Having looked at how to observe the Universe from Titan, I will consider how to depart from Titan and visit the rest of the Universe.
The first step in such a journey is to depart from Titan’s surface and go into orbit around that moon. Titan has a thick atmosphere and a low surface gravity, factors that act in opposite directions. Also, a low orbit would have an altitude of about 1000 km. Putting the factors together and doing some hand-waving estimates, I estimate a velocity change of about 2.3 km/s, compared to the Moon’s 2.0 km/s and the Earth’s 9.4 km/s.
Let’s see how much rocket fuel that one needs, how much mass of fuel for mass of payload, what one wants to get into orbit.
Around the turn of the twentieth century, Konstantin Tsiolkovsky worked out a lot of features of spaceflight from Newtonian mechanics and the like. In particular, he derived his famous rocket equation:
v = ve * log(mi/mf)
v = resulting velocity change (delta-V), ve = effective exhaust velocity (momentum / mass), mi = initial mass, mf = final mass.
So while v can be greater than ve, it can’t be much greater. I’ve collected some values of ve for various sorts of rockets:
- Hydrogen-oxygen: 4.5 km/s
- Hydrocarbons-oxygen: 3.25 km/s
- Hydrazine-family-nitrogen-oxides: 3.0 km/s
- Solid fuel: 2.75 km/s
- Ion engines (Dawn NSTAR: xenon): 30 km/s
The raw materials of all but the last one can readily be obtained on Titan.
Let’s now look at a vehicle that successfully did similar delta-V’s with the third kind of fuel: the Apollo Lunar Module.
- Descent stage: full 10.3 mt, empty: 2.1 mt, fuel: 8.2 mt
- Ascent stage: full 4.7 mt, empty: 2.15 mt, fuel: 2.55 mt
- Combined (descent phase): full 15.0 mt, empty: 6.8 mt, fuel: 8.2 mt
where mt = metric tons (megagrams).
This gives mass ratios: descent 2.21, ascent 2.19. By comparison, for departing from the Earth into low Earth orbit, the necessary mass ratio is typically around 40. So it should be fairly easy to depart from Titan by rocket.
Filed under: Sciences |