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!

Help with proving this Kepler's laws based theorem

  1. May 10, 2009 #1
    the proposition is that if r(t) has a constant length (//r(t)// is constant), then prove that r(t) is perpendicular to dr/dt at all t.


    I was thinking that if r(t) is just a function of the radius and its magnitude //r// is constant, then its basically talking about a circle? and by saying prove that r is perpendicular to dr/dt is basically asking to prove that the radius vector is always orthogonal to the velocity v(t) vector? I need a mathematical proof of this proposition using vectors. Thank, you
     
  2. jcsd
  3. May 11, 2009 #2
    What is d/dt(r.r)?

    BTW Kepler hasn't got much to do with it.
    Any wiggly path on the surface of a sphere has constant r.

    David
     
    Last edited: May 11, 2009
  4. May 11, 2009 #3
    I'm not sure what your asking. this is a general proof so i dont know what d/dt(r.r) is. all the information i have is posted.
     
    Last edited: May 11, 2009
  5. May 11, 2009 #4
    i got an idea... would the //x(t)// be equal to the length of the curve of x(t) and therefore //x(t)//=integral(x(t).dx/dt) and since //x(t)// is constant the indefinite integral can only=c if the x(t).dx/dt=0 and so they are perpendicular. Can someone plz tell me if all this makes sense?
     
  6. May 11, 2009 #5
    r^2 is constant:


    0 = d/dt r^2 = 2 r dot dr/dt
     
  7. May 11, 2009 #6
    oh ok so you're saying that since //r(t)// is constant then //r(t)//^2 is constant which means (r.r) is constant and that derivative which is a dot product is 0. yeah this was easier than i thought my integral solution also works i think
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with proving this Kepler's laws based theorem
Loading...