M
mlorrey
Guest
Stevehw33,<br />While I agree with you on nuclear, you are incorrect wrt electric propulsion. High Isp allows for constant acceleration that builds up over time. This isn't very important for lunar missions, but for any mission measured in tens of millions or hundreds of millions of miles, or longer, electric propulsion is far superior to chemical. It is so because with constant acceleration, your velocity is constantly building, and with extremely high Isp, you can keep it going for full missions. Here's the math:<br /><br />Lets say you have a 100 ton Mars mission, which has a choice of half its tonnage being a chemical booster, or a nuke plant and electric propulsion system. Lets assume the chemical system is 40 tons of fuel, 5 tons of structure, and 5 tons for a fuel plant to operate at Mars. If the fuel choice is LH2/LOX, the Isp is 450 secs, which means 450 lbs of thrust-seconds per lb of fuel. 40 tons is 80,000 lbs, which gives you a thrust budget of 36 million lbf-seconds of thrust. Lets assume you chose an engine that provides 100 tons of thrust (200,000 lbf), allowing you to boost at 1 g on your Hohmann transfer orbit to Mars. This allows 180 seconds of burn time with your booster, which equates to 5760 ft/sec or 2.25 km/sec, significantly less than the 5.6 km/sec needed to get to Mars from LEO. In fact, you'd need more than triple the 80,000 lbs of fuel, to 260,000 lb of LH2/LOX, to have sufficient delta-v to get from LEO to Mars orbit at Phobos altitude. <br /><br />Your return to Earth also demands 3.4 km/sec of delta-v, or 110,000 lb of fuel, (assuming you leave half your vehicle mass at Mars and come home with 50,000 lb of vehicle), which you either need to take with you, or produce on Mars from water you find there (and with what power supply for your electrolysis plant? Solar?). We'll assume that fuel for Mars descent and ascent are equally budgeted in the 100,000 lb of mission mass. If you carry that 110,000 lb of return fuel with you, you'll need an a