Wed Sep 01, 2010 8:52 am by RickLewis

Thanks, but I still don't quite get it.

I can see that in deep space, away from any significant gravitational pulls, if you take an object of mass M "at rest" and attach a rocket to it, and fire the rocket so you burn a certain amount of fuel, then you will give the object a bit of a shove. By the time the burn has finished, the object will be travelling at a velocity V, and because it is a vacuum with nothing to slow it down or pull it in any direction, it will then keep on going at that velocity V (however many km/s that is) indefinitely.

So I can see that you could express the amount of fuel needed to do that as V km/s, meaning "the amount of fuel needed to speed any mass up from rest to V km/s." Is this basically it? If so, then I do understand I think.

And the amount of fuel needed to bring the object back to a state of rest by firing a rocket in the other direction? Also V km/s, obviously.

So far so good, but what if gravity is involved, as it is in most of the calculations we're talking about here? If you start with an object on the surface of the earth, fire your rocket, and burn the same amount of fuel (V km/s) then the object will accelerate up away from the earth, but gravity is pulling it back again at 9.8 m/s/s. So when you switch off the rocket, the mass will slow and may fall back to earth. This is where I still have difficulty coping with quantities of fuel in km/s.