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Horizontal escape of a projectile

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem is attached

    2. Relevant equations

    L=mr^2(dr/dt)
    V_escape=(GM/R)^.5
    E=L^2/(2mR^2)-(GMm)/R=0

    3. The attempt at a solution
    after some work i got:
    dr/dtheta=((r^3/R)-r^2)^1/2
    dont know what to do from here.
    Thanks.
     

    Attached Files:

  2. jcsd
  3. Jun 17, 2011 #2

    gneill

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    Staff: Mentor

    I'm not completely clear on what it is you're trying to find. Is it the minimum velocity for escape for a horizontal launch?
     
  4. Jun 17, 2011 #3
    I'm trying to find r(theta)
     
  5. Jun 17, 2011 #4

    gneill

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    Staff: Mentor

    I see. Well, you're at the point where you'll have to integrate. The equation is separable into r and θ portions.
     
  6. Jun 17, 2011 #5
    then second part of my question i have to do is find the coordinates where r(theta) crosses the y-axis however, where i run into the problem is the integral i found doesnot cross the y-axis
     
  7. Jun 17, 2011 #6

    gneill

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    Staff: Mentor

    Hmm. What did your integration and its result look like?
     
  8. Jun 17, 2011 #7
    well seperating them it is dr/((r^3/R)-r^2)^1/2)=dtheta
    i used wolframalpha and got:
    what is attached and I used R= to 6.4E6m (radius of the earth)
     

    Attached Files:

  9. Jun 17, 2011 #8

    gneill

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    Staff: Mentor

    If you keep things in symbolic form a little longer, some good things might happen :smile:

    Attached is a snapshot of a Wolfram Integrator output where for the function I've pulled an r2 out of the radical before letting the integrator loose.

    Thus you have:
    [tex] \theta = 2 tan^{-1} \left( \sqrt{\frac{r}{r_o} - 1}\right) [/tex]
    which can be readily solved for r. Keep in mind the trig identity for tan(θ/2) :wink:
     

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