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A thrust of 15,000 pounds is about 70,000 newtons or 7,000 "kilograms" force. Multiply by 3 and it could lift 21 tonnes off the moon without too much trouble. Low lunar orbit requires a delta-v of 1.6 kilometers per second; lunar escape takes 0.7 km/sec more (roughly 3,600 miles per hour and 1,500 miles per hour respectively in the American system of measures). The vehicle would accelerate at 3.3 meters per second (a third of a gee) and the moon would slow it down by 1.6 meters per second, so it would gain velocity at the rate of 1.7 meters per second. Orbital insertion would therefore take 1,600/1.7 = 941 seconds or 16 minutes; doable. Total delta-v including gravity losses would be 3.3 x 941 = 3100 meters per second. With a specific impulse of 900, about 6 tonnes of the 21 would have to be hydrogen propellant; the rest would be everything else. The nuclear engine would be what? maybe 2 tonnes. The tank would be maybe 1.5 tonnes and avionics, etc., might be another tonne. So the vehicle could lift 21-6-2-2.5 = 10.5 tonnes of payload to low lunar orbit. The 6 tonnes of hydrogen would require 6 x 14 = 84 cubic meters of store space. If the tank were five meters in diameter, it would have to be 4.5 meters long.
A higher thrust hydrogen-oxygen engine would do as well because the gravity losses could be cut substantially, so the total delta-v would be maybe 1,800 meters per second rather than 3,100 meters per second, but the tanks would be smaller (because most of the fuel is high density liquid oxygen rather than low density liquid hydrogen) and one wouldn't need to haul the two-tonne reactor along.
So as GCNRevenger notes, to land stuff on the moon and fly back to orbit, one would need a higher-thrust engine.
-- RobS
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The vehicle would accelerate at 3.3 meters per second (a third of a gee) and the moon would slow it down by 1.6 meters per second, so it would gain velocity at the rate of 1.7 meters per second. Orbital insertion would therefore take 1,600/1.7 = 941 seconds or 16 minutes; doable. Total delta-v including gravity losses would be 3.3 x 941 = 3100 meters per second.
You are overestimating the gravitational losses. As the spacecraft accelerates into orbit(and as it looses mass), the gravitational losses decrease.
With a specific impulse of 900, about 6 tonnes of the 21 would have to be hydrogen propellant; the rest would be everything else. The nuclear engine would be what? maybe 2 tonnes. The tank would be maybe 1.5 tonnes and avionics, etc., might be another tonne. So the vehicle could lift 21-6-2-2.5 = 10.5 tonnes of payload to low lunar orbit. The 6 tonnes of hydrogen would require 6 x 14 = 84 cubic meters of store space. If the tank were five meters in diameter, it would have to be 4.5 meters long.
The engine is trimodal, and can produce 3 times as much thrust when it augments the NTR with a lox/lh2 chemical reaction. It should be able to lift at least 30 MT of payload.
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You're right, at the end of the flight the vehicle would be accellerating closer to 1/2 gee, so the gravity losses would be less. The LOX augmentation changes the entire equation because as I understand it, the augmented thrust is three times as much or 21 tonnes. That would make a huge difference.
-- RobS
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In LOX-boosted mode, you do get alot more thrust, but it will come at the expense of specific impulse, which probobly isn't as big a factor deep in a gravity well versus orbit, but I wonder if the trouble of adding the oxygen burning system is worthwhile.
In any event, a sizeable NTR vehicle is going to be pretty big (>6m dia at least, length determined by shielding and fuel requirements, probobly no less then 25-30m long) which will make landing reliably tricky.
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