Physics Help Forum potential energy function question- re added

 Apr 10th 2009, 10:26 AM #1 Member   Join Date: Apr 2009 Posts: 71 potential energy function question- re added Hi, I was hoping for some help with the following. This is the first question of this type I have attempted so, any help either with answering the parts of the question I could not do or if you notice any mistakes with the parts I have answered would be appreciated. A proton of mass m moves in one dimension. Its potential energy function is: U(x)=a/x^2 - b/x where a and b are positive constants. The proton is released from rest at x0=a/b. I also showed in earlier questions that that U(x) can be written in the following way: U(x)=(a/(x0)^2)[(x0/x)^2-(x0/x)] and that v(x) the speed of the proton as a function of position can be written as: V(x)=sqrt((2a/m(x0)^2)[(x0/x)-(x0/x)^2]) with x>x0 a. Give a short qualitative description of the motion in terms of the classification of Kepler orbits. I don't know what a kepler orbit is and my textbook (or the Internet) was not much help so, I could not do this part. What is a Kepler orbit? How are they classified? b. Show that at x=2x0 the speed of the proton is maximum and comute that speed. My attempt: Differentiating the first expression of U(x) with respect to x and setting the derivative equal to zero gives that: 2a/x=b therefore x=2a/b =2x0 as required. Subbing this into v(x) gives v=sqrt(a/2m(x0)^2) c. What is the force on the proton at the point x=2x0? My answer: F=0. d. Instead let the proton be released (from rest) at x1=3a/b. Derive an expression of v(x) for the new release point in terms of a,b and m, also give a qualitative description of the orbital motion. My attempt: Initially U(x)=U(3a/b)= -b^2/9a = Total energy So at any time: Kinetic energy + U(x) = -b^2/9a Hence 0.5mv(x)^2 +a/x^2 – b/x = -b^2/9a This is as far as I have got for this part of the question; I do not know how to get rid of the x’s. Any ideas? I also don’t know what it means by description of the orbital motion. Was it ok for me to use the same potential energy function for a different initial condition? e. for each release point, (x0 and x1) find the maximum and minimum x values reached during the motion. I did not even know how to start this part of the question, any guidance would be welcome.
 Apr 11th 2009, 04:18 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 Check out Classical Mechanics by Goldstein. It is a graduate level text book so the level is pretty high. It is considered a bible nevertheless. It has some chapter devoted to keplers laws i think.

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