[meteorite-list] Speed that meteors enter dark flight?

[MstrEman]

The simple answer is it depends on a lot of changing factors and
broadly ranging bounded assumptions. I can only share some of those
here to show why it is not a an easy answer.

<http://en.wikipedia.org/wiki/Drag_coefficient>

For one of those assumptions, you have to select amongst some drag
coefficients/cross-sections.  Typically: sphere, half sphere, cone and
streamlined.  One might assume that anything tumbling with sharp edges
is really facing high sheer forces and more likely to shear apart.
The drag cross section governs the air dampening of the gravity
acceleration which typically lies between 120-400 miles per hour
terminal velocity. This in itself requires choosing an assumption.

 The formulas in the link require an air density (aka fluid density)
average value as it isn't set up for a changing density which can be
considerable in a steep trajectory. There is a general acceptance that
the air is too thick below 5 miles/8 km above sea level for a meteor
below 1 meter to maintain incandescence velocities. ( 88% of the
atmosphere lies at or below 7 miles)   Air density is increasing at a
dramatically increasing rate.  In some respect so long as it is a deep
penetration to say under 12miles there should be ample distance to
travel to a point of all cosmic velocity being bled off and fall from
gravity acceleration alone. We'll assume a range of 4000-4500 kph for
retardation.

 One has to estimate a mass where total cosmic velocity can be
expended: which can be up to 10 tons/9000kg according to the AMS faq
page. I've also heard up to one meter but if you want to choose a
typical value pick 1kg.

One has to also integrate an acceleration factor as gravity is at work
even during retardation to extinction( note: retardation point is used
in our science but may be a misnomer but I won't get into a crust
argument).

So there is quiet a bit in the way of assumptions and perhaps a lot
more inaccuracy of accepted values.

Since we are in the "I wonder" mode-- lets choose a surrogate
meteorite/oid which has more data:
<http://hypertextbook.com/facts/JianHuang.shtml>   Freefall
parachutist records.
"Captain Joseph Kittinger entered the record books when he stepped
from the gondola of a helium balloon floating at an altitude of 31,330
m (102,800 feet) and took the longest skydive in history  He fell for
four minutes and 36 seconds, reaching a maximum speed of 614 miles per
hour (988 km/h before opening his parachute at 18,000 feet (5,500 m)".
   It takes an average sky diver 14±1 to fall one mile. according to a
graphic (pdf) on this page:
<http://www.greenharbor.com/fffolder/math.html>  and a skydiver will
fall about 10,000ft. in one minute including the 12 seconds to reach
terminal velocity of 120mph.

All that said it Chris's answer is pretty much within a 3-10 second
limit and impact in under 2 minutes max 95% of the time.

Elton

Sorry for all the co-mingling of metric and SAE values.

On Wed, Mar 7, 2012 at 1:58 PM, Chris Peterson <clp at alumni.caltech.edu> wrote:
> It depends on the mass of the body. But realistically, under "typical"
> conditions that might lead to meteorite production, I think it's safe to say
> that this happens almost instantly.
>
> For example, a 100 kg stone that survives to 20 km height will be
> experiencing a deceleration of ~1500 m/s^2. A 10 kg stone will experience
> ~4000 m/s^2. Of course, no stone is likely to survive the forces that would
> result without breaking up. You need to play all sorts of games with
> different parameters for mass, speed, and height to find survivable
> scenarios. They all produce a very short period of dark flight before
> terminal velocity.
>
> This is why the retardation point is typically overhead any strewn field,
> and you don't usually have meteorites significantly down field from the
> retardation point. In fact, wind during dark flight may move meteorites
> farther than their last bit of momentum did- and that can be in any
> direction.
>
>
> Chris
>
> *******************************
> Chris L Peterson
> Cloudbait Observatory
> http://www.cloudbait.com
>
> On 3/7/2012 11:45 AM, Mike Hankey wrote:
>>
>> the follow up to this question/answer I still wonder about is:
>>
>> after dark flight begins, how many seconds will it take to completely
>> decelerate so that all forward momentum is lost after dark flight
>> starts.
>>
>> for example: if the meteor goes dark at 4km/s how many seconds before
>> it will be at 0km/s and/or what does that deceleration curve look
>> like?