black_stallion Posted May 25, 2011 Share Posted May 25, 2011 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) Link to comment Share on other sites More sharing options...
Dave Posted June 9, 2011 Share Posted June 9, 2011 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. Link to comment Share on other sites More sharing options...
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