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Kepler's Third Law and centripetal acceleration

  1. Sep 13, 2010 #1
    I'm not getting very far with these problems. Don't know what my problem is.

    Given a circular orbit, the centripetal acceleration is given by the gravitational force between the two stars. I chose the center to be halfway between the stars d/2.

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-13225621.jpg?t=1284437088 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-13225642.jpg?t=1284437089 [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 14, 2010 #2


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    Try writing v in terms of d and τ.
  4. Sep 14, 2010 #3
    pi*d = vt

    (1/t)*pi*d = v

    v = pi*d*τ

    That doesn't get me anywhere.
    Last edited: Sep 14, 2010
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