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Finding the gravitational field of a rod

  1. Apr 18, 2012 #1

    jjr

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    1. The problem statement, all variables and given/known data
    A homogeneous rod with even thickness has mass M and length L.
    a) Find the gravitational field g(x) at a distance x from the rod on an axis normal to the midpoint of the rod.
    b) Find the gravitational field g(y) at a distance y from the rod's midpoint on an axis through the rod in the direction of its length.


    2. Relevant equations

    Gravitational potential V(r) = -GM/r (1)
    Gravitational field g(r) = -∇V(r) (2)

    Where G is the gravitational constant, M is the mass of the relevant object and r is the distance from the object.

    3. The attempt at a solution

    a) My plan was to first work out the gravitational potential through integrating the contributions from infinitesimal lengths of the rod (from -L/2 to L/2) and then use equation (2) to find the field. The mass per length of the road is μ = M/L, so an infinitesimal piece of the rod dm = μ dl, where dl is an infinitesimal length. I got dV(r) = -[itex]\frac{G dm}{r}[/itex] = -[itex]\frac{G μ dl}{r}[/itex]. The r is (pythagoras) √(x2 + l2), where l is the distance from the midtpoint of the rod. The problem arises when I try to integrate this function dV(r) = -[itex]\frac{G μ dl}{√(x^2 + l^2)}[/itex] from -L/2 to L/2. Here's when I get in trouble. This integral is a bit hairier than expected, and I find it hard to get a reasonable answer. Could someone point me in the right direction here?
    I'll post my attempts at the solution for b) if I can't figure it out after getting help with a).

    Thanks,
    J
     
  2. jcsd
  3. Apr 18, 2012 #2

    Curious3141

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    I didn't really check the rest of your work, but if you want to integrate [itex]\frac{G μ dl}{√(x^2 + l^2)}[/itex],

    try expressing that as [itex]\frac{\frac{G μ}{l} dl}{√(1 + \frac{x^2}{l^2})}[/itex] then making the substitution [itex]\frac{x}{l} = y[/itex] to get it into a simpler form, then substitute [itex]y = \sinh{u}[/itex].
     
  4. Apr 19, 2012 #3

    jjr

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    Tried solving the integral and evaluating it from -L/2 to L/2, but it doesn't look legit. I think my reasoning has failed me at some point while setting up the expression for dV(r). The answer is suppose to be g(x) = -2 [itex]\frac{G M}{L x}[/itex] * [itex]\frac{1}{√(1+(2x/L)^2)}[/itex] i. Suggestions?

    J
     
  5. Apr 19, 2012 #4

    ehild

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    I think the integral is easier if you determine the field intensity directly.

    ehild
     
    Last edited: Apr 19, 2012
  6. Apr 19, 2012 #5

    jjr

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    Right you are, I was perhaps a bit vague at this point; if I can find the scalar gravitational potential field the grav. vector field is given to be its gradient.
     
  7. Apr 19, 2012 #6

    ehild

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    Yes, I noticed it at the end. Try to find the field directly. It leads to an easier integral.

    ehild
     
  8. Apr 19, 2012 #7

    ehild

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    And calculate with theta, using y=xtan(θ), r=x/cos(θ).

    ehild
     

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