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Force on a mass by two fixed masses

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Two identical point masses, each of mass ##M##, always remain separated by a distance of ##2R##. A third mass ##m## is then placed a distance ##x## along the perpendicular bisector of the original two masses. Show that the gravitational force on the third mass is directed inward along the perpendicular bisector and has a magnitude of
    $$F = \frac {2GMmx}{(x^2+R^2)^\frac{3}{2}}$$
    2. Relevant equations
    $$F = G\frac {m_1m_2}{r^2}$$
    3. The attempt at a solution
    20161024_185403.jpg

    Am I on the right track? If so, what do I do from here? If not, where did I go wrong?
     
  2. jcsd
  3. Oct 24, 2016 #2

    TSny

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    You're on the right track. In your expression for the force, FM on m, you wrote the denominator as (R2 + x2)1/2. The power of 1/2 is not correct.

    You can simplify cosθ. There is no need to write it as cos(tan-1(R/x)). You should be able to "read off" cosθ from one of your triangles in your picture.
     
  4. Oct 24, 2016 #3
    Why is ##(R^2+x^2)^{1/2}## not correct? Isn't that the distance between ##M## and ##m##?
     
  5. Oct 24, 2016 #4

    TSny

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    Yes, it's the distance between M and m. But what is the denominator in Newton's law of gravity?
     
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