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Gravitational force between two object

  1. Feb 20, 2007 #1
    Use the fact that the gravitational attraction between particles of mass mi and m2 at a distance r apart is Gm1m2/r^2. Slice the object into pieces, use this formula for the pieces, and sum using a definite integral

    What is the force of gravitational attraction between a thin uniform rod of mass M and length l and a particle of mass m lying in the same line as the rod at a distance a from one end
     
  2. jcsd
  3. Feb 20, 2007 #2
    I dont know where to start this problem
     
  4. Feb 21, 2007 #3

    arildno

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    Start by defining where the rod lies, in terms of COORDINATES.
     
  5. Feb 21, 2007 #4
    what do you mean?? well I just tried solving this and is it the integral of GMm/(l-a)^2?
     
  6. Feb 21, 2007 #5

    HallsofIvy

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    Since everything there is a constant, it's not much of an integral is it? Also there is no "dx" or "dy" or whatever specifying the variable of integration.

    Arildno said
    Since this is a linear problem (the point is in the same line as the rod), you can use a number line rather than Cartesian coordinates.

    Set up coordinates so one end of the rod is at 0 and the other is at l. The "test object" will be at -a. Now imagine a tiny piece of that rod between x and x+dx. What is the mass of that piece? The distance from that tiny piece to the test mass is a+ x (for "infinitesmal" dx you can treat that as the single distance for the whole piece- that's the whole point of this method). What is the force that tiny piece on the test mass? Now "add" that up (integrate it from x= 0 to x= l) to find the force of the entire rod on the test mass.
     
  7. Feb 21, 2007 #6
    What is the mass of that piece?

    Did you miss typed this?? Should this be force?? So the answer is the integral from 0 to 1 of Gmm/(a+x)^2
     
  8. Feb 22, 2007 #7

    arildno

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    Sigh.
    Not even the modelling of a straight line segment does it seem you are able to accomplish.


    I'll do this problem for you; pay CLOSE attention to the structure of the solution procedure:

    1. Modelling of a straight line segment:
    We have only need of a single coordinate variable (or ordinate, strictly speaking) to describe the linie segment:
    We let one end point of the rod lie at x=0, whereas the other end point lies at x=L. (L is the length of the rod)
    Thus, the rod is modelled by the interval [itex]0\leq{x}\leq{L}[/itex], where any point ON the rod is assigned its own ordinate number "x", lying between 0 and L, where the interpretation of its specific x-value is the distance between the point on the rod and the end-point that has been assigned x-value 0.

    2. Positioning of separate point mass:

    We let the point mass be placed a distance "a" to the left of the end point at x=0, i.e, the point mass has the position x=-a

    3. Uniform density:
    The mass density at any point on the rod is M/L.
    Therefore, a portion of the rod with length I has mass (M/L)*I.
    In particular, an infinitesemal portion of length dx has (infinitesemal) mass dM=(M/L)*dx

    4. Gravitational attraction on point mass:
    This is quite simply the sum of the attractions it experiences from each point of the rod (each with a location between 0 and L). Thus, summing over the attractions of these mass points (individually called dF)will give us the total force F:
    [tex]F=\int_{rod}dF=\int_{rod}\frac{GmdM}{(x-(-a))^{2}}=\int_{rod}\frac{GmMdx}{L(x+a)^{2}}=\frac{GMm}{L}\int_{0}^{L}\frac{dx}{(a+x)^{2}}=\frac{GMm}{L}(-\frac{1}{a+L}+\frac{1}{a})=\frac{GMm}{L}\frac{L}{(a+L)a}=\frac{GMm}{a(a+L)}[/tex]

    A somewhat interesting observation can be made of this:
    If a>>L, then the gravitational force acting upon the mass point is the same as if the rod's mass was concentrated at its center of mass, whereas if L<<a, then the gravitational force acting upon the mass point is the same as if the rod's mass was concentrated at the geometric mean between "a" and "L".
     
    Last edited: Feb 22, 2007
  9. Feb 22, 2007 #8
    if you put a in the right of the rod then we would get the same answer right??
     
  10. Feb 22, 2007 #9

    arildno

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    Presumably, you mean by "in the right of the rod" "beyond the rod at its right-hand side".

    The MAGNITUDE of the force acting upon the mass point would certainly be the same, but what would be the DIRECTION of this force?
     
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