1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Kepler Orbits (algebra manipulation)

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    We have proved that any Kepler orbit can be written in the form of r([itex]\phi[/itex]) = [itex]\frac{c}{1+\epsilon*cos(\phi)}[/itex] where c>0 [itex]\epsilon[/itex][itex]\geq[/itex] 0. for the case that 0 [itex]\leq[/itex] [itex]\epsilon[/itex] < 1, rewrite this equation in rectangular coordinates (x,y) and prove that the equation can be cast in the form 8.51 which is the equation of an ellipse. Verify the values of the constants given in 8.52

    2. Relevant equations

    8.51: 1 = [itex]\frac{(x+d)^{2}}{a^{2}}[/itex] + [itex]\frac{y^{2}}{b^{2}}[/itex]

    8.52: a = [itex]\frac{c}{1-\epsilon^{2}}[/itex] ; b = [itex]\frac{c}{\sqrt{1-\epsilon^{2}}}[/itex] ; d = a[itex]\epsilon[/itex] or d = [itex]\frac{c\epsilon}{1-\epsilon^{2}}[/itex]

    3. The attempt at a solution

    i plugged in equation 8.52 into 8.51 and started simplifying stuff... I ended up with x[itex]^{2}[/itex]+y[itex]^{2}[/itex] = c[itex]^{2}[/itex] + [itex]\epsilon^{2}[/itex]x[itex]^{2}[/itex] - 2xc[itex]\epsilon[/itex]

    i know the left hand side is correct because it equals r[itex]^{2}[/itex]. So i know i have to eventually take the square root of each side, however, I have no idea how to get the right hand side of my equation to look like the right hand side of the equation listed in the problem.

    any ideas?
     
  2. jcsd
  3. Apr 12, 2012 #2
    i just simplified it down more so that

    [itex]\sqrt{x^{2}+y^{2}}[/itex] = c - [itex]\epsilon[/itex]x

    because the right hand side was a perfect square and could be made into (c-[itex]\epsilon[/itex]x)[itex]^{2}[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The Kepler Orbits (algebra manipulation)
  1. Algebraic manipulation (Replies: 1)

Loading...