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Gravitational field

  1. Apr 23, 2005 #1
    the question is here

    [tex]g=\frac{GM}{x^2}[/tex]
    [tex]x=\sqrt{r^2+a^2}[/tex]
    [tex]g=\frac{GM}{(\sqrt{r^2+a^2})^2}[/tex]
    [tex]g=\frac{GM}{r^2+a^2}[/tex]

    since there are 2 masses...
    [tex]g=2 \frac{GM}{r^2+a^2}[/tex]

    my book's answer is:
    [tex]g=\frac{2MGr}{(r^2+a^2)^{3/2}}[/tex]

    what did I do wrong?
     
  2. jcsd
  3. Apr 23, 2005 #2

    Doc Al

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    Staff: Mentor

    OK, this is the field due to one of the masses. Note that it's a vector. What is its direction?

    Since the field contributions are vectors, they must be added as such. (You can double it only if the vectors were in the same direction.) Hint: The vertical components will cancel.
     
  4. Apr 23, 2005 #3
    horizontally:
    [tex]g=\frac{GM}{r^2}+\frac{GM}{r^2}[/tex]
    vertically:
    [tex]g=\frac{GM}{a^2}-\frac{GM}{a^2}=0[/tex]

    so then g would be: [tex]g=2\frac{GM}{r^2}[/tex]

    not quite sure what to do
     
  5. Apr 23, 2005 #4

    VietDao29

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    Homework Helper

    That's WRONG.
    Vertically:
    [tex]\frac{GM\sin{\alpha}}{a^2 + r^2} - \frac{GM\sin{\alpha}}{a^2 + r^2} = 0[/tex]
    Horrizontally:
    [tex]\frac{GM\cos{\alpha}}{a^2 + r^2} + \frac{GM\cos{\alpha}}{a^2 + r^2} = \frac{2GM\cos{\alpha}}{a^2 + r^2}[/tex]
    Find [itex]\cos{\alpha}[/itex]. Can you handle it from here?
    Viet Dao,
     
  6. Apr 23, 2005 #5
    [tex]\cos{\alpha}=\frac{r}{a^2+r^2}[/tex]

    [tex]g=\frac{2GM\frac{r}{a^2+r^2}}{a^2 + r^2}[/tex]

    ???
     
  7. Apr 23, 2005 #6

    VietDao29

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    Homework Helper

    That's not correct. Recheck your [itex]cos{\alpha}[/itex].
    Viet Dao,
     
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