[TML] Asteroid Mining (was Molding Ships)

[Cyberia]

Timothy Little wrote:
> On Thu, Sep 27, 2007 at 05:59:44PM -0700, shadow at shadowgard.com wrote:
>   
>> You can't drill without matching velocity (including spin) *and* 
>> anchoring yourself to the rock. Otherwise the pressure of the drill 
>> will push the ship and rock apart. 
>>     
>
> Unless you have thrusters - which do happen to be cheap, powerful, and
> don't require reaction mass in Traveller.
>
>
>   
>> There's an awful lot of energy tied up in that spin. 
>>     
>
> Not as much as you might think.  A typical spin for a small asteroid
> is a few rotations per day.  Even fast-spinning ones almost all have
> periods greater than two hours.  For the sizes of asteroids we're
> talking about here (say at most 100m in diameter), that's on the order
> of a few centimetres per second.  Even for an asteroid mass of a few
> million tonnes, the kinetic energy of rotation at that speed is only a
> few megajoules - about the same as a bus on a highway.
>
> It may be worth noting that almost all asteroids rotate slowly enough
> that even their otherwise negligible gravity is sufficiently strong to
> counter centrifugal acceleration.
>
>   
This is quite an interesting thread, so here's my 2% of 0.02Cr.

Assume to first order that the asteroid is approximately a solid sphere 
of uniform composition, and
that it consists essentially of nickel iron.
Thus density works out to approx 7850 kg/m^3, from "Energy Coupling to 
Asteroids and Comets",
B.P. Shafer et al.
100m diameter works out to 50m radius, and a volume of 523,598.7756 m^3, 
rounded to 520k m^3.
Thus the mass is approximately 4,080,000 tonnes, or roughly 4E9 kg.
 From "Physics for Scientists & Engineers with Modern Physics", 3d ed, 
D. Giancoli, the moment of inertia I
of a solid sphere of radius R, mass M is 0.4 * M * R^2 kg m^2, or approx 
1E13 kg m^2.

 From Giancoli again, rotational energy E of a body with angular 
velocity omega is 0.5 * I * omega^2 J.
A rotational period of 2 hours implies 1/7200 * 2 pi rad/s, or pi/3600 
rad/s.
E thus becomes approx 3,808,000 J, or 3.8 MJ.  So it looks like Mr 
Little was being overly generous in
his estimates of such a rock's rotational energy.