Newsgroups: sci.space
Path: utzoo!utgpu!watserv1!watdragon!watyew!jdnicoll
From: jdnicoll@watyew.uwaterloo.ca (James Davis Nicoll)
Subject: Re: Mining El Dorado
Message-ID: <1991Jun17.144137.23456@watdragon.waterloo.edu>
Sender: news@watdragon.waterloo.edu (News Owner)
Organization: University of Waterloo
References: <1991Jun16.195153.9959@stb.info.com> <1991Jun17.052228.8112@sequent.com>
Date: Mon, 17 Jun 1991 14:41:37 GMT
Lines: 57

In article <1991Jun17.052228.8112@sequent.com> szabo@sequent.com writes:
>
>Some issues here:
>
>* We can't use Earth gravity assist or aerobraking for such a large
>  asteroid and unpredictable propulsion scheme.  Therefore, the delta-v
>  is likely to be quite large (several km/s).
>
>* The energy required to achieve that delta-v would likely disrupt or
>  vaporize the asteroid.  For example, to achieve 4,000 m/s the power
>  output is (1e10 kg)(4,000 m/s)^2/(.01 s) = 1.6e19 watts assuming a 
>  propogation time of .01 second.  Of course, we could do multiple
>  explosions, but hundreds to thousands of small nuclear warheads starts
>  to get expensive.

	This is probably a stupid question, but is the change in kinetic
energy (and a power rate based on a time value that appears to me to be
arbitrary, unless I missed something) a particularly useful number to
look at? Hmmm. That is phrased poorly. Maybe an example:

	Take a 1 kilo object and change its velocity by 1000 m/s.
That's a change in Ek of 0.5(1 kg)(1000 m/s)**2 or 5 x 10**5 Joules.

	Now, let's assume we're using a rocket to cause the delta vee
which has an Isp of 400. Mass ratio will be something like: 
		
		M1/M0 = e**delta vee/exhaust velocity
		      = e**(1000 m/s)/(4000 m/s)
		      = 1.28 
		
		If M0 = 1 kg, then
		   M1 = 1.28 kg, so we have to throw .28 kg to get our 
		delta vee.

		Ek = .5(.28)(4000)**2
		   = 2.24x10**6 Joules.

	To power the system we're using, we have to come up with 2.24x10**6 J,
but the change in Ek in the object is only 5x10**5 J, a bit more than a fifth
of the energy we actually use. The ratio between the delta Ek of the object
and the energy we actually use for a given propulsion system goes up as Isp
increases, of course. In your example up above, while the delta Ek is 
1.6x10**17 J, the actual energy we'd need would be higher (How much higher
depends on the Isp of the system we use). As an aside 1.6x1017 J is about
the amount of energy we could produce using all the the world's current off-
the-shelf supply of nuclear explosives (Actually, it's two or three times as
much), so we'll need to build more nuclear explosives to move the above rock,
perhaps much more, depending on hw we use them.

	Anyone out there have an idea how much nukes cost per megatonne?

	If there's some painfully obvious reason you used the delta Ek and 
the .01 second values you did, take it as written I pound my head shapeless
in contrition.

								James Nicoll