Okay, in the Escape Velocity Equation, with the usual notations,

v²=2gR²/r + (v₀²-2gR)

 

a few articles like www.math.binghamton.edu/erik/teaching/02-separable.pdf and http://www.uv.es/EBR...5000_17_84.htmlgive the following explanation:

 

In

review, we know that at the surface of the earth, i.e., at r = R, the velocity is positive, i.e., v = v₀. Examining the right side of the Velocity Equation reveals that the velocity of the object will remain positive if and only if:

 

(v₀²-2gR)>=0

...

 

...

Hence, the minimum such velocity, i.e. v₀=√2gR is the escape velocity

...(implying that v₀>=√2gR)

 

 

That raises a few doubts in my mind:

(1)Does the term 2gR²/r become zero?

 

(2) If the answer to (1) is yes(i.e. 2gR²/r=0), then how come v₀ is allowed a value equal to √2gR?(since, substituting v₀=√2gR in the original equation(i.e. v²=2gR²/r + (v₀²-2gR)) will yield v=0(which is obviously not desired if a particle has to escape the Earth)

 

(3)If the answer to (1) is no(i.e. 2gR²/r>0), then what is the fuss about (v₀²-2gR) being required to be positive at all? I mean, if it were negative, does it affect the original equation so much that v becomes zero?

 

Question (3) could be rephrased in this way:

If (v₀²-2gR) becomes negative, does it always become less than (2gR²/r)?(thus giving v as negative)

I think that they are making the observation that r is very large (and much bigger than [math]R^2[/math]). Therefore the first time is very close to zero and thus ignored.