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Derivation of Newton Gravitation problem application

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to show that the first equation becomes the last. First sub in for r1 and r2 and then use the binomial theorem to expand to first order in d. Then use the assumption that d<<r1 and r2. To show it reduces to the last eqn

    **In the first eqn it is NOT r3 is should be r2 sorry**
    http://img43.imageshack.us/img43/9427/fffgb.gif [Broken]
    2. Relevant equations

    Binomial theorem

    I forgot to say d is the abs value of r2 -r1

    3. The attempt at a solution

    I know how to apply the binomial theorem. My first instinct was to plug the vectors r1 and r2 into the first equation. Then expand the denominators using the binomial theorem. But the denominators are the magnitude cubed so I am not sure that makes sense. I dont want anyone to just give me the answer because I want to figure it out. But I need a hint on how get this into a form where i can apply the binomial theorem and the ignore the higher order terms of d because it is very small.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 4, 2012 #2

    vela

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    Sounds like a good plan. What about having the magnitude cubed in the denominator is confusing you?
     
  4. Feb 4, 2012 #3
    http://img651.imageshack.us/img651/1681/uploadthis.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Feb 4, 2012 #4

    vela

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    Use the fact that ##\| \vec{r} \|^2 = \vec{r}\cdot\vec{r}##.
     
  6. Feb 4, 2012 #5
    http://img821.imageshack.us/img821/7811/uploaddd.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Feb 4, 2012 #6

    vela

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    You have to multiply it out:
    $$(\vec{x}+\vec{y})\cdot(\vec{x}+\vec{y}) = \vec{x}\cdot\vec{x} + 2\vec{x}\cdot\vec{y}+\vec{y}\cdot\vec{y}$$
     
  8. Feb 7, 2012 #7
    Also in the denomiantor there was 3 terms when I did the multipliction but the 3rd term was somthing d^2 so since we are using the assumtion d is small I ignored that term



    http://img708.imageshack.us/img708/2586/57716226.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  9. Feb 7, 2012 #8

    vela

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    The denominator should be
    $$(r_c^2 \pm 2\frac{m_i}{M}(\vec{r}_c\cdot \vec{d}))^{3/2} = (r_c^2 \pm 2\frac{m_i}{M}r_c d \cos\theta))^{3/2}$$ where ##\theta## is the angle between ##\vec{r}_c## and ##\vec{d}##. You can't just have two vectors in the expression like you did.

    Pull ##r_c^2## out to get a factor of the form (1+x)3/2. Then apply the binomial expansion to $$\frac{1}{(1+x)^{3/2}} = (1+x)^{-3/2}.$$
     
  10. Feb 8, 2012 #9
    Thanks I finally got it. This make me realize I need some more practice with vectors. But thanks for all the help!
     
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