Asteroid Composition and Temperatures

Several questions about the following asteroid:

*Main belt (2.5AU, pretty circular orbit)
*Stony-iron composition, with some volatiles (3.5g/cc density)
*100 miles diameter (more or less)
*Not (very) differentiated internally (did not melt after
formation)
*8-hour rotation

Questions:
1) Is it reasonable for an asteroid that large to have not
differentiated (melted and settled out) after formation?
2) If it is reasonable for the asteroid to not differentiate
into a metallic core/silicate mantle, would the resulting
undifferentiated asteroid be basically a giant "dust ball,"
or would material strength be higher due to impact welding,
compression, etc?
3) What would the approximate core temperature be? About
"ambient," a balance of day and night side temps at 100-200K,
or significantly hotter, like 500-1000K?
4) What would the pressure be in the core? I ran a simple
estimate for a column of 3.5g/cc soil 50 miles high in the
asteroid's gravity, but I figure that was pretty conservative -
I didn't allow for the drop in gravity closer to the core.

Mike Miller, Materials Engineer

Mike Miller wrote:

4) What would the pressure be in the core? I ran a simple
estimate for a column of 3.5g/cc soil 50 miles high in the
asteroid's gravity, but I figure that was pretty conservative -
I didn't allow for the drop in gravity closer to the core.

I've been told that, for an asteroid of of uniform density

P(cent) = 4 pi G rho^2 r^2/9

P(cent) is pressure at center, rho is density and r and is radius of
asteroid.

Regards,

Hop
http://clowder.net/hop/index.html

JRS: In article , seen
in news:sci.space.science, Mike Miller posted at
Wed, 23 Jul 2003 03:20:14 :-
Hop David wrote in message
news:3F1B4E7B.90107
...

I've been told that for an asteroid of uniform density

P(cent) = 4 pi G rho^2 r^2/9

Where rho is density, r is radius and G gravitational constant.

Where density is in kilograms per cubic meter, radius is in meters,
and pressure is in...pascals?

If the density is in kilograms per cubic metre, the radius is in metres,
and G is compatible, then dimensionally the result is in newtons per
square metre, which are pascals.

One Atmosphere is about 10^5 pascals.

The proportionalities to G and rho^2 are reasonably obvious; the only
other thing likely to matter is r, which dimensionally must appear as
r^2.

The field within a sphere is proportional to the radius, and the amount
of material within a column is proportional to its length, which
bolsters the r^2 theory.

I can't vouch for the numerical constant, although the 4 pi part seems
likely.

The formula must assume a mechanically weak body.

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