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Circular Motion - Help

  1. Dec 17, 2003 #1
    I'm having troubles with one of my homework questions, and I was wondering if someone could help me with it.

    okay here it is...

    Q: Two masses, object A and object B are located 2m apart from eachother, the mass of object a is m and the mass of object b is 4m.

    Showing your calculations find the point between these two objects where a third object would experiance no gravitation force.

    I'm really stuck on this question, any help would be very much appreciated.
     
  2. jcsd
  3. Dec 17, 2003 #2
    "a third object would experiance no gravitation force."
    what they mean is, no net gravitation force
    imagine a particle C lying on the line connecting A and B - what is the force that C feels from A? what is the force that C feels from B?
    What is the net force?
     
  4. Dec 17, 2003 #3
    I dont understand that at all...

    I would have thought I'd try and figure out what m was = to then, maybe do that... because from what I know I dont think I can figure out the net gravititional force from just what I was given...

    A little more help? heh
     
  5. Dec 17, 2003 #4
    Are you aware that the force for gravitation is
    [tex]\vec{F}=-\frac{Gm_1m_2}{r^2}\hat{r}[/tex]

    You have two masses m and 4m and some unknown mass m_2, so the magnitude of the gravitational attraction between m and m_2 is
    [tex]F_1=\frac{Gmm_2}{r_1^2}[/tex]
    where r_1 is the distance between m_2 and m. The strength of the gravitational attraction between 4m and m_2 is
    [tex]F_2=\frac{G(4m)m_2}{r_2^2}[/tex]
    where r_2 is the distance between 4m and m_2. If m_2 lies along the line between 4m and m and the distance between m and 4m is R, then
    R=r_1+r_2, so the second equation becomes
    [tex]F_2=\frac{G(4m)m_2}{(R-r_1)^2}[/tex]
    If m_2 is right between A and B, then the forces act in opposite directions and so cancel each other out. When F_1-F_2=0, m_2 will feel no force.
     
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