# Debris field for a broken meteor

I happened to catch two parts of two different episodes of Meteorite Men - a show about two guys that look for meteorites. In both of the snippets I saw, they were talking about a debris field for a meteor that breaks up. In these fields, the larger chunks of the meteorite are further down in the field. Why is this?

Let me approach this first from a terminal velocity view. This requires a model for air resistance. I will use the following: Where:

• rho is the density of air
• A is the cross sectional area of the object
• C is a drag coefficient that depends on the shape of the object
• v is the speed of the object
• And this gives a force with a direction opposite of the velocity vector

Let me assume that all the pieces of a meteor have the same density and shape - for simplicity, I will assume a sphere. Here is a diagram for two different sized pieces falling (straight down) at the same speed. Meteor A (the big one) has a greater gravitational force because it has more mass. It also has a greater air resistance because it's cross sectional area is larger. I picked a speed so that meteor B would be at terminal velocity. This is when the air resistance has the same magnitude as the gravitational force. If I assume that meteor B has a radius of rB and a density of rhom then: Where vT is the terminal velocity. If I solve for this value, I get: Here you can see the key point. The terminal velocity depends on the size. This is because the air resistance is proportional the area (r2) and the weight is proportional to the area volume (r3). These two things do not cancel.

## Modeling a debris field

I have created a python model for shooting bullets. I can simply modify this to calculate the trajectory of a dozen or so different sized (but same shape and density) meteor pieces.

The following is a plot of the trajectory of a few pieces of a meteor. I (for random reasons) started the model at 5,000 meters above the ground moving at 350 m/s aimed 30 degrees below the horizontal. Here is what I get: So, the bigger the piece, the farther it will go. My biggest piece was 1 meter.

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Hi Rhett,

I think that in the paragraph under your bottom most equation (the one the gives v sub T) you meant that weight is proportional to volume, not area.

I wonder what a realistic speed for a meteor is? I would imagine that they generally exceed the speed of sound so I'm not sure the naive equation for drag would apply. They typically burn so both the cross sectional area and volume would be decreasing as they descend.

@the other Rob,

Thanks for catching my mistake - I fixed it.

I really don't know about the speed of a meteor - I guess I could estimate this by modeling this as a rock coming from Jupiter or something. But, for this case, I just need the speed and height when it breaks up to get the debris field.

According to this site, a medium speed meteor has a speed of ~40 km/s. I remember reading (though I don't remember where) that large bolides are moving so fast that air doesn't have time to move around them and the huge pressure difference between the front side and the rear side is what can cause them to fragment.

I don't think your model of air resistance applies at all.

By Grep Agni (not verified) on 24 Mar 2010 #permalink

@Greg,

Yes - at 40 km/s, my model is fubared. But...the meteor will eventually slow down and then I can use the v^2 model of air resistance. The key is to consider when the thing breaks up. Either way, I assume the air resistance model for high speeds would still be proportional in some way to the cross sectional area and the weight is proportional to the volume.

Hello Rhett,

I am a proffessional meteorite hunter (semi-retired) and I know Mr. Notkin and Mr. Arnold (the Meteorite Men) very well. Your analysis is pretty much right on; I'm going to restate the facts in a way that may help crystalize the dynamic creation of strewnfield distribution.

There are two basic levels of velocity that matter, each with thier own set of physics; what we call the meteor's cosmic velocity (starting @ 72 to 12 km/sec), this velocity in a very general sense operates independent of gravity and air resistance and is referred to as a vector of that velocity. As the body transitions to terminal velocity then gravity and air resistance (winds aloft) become "everything" to the movement behavior of the body(s).

In 2003 when the Park Forest meteorite fell on Chicago, we learned just how much different the two modes of flight are and how counterintuitive a strewnfield distribution can seem w/o understanding them. We plotted the velocity vector expecting large stones farthest along the vector which was exactly what we found ... except the successively smaller bodies weren't on or along the vector(SSW-NNE) but extended out perpendicular to the vector (W-E) from the heavy impacts. The smallest were several kilometers away from that velocity vector. Why on earth (pardon the pun) was this distribution so skewed from the vel vector? The answer we found in the upper level winds; the jet stream was over Chicago that night and it's 150 mph west to east winds stretched the whole strewnfield to the east of the heavy stones. So there are two paradigms that follow the heaviest meteorite; they will be found furthest along the velocity vector AND closest to the velocity vector.

BTW all of this fluid and gravitational dynamics only helps when a body falls that is observed. If one happens to find a strewnfield that was not observed to fall, the only paradigms that help are the ellipsis rule and the general weight distribution. Strewnfields of any type tend to fall in an elliptical pattern.

PS: I almost forgot the third phase of strewnfield distribution TERRESTRIAL (erosional, wind, alluvial). Needless to say, this phase can obliterate any trace of its cosmic beginings if left to the job for long enough.

Take care and good physics!

By Mark Jackson (not verified) on 26 Mar 2010 #permalink