Why Is Zero the Upper Limit in Escape Velocity Derivation?

[shaan_aragorn]

escape velocity derivation

The problem is thus:
The acceleration of a satellite is given by -gR^2/r^2 where R= radius of Earth and r = distance of satellite from center of the earth. Find escape velocity.

Now I have read the solution to this problem and in it the author at one point has integrated v dv = -gR^2/r^2 with the upper limit for vdv being 0 and the lower limit being Ve(Escape velocity).

Now my problem is, why is 0 the upper limit and Ve the lower limit if in magnitude Ve>0?

 

[dicerandom]

Think of it this way: the escape velocity is the bare minimum velocity needed to always move away from the planet, so you could say that "at infinity" (which is a strange concept, let's just say "very far away") the satelite's velocity is "zero" (really close to zero). Thus the satelite starts out near the Earth at the escape velocity Ve and ends up out at infinity with zero velocity.

 

[kyle8921]

The escape velocity derivation is a mathematical process used to determine the minimum velocity needed for an object to escape the gravitational pull of a larger body, such as a planet or star. In this case, the problem is asking for the escape velocity from the Earth's gravitational pull.

The reason for setting the upper limit as 0 and the lower limit as Ve is due to the direction of the velocity. When integrating, the upper limit represents the initial velocity (in this case, 0) and the lower limit represents the final velocity (in this case, the escape velocity).

Since the escape velocity is in the opposite direction of the gravitational pull, it is considered negative in this equation. This is why the lower limit is set as Ve (a negative value) in the integration.

In summary, the escape velocity derivation takes into account the direction of the velocity and sets the upper and lower limits accordingly to properly calculate the minimum velocity needed for escape.