STUDYING FOR TEST, need a tutortial

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Homework Help Overview

The discussion revolves around understanding the concept of escape velocity in the context of a rocket launched from Earth. The original poster expresses confusion regarding the derivation of the escape velocity formula and the application of energy conservation principles.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between kinetic and gravitational potential energy, questioning the assumptions made in the equations presented. They discuss the implications of setting certain energy values to zero and the cancellation of mass in the equations.

Discussion Status

Some participants have provided insights into the derivation of the escape velocity formula, explaining the reasoning behind setting certain terms to zero and the steps taken to simplify the equations. However, there remains uncertainty among some participants regarding specific steps in the derivation process.

Contextual Notes

The original poster mentions having access to the answers in the book but seeks clarification on the derivation process, indicating a focus on understanding rather than simply obtaining results.

ElectricMile
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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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