Help Applications of Differential Equations to Escape velocity

[black_stallio]

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"[/URL]give the following explanation:

[QUOTE]In
review, we know that at the surface of the earth, i.e., at r = R, the velocity is positive, i.e., v = v[SUB]0[/SUB]. Examining the right side of the Velocity Equation reveals that the velocity of the object will remain positive if and only if:

(v[SUB]0[/SUB][SUP]2[/SUP]-2gR)>=0[/QUOTE]...

...[QUOTE]Hence, the minimum such velocity, i.e. v[SUB]0[/SUB]=√2gR is the escape velocity[/QUOTE]...(implying that [B]v[SUB]0[/SUB]>=√2gR[/B])


That raises a few doubts in my mind:
(1)Does the term 2gR[SUP]2[/SUP]/r become zero?

(2) If the answer to (1) is yes(i.e. 2gR[SUP]2[/SUP]/r=0), then how come v[SUB]0[/SUB] is allowed a value equal to √2gR?(since, substituting v[SUB]0[/SUB]=√2gR in the original equation(i.e. v[SUP]2[/SUP]=2gR[SUP]2[/SUP]/r + (v[SUB]0[/SUB][SUP]2[/SUP]-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. 2gR[SUP]2[/SUP]/r>0), then what is the fuss about (v[SUB]0[/SUB][SUP]2[/SUP]-2gR) being required to be positive at all? I mean, if it were negative, does it affect the original equation [I]so[/I] much that v becomes zero?

Question (3) could be rephrased in this way:
If (v[SUB]0[/SUB][SUP]2[/SUP]-2gR) becomes negative, does it [I]always[/I] become less than (2gR[SUP]2[/SUP]/r)?(thus giving v as negative)

 

[henry_m]

Think a little about what that first equation means, and what escape velocity means and it should become clearer.

Your first equation expresses the speed v of an object, launched from the Earth with speed v₀ and moving freely under gravity, when it is a distance r from the centre. The further from the earth, the slower it will be going, so larger r means smaller v.

Escape velocity means the minimum speed v₀ such that the object will never return. The object has two possibilities.

Firstly, it could stop at some r_max. This means that as you increase r in your equation, v will reduce to zero at r_max, and for r any larger, the equation stops making sense (v² will be negative). This indicates that the object can never get to this distance. But for small enough r, the equation will still give a sensible speed. Hopefully this answers (3).

On the other hand, you could increase r indefinitely, and v will never reach zero. The object carries on forever, never stops, and never returns: it escapes. Which of the scenarios occurs depends crucially on v₀; specifically the sign of the second term. Try graphing speed against distance for different values of initial speed.

 

[black_stallio]

Yes, I found the original explanation satisfactory after thinking on a more general level rather than looking at it in a-little-too-much detail. Sometimes, all you need is a different perspective to find an answer. Anyway thanks!