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Acceleration due to gravity of rod

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a thin rod of mass M and length L. Determine the gravitational potential at a point which is a distance r from the center of the rod and which lies on the midplane of the rod.

    Now suppose M is equal to the mass of the Earth and L is equal to the radius of the Earth.

    Determine the acceleration due to gravity (of the rod) at a distance r = 0.86 L from the center of the rod.


    2. Relevant equations

    [itex]\Phi[/itex](r)=[itex]\int[/itex]L/2-L/2[itex]\frac{-G(M/L)}{sqrt(z^2+r^2)}[/itex]dz

    U=mgh

    3. The attempt at a solution

    I calculate the gravitational potential to be -6.91657*10^7 Joules.

    mgh=-6.91657*10^7 Joules

    g=[itex]\frac{-6.91657*10^7}{m*h}[/itex]

    The distance from the rod is r=.86*L but what is m? If I ignore m I get -12.62 m/s^2 which is incorrect.

    Suggestions? Thank you!
     
  2. jcsd
  3. Nov 1, 2013 #2

    ehild

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    Correct so far... What did you get for the integral?

    mgh is the gravitational potential energy of a mass m at height h above the Earth surface. It is irrelevant here.

    The [STRIKE]potential[/STRIKE] force in the gravitational field of the rod is the force acting on unit mass. It depends on the mass of the rod and contains the gravitational constant G.

    That is all wrong. Determine the integral, and substitute r=0.86 L, M= M(Earth) and L= R(Earth)

    Edit:
    The [STRIKE]potential [/STRIKE] negative gradient of the potential is the magnitude of the gravitational force on unit mass, and the gravitational force exerted on mass m is F=-gradΦ m, and F=ma . The magnitude of the gravitational acceleration is equal to the gradient of the potential at the given point.

    ehild
     
    Last edited: Nov 1, 2013
  4. Nov 1, 2013 #3

    haruspex

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    No, that's the field, the grad of the potential.
    Gravitational potential has dimension L2T-2. So J/kg here.
    No, you cannot equate the potential to mgh. First, you don't want the m at all. The potential is a property of the field, independently of any mass placed in it. Ditching the m at least makes the dimensions right. But neither is it equal to gh. That would be the potential difference when moving with or against a uniform field for a distance h. The potential is zero at infinity and negative everywhere else.
    For the acceleration you need to differentiate the potential wrt r.
     
  5. Nov 1, 2013 #4

    ehild

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    Huh, I made a big mistake. Thank you, haruspex that you warned me.

    ehild
     
  6. Nov 1, 2013 #5
    Ah, not sure what I was thinking. It was a long night. Thank you!

    Evaluating the integral I get:

    [itex]\frac{-2GMsinh^{-1}(L/2r)}{L}[/itex]

    Taking the derivative of that with respect to r I get:

    [itex]\frac{2GMr}{(r^2)^{3/2}sqrt{L^2/r^2+4}}[/itex]

    Plugging in my known values I get 11.45 m/s2

    Thank you both for all your help!
     
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