STUDYING FOR TEST, need a tutortial

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The discussion revolves around understanding the derivation of the escape velocity formula for a rocket. The key equation K2 + U2 = K1 + U1 simplifies to (1/2)m*v^2 = GMm/R, where K2 and U2 are zero at infinity, indicating that the rocket must have enough kinetic energy to escape Earth's gravitational pull. By dividing both sides by m and manipulating the equation, the escape velocity is derived as v = sqrt(2GM/R). The participants clarify that the mass m cancels out, allowing the equation to focus solely on gravitational constants and distance. This discussion emphasizes the importance of understanding energy conservation in gravitational fields to grasp escape velocity concepts.
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ok..haveing trouble understanding some examples in the book...

1) 1000kg rocket is fired striat up, need rocket to escape and never return. what speen does rocket need to escape from grav pull to never come back?(non rotating earth)

i know rocket can't stop so or it will be drawn back into earth. so...
K2+U2=K1+U1 is:
0+0=(1/2)m*v^2 - (GMm)/R)

whats with the 0+0?

then it goes to:
Vescape= v1 = SQRT(2GM/R)

howd they get to that equation??
i know the answers and stuff its in the book, i just don't know how they got to those equations
 
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For escape velocity, you only need enough speed so that you escape the gravitational potential of the object. Therefore, any kinetic energy at infinity would be wasteful. So K2 = 0. U2 = 0 because the gravitational potential falls off like 1/r which goes to zero at infinity.

So we have (1/2)m*v^2 = GmM/R

so v^2 = 2GM/R

v = sqrt(2GM/R)
 
still not sure how you got rid of the 1/2...and did the m just cancel out on each side?
 
ElectricMile said:
still not sure how you got rid of the 1/2...and did the m just cancel out on each side?

(1/2)m*v^2 = GmM/R

1] Divide both sides by m

(1/2)*v^2 = GM/R

2] Times both sides by 2

v^2 = 2GM/R

3] Take the root

v = sqrt(2GM/R)
 
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