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:
...
...
...(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)