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Gravity outside a cylinder.

  1. Apr 12, 2008 #1
    With reference to this web page http://principles.ou.edu/earth_figure_gravity/index.html

    I calculate the force of gravity tangential to the long axis of a cylinder to be proportional to 4GM/R where M is the mass per unit length of the cylinder. Does that seem correct?
     
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  3. Apr 12, 2008 #2

    pam

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    F is perpendicular to the long axis of a cylinder.
     
  4. Apr 12, 2008 #3

    ZapperZ

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    No it isn't.

    Did you use the Gauss's Law equivalent to calculate g?

    Zz.
     
  5. Apr 12, 2008 #4
    I was trying to use the method demonstrated on that web page using an an imaginary Gaussian surface and the integral form of Poisson's equation. M is the mass contained within the imaginary volume of the Gaussian surface. For a cylinder I assumed the mass is proportional to [tex] 2 \pi R^2 p L[/tex] where L, R are length and radius of the cylinder and p is the density per unit volume of the cylinder. The integral of the Gaussian volume is [tex] \pi r L [/tex].

    Using the formula given,

    [tex] g(\pi r L) = 4\pi GM [/tex]

    [tex]g = \frac{4GM}{r L} = \frac{4G (2 \pi R^2 p L)}{ r L} = \frac{8 \pi G R^2 p}{r} [/tex]

    Where R and p are constants.

    I am not sure if that is correct. I was just trying to interpret the method shown. I would like to know what the correct solution is.

    What I am trying to establish is if the force of gravity perpendicular to the long axis of a cylinder is proportional to 1/r rather than the usual 1/(r^2) when considering a spherical massive body.
     
    Last edited: Apr 12, 2008
  6. Apr 12, 2008 #5
    This paper seems to agree that the force of gravity perpendicular to a cylinder is proportional to 1/R.


    http://www.pgccphy.net/103/gravity.pdf

    Interestingly the force equation is independent of the length of the cylinder and so applies equally to the force of gravity perpendicular to the axis of symmetry of a thin disk. According to this, an object leaving a solar system that had a lot of mass at the perimeter of the system (in the form of a disk of meteorites for example) would experience greater gravitational acceleration than GM/(R^2) where R is the total mass of the solar system.

    For a disk of uniform density the force of gravity on a test particle at a radius that is less than the radius of the disk appears to be independent of the distance from the centre of the disk. For a galaxy that happened to have an ideal disk shape and an even distribution of mass would have much greater orbital velocities consistent with F= GMm. For a more realistic galaxy with less than a perfect disk shape and greater density towards the centre the orbital velocities would be consistent with GMm/R > F > GMm/(R^2).
     
    Last edited: Apr 12, 2008
  7. Apr 12, 2008 #6

    rcgldr

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    For an infinitely long cylinder, gravity diminishes relative to 1/r. For a infinitely large plane, gravity remains constant regardless of position. A long cylinder or a large plane closely approximate this if distances from the cylinder or plane are relatively small compared to the size of the cylinder or plane.
     
  8. Apr 14, 2008 #7
    As an extension to the question in the OP what is the gravitational potential for a hypothetical body that has a gravitational field that attenuates relative to 1/r ?
     
  9. Apr 16, 2008 #8
    I read somewhere that there may be difficulties with using the gaussian method to determine the force of gravity. Is it possible to calculate the gravity/distance relationship for a flat slab by another method such as integrating the total forces of infinitesimal "point" masses that make up up the slab?

    I believe Newton used a similar method to calculate the gravitational forces of spherical shell when he concluded that the gravity of a large sphere is equivalent to that of a point particle of the same mass.
     
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