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Ciruclar motion and gravitation question

  1. Mar 5, 2006 #1
    Q:Two objects attract each other gravitationally with a force of 3.3 10-10 N when they are 0.25 m apart. Their total mass is 4.0 kg. Find their individual masses.

    Now my professor only told us to use F= Gm1m2/r2
    Force in this case would be the 3.3e-10 Newtons, the radius squared = 0.0625 m and G is a the gravitational constant 6.67e-11 N so when i solve for m1 and m2 the larger mass come out to 4.2 kg and thats impossible
     
  2. jcsd
  3. Mar 5, 2006 #2

    arildno

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    Hence, you haven't done your arithmetic right.
     
  4. Mar 5, 2006 #3

    Hootenanny

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    How did you solve for m1 and m2?
     
  5. Mar 5, 2006 #4

    arildno

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    You have:
    [tex]m_{1}m_{2}=\frac{r^{2}F}{G}[/tex]
    withe the condition [tex]m_{1}=4-m_{2}[/tex]
    you get a quadratic equation to solve for [itex]m_{2}[/itex]
     
  6. Mar 5, 2006 #5

    Hootenanny

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    That's how I would do it.
     
  7. Mar 5, 2006 #6
    ok so what i come up with is :

    4m-m2^2=.25^2(3.3e-10)
    ---------------
    6.67e-11

    so when i solve i get :(Mass 2)^2-4(mass2)-.309 m=0
    but when i solve for mass to i get 4.02 kg.. what am i doing wrong?
     
  8. Mar 5, 2006 #7

    Hootenanny

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    You should get:
    [tex]4m_2 - m_{2}^{2} = \frac{r^2 \cdot F}{G}[/tex]
     
  9. Mar 5, 2006 #8

    arildno

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    First of all:
    DO NOT ENTER NUMBERS BEFORE THE VERY LAST MOMENT!! :grumpy:

    You have:
    [tex]m_{2}(4-m_{2})=\frac{r^{2}F}{G}[/tex]
    which you rewrite as:
    [tex]m_{2}^{2}-4m_{2}+\frac{r^{2}F}{G}=0[/tex]

    This has the solutions:
    [tex]m_{2}=\frac{4\pm\sqrt{16-\frac{4r^{2}F}{G}}}{2}[/tex]
    This yields ALWAYS positive answer(s), in so far as the solution exists.

    Note that the sum of your two solutions equals 4. What does that mean?
     
    Last edited: Mar 5, 2006
  10. Mar 5, 2006 #9
    thank you very much.:-)..i realize where i was going wrong.

    So after my calculations mass 1 =.078 kg and mass 2= 3.92 kg
     
  11. Mar 5, 2006 #10

    arildno

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    I haven't the slightest idea what plug&chug would yield. Your numbers sort of add up to 4, though, which isn't bad at all..:smile:
     
  12. Mar 5, 2006 #11

    Hootenanny

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    plug&chug?
     
  13. Mar 5, 2006 #12

    arildno

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    Plug into the formula, chug out the answer..
     
  14. Mar 5, 2006 #13

    Hootenanny

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    ahhhh... It's all starting to make sense...:wink:
     
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