Range of the gravitational field

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SUMMARY

The discussion centers on the concept of escape velocity, specifically the velocity required (11.2 km/s) for an object to theoretically leave Earth's gravitational influence. The formula used, 2gh = Vf² - Vi², indicates that at this velocity, the object reaches a height of 6400 km, which is equivalent to Earth's radius. However, participants clarify that the gravitational field extends infinitely, meaning an object never truly escapes but rather experiences diminishing gravitational effects. The total energy of the object remains positive, ensuring it continues to move indefinitely without returning to Earth.

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  • Understanding of escape velocity and its implications.
  • Familiarity with gravitational potential energy concepts.
  • Knowledge of Newtonian physics, particularly gravitational force equations.
  • Ability to apply kinematic equations in physics.
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  • Study the full Newtonian potential equation: U = -GMm/r.
  • Research the implications of gravitational fields extending to infinity.
  • Explore the relationship between kinetic energy and gravitational potential energy.
  • Learn about the effects of varying distances from Earth on gravitational force.
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Students of physics, aerospace engineers, and anyone interested in understanding gravitational dynamics and escape velocity concepts.

mars shaw
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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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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 don't 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.
 
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).

Now, let's see about your equation. You have:
2gh=v_f^2-v_i^2
Now, you're obviously using the gravitational potential of the Earth in the h<< radius of Earth approximation. This works fine for throwing up balls and such, but is certainly incorrect for an object with escape velocity. In the cases where the distance from the center of the Earth is comparable to the radius of the earth, you need to use the full Newtonian potential:
U=-\frac{GMm}{r}

I won't go through any more of the treatment of the escape velocity problem, but you can see that while this potential goes to zero as r approaches infinity, it is NEVER zero. Therefore, an object will never leave the gravitational field of the earth, no matter what velocity it initially has.
 

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