I’m afraid your definition of “escape velocity” is off a bit, Beeruser, though this is quite understandable since a number of web sites also get it wrong. For example, “Sea and Sky’s” Glossary of Astronomy Terms mistakenly (or at least misleadingly) defines it as: “The speed required for an object to escape the gravitational pull of a planet or other body.”
But escape velocity is not the speed at which the planet’s gravitational attraction is zero. It’s the velocity at which your kinetic energy is equal to your gravitational potential energy. It’s the speed you must reach in order to counter the body’s gravitational effects, not eliminate them.
You can see from the real definition that escape velocity exists as a finite quantity and is an entirely valid concept.
<BLOCKQUOTE><font size=“1” face=“Verdana, Arial”> quote:</font><HR> since gravity is theoretically infinite, it will still have a net force (and net acceleration) pointing back to the earth, no matter how many googles of light years it is away.
Though this is more or less correct in the context you used it, remember that gravity is not really infinite, since it propagates at the speed of light (though you’re absolutely right that no escaping object can outrace gravity). Outside the light cone of the originating astronomical body (i.e., googles of light years away), its gravity is zero.
Much more to the point, your statement is really only valid if you and the Earth were the only two bodies in the universe. I think it would be more accurate and far more useful to say that the vector pointing back to your origin is effectively zero as soon as you enter another gravitational gradient that counter-balances it. This is the way vectors normally work in physics.