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Geodesic deviation equation

  1. Mar 14, 2008 #1
    I'm a bit stuck with trying to interpret this for acceleration in earths gravity. The question is:

    [tex]D^2 Y^d = R_{abc}^dV^aY^bV^c[/tex]

    Let (t, x, y, z ) be the natural coordinates for an observer at a point P just above the
    surface of the Earth, i.e. with z measuring height. Explain why at P, [tex]R_{0101}[/tex] and [tex]R_{0202}[/tex] are approximately −g/RE , and [tex]R_{0303}[/tex] is approximately 2g/RE , where RE is the radius of the Earth and g the acceleration due to gravity in non-relativistic theory.

    Ok so I assume [tex]Y = (0, x, y, z) [/tex] describe the observer and [tex]X = (1,0)[/tex]
    describes the earth. Then we get

    [tex] a_x= g/R_E x a_y= g/R_E y [/tex]

    and [tex] a_z = -2gz/R_E[/tex]

    but this doesnt seem to make sense
     
    Last edited: Mar 14, 2008
  2. jcsd
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