Zero gravity area How can I find the radius of the circle (average) centered at the center of the earth where the gravitational force of the earth cancels the gravitational force of the other celestial bodies?
Will the velocity which is required to reach the circumference of the circle be the escape velocity of earth? I think because the net acceleration on the object outside the circle will be directed away from the earth.. so that the body has truly escaped the earths gravity..?
Sorry for my non scientific language im not even a layman.
 A: Your question is a bit vague. There are a lot of objects moving around the solar system, and they are not always in the same place.
That said, there are two objects that affect the Earth in a more or less constant way: the Moon and the Sun.
The gravity of two massive bodies rotating each other, provided that one of them is quite bigger than the other (such as the Earth-Moon or the Sun-Earth systems), cancel at several points: those are the Lagrangian points.
The nearest one, the Earth-Moon L1 is about 326000 km high.
About your question:

Will the velocity which is required to reach the circumference of the circle be the escape velocity of earth?

No, the escape velocity of Earth is (by definition) the velocity needed to reach an infinite distance from the surface of the Earth. To reach the Earth-Moon L1 you need less speed because:


*

*It is nearer than infinity (obviously).

*You have help from the gravity of the Moon.


That said, there are another point where the Earth and Moon gravity cancel out, deep inside the Earth. Since the gravity as you go deep inside the planet decreases (proportional to the distance to the center) while the Moon gravity changes little (350000 km vs 356000 km is not a big deal), there is a point half-way (I didn't do the calculations) where the gravity of both the moon and the Earth would cancel out.
