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Gravitational field due to a cylinder

  1. Apr 10, 2005 #1
    Calculate the gravitational field due to a homogeneous cylinder at an exterior point on the axis of the cylinder. Perform the calculation by computing the force directly.

    I'm not sure if I did this right, here's what I did

    z = distance from point to center of cylinder
    r = distance from point to vertical center of cylindrical shell element
    a = radius of shell element

    (these three lines form a triangle with r as the hypotenuse)

    [tex] \theta [/tex] = angle between r and z
    R = radius of cylinder M = mass of cylinder [tex] \rho [/tex] = density of cylinder V = volume of cylinder L = length of cylinder

    [tex] dg = \frac {-GdM} {r^2} = \frac {-G \rho dV} {r^2} [/tex]

    [tex] dV = 2 \pi aL da [/tex]

    [tex] g_x = g_y = 0 [/tex] [tex] g_z = gcos( \theta) [/tex]

    [tex] cos( \theta) = \frac {z} { \sqrt {z^2 + a^2}} [/tex]

    [tex] g = \int_{0}^{R} dgcos( \theta) = -2 \pi G \rho l \int _{0}^{R} \frac {a z da} {(z^2 + a^2)^{3/2}} [/tex]

    [tex] g = 2 \pi G \rho lz ( \frac {1} { \sqrt {z^2 + a^2}})_{0}^{R} [/tex]

    [tex] g = 2 \pi G \rho l (\frac {z} { \sqrt {z^2 + R^2}} - 1) [/tex]

    Is this right?
  2. jcsd
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