[ExI] Re :rocket fuel was Re: SpaceX launch
hkeithhenson at gmail.com
Sat Dec 11 17:31:21 UTC 2010
On Sat, Dec 11, 2010 at 5:00 AM, Eugen Leitl <eugen at leitl.org> wrote:
> In terms of economy you're likely to employ a hybrid design.
> A maglev launch stage (e.g. up Mount Chimborazo) instead of a first
> chemical stage, second stage air-breathing scramjet/rocket hybrid
> or simple chemical rocket, or laser driven ablation (not sure this
> will ever work, though current prototypes are sure nifty).
If you have never looked at the power requirement for maglev, you should.
> Also, there's nothing fundamentally wrong with chemical rockets,
> as long as your transport rate is limited, e.g. for a lunar bootstrap
> using in-situ resources. Suited monkeys are quite pointless, so you'll
> proceed to teleoperated robots shortly, and whether you teleoperate
> them from a pressurized can nearby or from rotating ground centers (with
> the advantage of 24/7 operation) only adds about 2 seconds of latency.
Dr. Peter Schubert is the only one I know of who tried to put numbers
on a lunar bootstrap. He got $2 T and 20 years.
Adrian Tymes <atymes at gmail.com> wrote:
> On Fri, Dec 10, 2010 at 2:46 PM, Keith Henson <hkeithhenson at gmail.com> wrote:
>> The key concepts are mission velocity and rocket exhaust velocity. ?To
>> get into LEO takes around 9km/sec delta V. ?Orbital is only 8 km/sec
>> but you burn somewhat more because of the time gravity is accelerating
>> you downwards and from air drag.
>> In simple terms, going to twice the exhaust velocity (which is what
>> you have to do with LH2/LOX), the mass of the fuel has to be about 85%
>> of the takeoff mass. ?A reusable rocket is about 15% structure, which
>> leaves no payload using LH2/LOX.
>> Laser heated hydrogen at 3000 deg K has an exhaust velocity of 9.8
>> km/sec. ?That means the structure plus payload can be 1/3rd of the
>> takeoff mass. ?I.e., a 300 ton vehicle could reach orbit with 50 tons
>> of structure and 50 tons of payload.
>> To get in excess of one g takes around 6 GW. ?The only way this makes
>> economic sense is if the lasers are run 90% of the time.
> If it's just a matter of exhaust velocity, what about ejecting stuff
> at 100+ km/s?
As other people mentioned later, energy.
mv =MV, but Ke = 1/2mv^2
Ten times the exhaust velocity takes 100 times as much energy. We are
already at $60 B for the lasers, 100 times as much is $6 T. It would
take 1.2 TW to power the lasers. That's more than the installed
electric power for the US.
PS. It would, however, make a hell of a weapon. 6GW is 1.5 tons of
TNT per second. 600 GW would be 150 tons of TNT per second, less than
two hours for a megaton.
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