# Range of the gravitational field

1. Oct 13, 2009

### 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?

2. Oct 13, 2009

### Molydood

I am interested in the answer to this as the way I see it, it escapes as soon as it is fired if it is fired with 'escape velocity'
I don't see that there is a real answer to the question as it is never completely out of the gravitational field, it's just a diminishing relationship of gravity versus velocity, and the balance is determined right from the very start as being in favour of the projectile, and although the velocity of the projectile and gravitational pull from the planet both decrease with time (or distance away form the planet) and dont ever reach zero, the balance of 'power' doesn't change from t=0. Just my view on it, looking forward to somebody providing a better answer.

3. Oct 13, 2009

### Nabeshin

Molydood is right, the range of the gravitational force is infinite. Therefore, it is technically never correct to say something "leaves" the gravitational field, even if it is launched with escape velocity. Escape velocity simply means that the object will never stop moving, i.e, will never fall back towards the Earth, in this case. That is, the total energy of the body will always be positive (K+U > 0 for all time).

$$2gh=v_f^2-v_i^2$$
$$U=-\frac{GMm}{r}$$