How high would you have to jump to loose gravity on the moon?

2.38 kilometers per second is the Moon's escape velocity.

If you fired a high velocity rifle straight up from the Moon's surface the bullet would be traveling far too slowly to leave permanently.
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It's not a question of "how far." Gravity extends outwards infinitely far from every object. It's a question of "how fast" must you throw it so that gravity can never make it return.

It's not really a question of how high, it's a matter of speed.  Nonetheless, if you can get that apple up 61,500 km, then it will outside the range where the Moon's gravity is the dominant factor.  The easy way to do this is to launch it with a speed of 8568 km/hr.  The apple would then orbit the Earth or, if it had enough speed, it would orbit the sun.

The formula is r = a cuberoot( m/(3M) )
a is the Earth-Moon distance = 384,000 km
m = mass of moon
M = mass of Earth
m/M = 1/ 81.3

The question is not how high, but how fast.

You have to imagine the gravity field as some sort of hilly terrain. The apple falls down to the moon just like it would roll down the hills. Every planet and every moon leaves a more or less deep depression into the "gravity potential". if you know remember that the solar system is in 3D and gravity acting in all three dimensions, you might feel that my hill example is in reality in the forth dimension.  The further you get away from the planet or moon, the weaker its pull get gets: The slope of the "hills" gets less.

See here for a nice visualization of the concept: http://xkcd.com/681/

The real gravity field of the solar system is a very complex and chaotic thing, but we already know a lot about how to navigate it for our advantage (In your example: We can't even just make the apple leave the moon, but afterwards drop on Mars after some decades of travel, by exploiting that moons and planets move)

The trick is now, to throw the apple so strong, that, while it gets slowed down by gravity, it still manages to reach the top of the depression left by the moon and does not fall back. The velocity that you need for that is called escape velocity or second cosmic velocity. It depends on your distance to the moon, since, as you can imagine, the further you are away from the moon, the closer you are to the borders of its gravity well. Yes, this is also equal to just throwing the apple to the borders of the gravity well...about 96,000 km away from the moon, but the distance varies with time and the positions of the other objects in the solar system. The region at the edge of the gravity well of an object is called "weak stability boundary", and weak stability is an understatement. Just a few kilometers error in navigation can mean you get to a completely different place after some days.

The escape velocity of the moon is "just" 2.38  km/s at its surface. That is about 3 times faster than a rifle bullet. So, just throwing isn't enough and even a normal cannon would be able to fire the apple away from the moon. A light-gas cannon, like used for testing meteoroid shields for spacecraft, would be able to do this.

Just putting the apple into orbit, at a lower velocity, making it loop the moon like a satellite, wouldn't be enough. First of all, the apple would always get as close to the surface of the moon, as you are when you "throw" it away. So, you have a high chance of hitting a high mountain one day. Next, the moon has a very complex gravity field because of mass concentrations below the surface, which make satellites close to the moon slowly get into a more elliptic shaped orbit, until the perilune, the point closest to the moon, is below the surface and your apple impacts on the moon.