## Friday, March 12, 2004

### Using ion propulsion

So supposing I've potentially got 20 tonnes into LEO, but I actually want to get something to the Moon, what's the fraction of that 20 tonnes that has to be propellant?

I'm going to assume that I can have my 20 tonnes in an orbit parallel with that of the Moon, at an altitude of 1000 km. I'm also going to assume that the only problem is getting out of the Earth's gravitational field, to a target orbit the same as the Moon's (i.e. an orbital radius of 384000 km). So from the equations for a circular orbit (see for instance an A-level Physics textbook) initial velocity v0 = 7.35 kms-1 and final velocity vf = 1.02 kms-1, giving Δv = 6.33x103 ms-1.

I'm going to make the blatant assumption that as much electrical power is available as I require, and use a high-powered ion thruster. Now, the example I looked up on the Internet [1] had a quoted specific impulse Isp of 3800 s. I can use the fact that exhaust velocity is equal to specific impulse divided by the gravitational field strength on the Earth's surface (g = 9.81 ms-2) to find exhaust velocity ve = 37240 ms-1. Then from the rocket equation in the form e-Δv / ve = mf / m0 the ratio of final mass to initial mass is 0.843.

This is interesting, because it implies that out of 20 tonnes in LEO 16 tonnes will make it to lunar orbit: a much better ratio than for a conventional booster! Assuming, however, the payload is destined for the Moon's surface, that sixteen tonnes must include: the ion engines themselves and fuel tankage; the solar panels required to power the ion engines; the landing retrorockets and enough fuel to land the payload; and the landing gear. So it's probable that less than half of the original 20 tonnes would consist of non-propulsion payload. But the setup would probably still be an improvement on a conventional chemical-rocket-only system.

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