the graph? It's the diameter of the object on the horizontal, and then the exponent of the relative probability distribution function (PDF) on the vertical. That's a little more confusing than it needs to be, but the gist is PDF = CD^-2.3 or CD^-4. The source I used specifies these are "Main Belt asteroids". $\endgroup$ – AlanSE Jul 22 '13 at 14:18
  • $\begingroup$ See if I understand this correct: I look at at D=10, logically log(n/n_10)=0, I look at D=1, log(...) is ~3 so a thousand times more rocks than at D=10 - correct? $\endgroup$ – mart Jul 22 '13 at 14:47
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This answer and question on physics cover a similar topic, I will link to it here and block quote the most relevant parts.

The most notable part of this answer is most likely going to be:

Asteroids are not distributed uniformly in the asteroid belt, but could be approximated to be evenly spaced in a region from 2.2 AU (1 AU is 93 million miles, or the average distance between Earth and the Sun) to 3.2 AU from the Sun and extending 0.5 AU above and below the ecliptic (the plane of Earth's orbit, which is a convenient reference for the solar system). That yields a volume of roughly 16 cubic AU, or about 13 trillion trillion cubic miles. (Note: space is big!)

So you can see that there is a huge gap in the asteroids, bigger than the distance between the earth and the sun by at least double!

As referenced by the answer on physics there is currently a NASA mission underway to venture to the asteroid belt to view it in more detail, relevant information for this mission can be found at:

http://dawn.jpl.nasa.gov/mission/

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To plug the numbers given into a simple calculation: The asteroid belt has a volume of 4,35E25 km^3. If we assume for a moment that the asteroids are evenly spaced, and there are 15 billion asteroids (estimate given above for sizes 0,1km+), we arrive at roughly 180km distance from asteroid to asteroid. 3 Orders of magnitude in the number of asteroids give us one order of magnitude in the distance - if there were only 15 million asteroids, we'd have a mean distance of 1800 km.
Actually, a few hundred kilometers between every object >100m does not seem that much.

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Apart from the points were an actual asteroid is, the density is very low.

"Contrary to popular imagery, the asteroid belt is mostly empty. The asteroids are spread over such a large volume that it would be improbable to reach an asteroid without aiming carefully." (source)

Furthermore the distribution of the asteroids is not uniform, thus the mean density might not be very characteristic.

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