Escape Velocity

OK, I'm glad we've got that little issue cleared up.

Ok, so if I am in a stationery craft, how far from the Earth would I have to be to not be dragged back to the Earth.

In theory you need to be infinitely far away to be stationary and never fall back. Basically that is the definition of escape velocity. In real life you only need to be far enough away to fall under the influence of some other planet's gravity. Then you fall there instead of back to Earth. And in real REAL life it is way more complicated, because you are never stationary. All planets orbit the sun, the sun orbits the galaxy, the galaxy moves in space relative to other galaxies, and so on.

campbelp2002 is right. There is no such distance, but there are other objects you can get influenced by. Let's get "realistic" and see what distance you need to be for an escape velocity of 10^-3 m/s (that's 1 mm/second, not much right?)

M (Mass of earth ) = 5.9736×10^24 kg
G (Gravity constant) = 6.67 × 10^-11 N m^2 kg^-2

Earlier, I demonstrated the relationship between the escape velocity V and the distance to the centre of the earth r (see earlier in this thread). That relation is:

1) V = sqrt( 2 * M * G / r)

We need the oposite relationship. Rewriting gives:

2) V = sqrt( 2 * M * G / r) <=> V² = 2 * M * G / r <=> r = 2 * M * G / V²

Filling in M and G, and using 10^-3 for V, this results in

r = 7.97 *10^20 meter, or roughly 135 million times the (average) distance between the sun and Pluto.

Please note that this does by no means set you free from the gravitational pull of the sun, let alone from the galaxy...
I'll leave the question how fast you'd have to go from earth in the direction of the sun to get free of the influence of earth and be captured by the sun as an exersise to the reader

BTW: I just discovered http://en.wikipedia.org/wiki/Escape_velocity It seems my mathematical derivation was correct