At what height body body escapes gravitational fields

When a body is fired upward with escape V (i.e 11.2 km/s) then what will be the height when the body leaves the gravitational field?
I found by using formula
2gh=Vf^2 - Vi^2 keeping vf= 0 and vi=11200m/s then I got h=6400km= radius of the earth.
Is it range of the gravitational field?
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 Recognitions: Homework Help Science Advisor There is no limit to the range of the gravitational field. What you calculated was the kinetic energy an object gains from falling from a very large distance (where it's speed is zero) to the surface of the earth. This is pretty much the definition of escape velocity. Remember that although the gravitational field goes on forever the gravitational potential energy reaches a limit, because as you go further away the field gets weaker. So if you launch an object with enough speed then when it reaches an infinite distance it will still have some ke. This limiting speed is the escape velocity.

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 Quote by mars shaw When a body is fired upward with escape V (i.e 11.2 km/s) then what will be the height when the body leaves the gravitational field? I found by using formula 2gh=Vf^2 - Vi^2 keeping vf= 0 and vi=11200m/s then I got h=6400km= radius of the earth. Is it range of the gravitational field?
As mgb_phys says, there's no limit to the extent of the earth's gravitational field. The equation you are using is only valid in regions close to the earth's surface where the acceleration due to gravity is 9.8 m/s^2. For other distances, you need to use a different formula for the gravitational potential energy: PE = -GMm/r, where r is the distance from the earth's center. Using that corrected expression you'll get r = ∞. (Which is no surprise, since this formula is used to calculate the escape velocity in the first place.)

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At what height body body escapes gravitational fields

 Quote by Bob_for_short No, it is the Earth radius or so. The gravitational force decreases as 1/r2. When you throw a stone at some velocity horizontally, it falls on the ground somewhere. The distance increases with the initial velocity. At v = 11.2 km/s the distance is equal to the Earth radius so the stone cannot land because the land surface "goes down". The stone misses the planet at such and higher horizontal velocities.
Nope. 11.2 km/s is the escape speed, not the speed you need to throw a stone horizontally so that it just orbits the earth without landing. (Given the usual assumptions of a smooth, symmetrical earth and an unobstructed path, of course.) That speed is less by a factor of √2--about 7.9 km/s.
 Yes, that's right. I made a mistake. This is a vertical velocity to fly away from the Earth surface R=6400 km to infinity.

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