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Rearranging equation with a square root

  1. Sep 15, 2011 #1
    " eebf9b13b294b1ac360379067a6d8e78.png
    Solving this for κ, substituting into (1), and rearranging for G, the result is:
    bd6b26bec3fae054fc52dbe55b23f4ad.png "

    I am trying to rearrange the first equation to make κ the subject and I get:
    [URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(6).gif[/URL]

    [URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(5).gif[/URL]

    [URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(4).gif[/URL]

    [URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(3).gif[/URL]

    [URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(2).gif[/URL]

    [URL]http://www.adamrapley.com/eqn1/CodeCogsEqn%20(1).gif[/URL]

    But somehow I think I've done something wrong :S Please could someone point out where I've gone wrong (or if I am indeed actually right, then how I substitute it into
    equation (1) which is [PLAIN]http://upload.wikimedia.org/math/2/c/a/2ca1bb59fb981ea80bd5c3d642d26949.png) [Broken]

    Thanks :)
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 15, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, what you have done is perfectly correct. Now, substituting that into your "eq. 1", you have
    [tex]\frac{mL^2(2\pi \theta^2)}{T^2}= \frac{LGmM}{r^2}[/tex]

    Solve for G by multiplying both sides by [itex]r^2/(LmM)[/itex]
     
    Last edited by a moderator: May 5, 2017
  4. Sep 15, 2011 #3
    Sorry, but I'm not totally sure how you substituted that in because when I tried it, somehow I managed to get this:

    [tex]\frac{mL^2(2\pi^2)\theta}{2T^2}=\frac{LGmM}{r^2}[/tex]

    :\
     
  5. Sep 15, 2011 #4
    The result it says I should get once "Solving this for κ, substituting into (1), and rearranging for G" is:

    bd6b26bec3fae054fc52dbe55b23f4ad.png

    But I'm not sure how to get there...
     
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