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Variation of gravity with height

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the variation of gravity with height can be accounted for approximately by the following potential function:

    V(z)=mgz(1-z/R)

    Where R is the radius of the earth and z the height above the surface.


    2. Relevant equations
    r=R+z
    V=-GmM/r
    F=GmM/r^2


    3. The attempt at a solution

    First, I said define z such that r = R + z. We have the potential energy function for the earth as V(r)=-GmM/r, so V(z)=-GmM/(R+z). I then expanded this around z=0 and took the first two terms in the series to get:

    V=-GmM/R + GmMz/R^2

    and you can factor this to get

    V=GmM/R * [1-z/R]

    but the force F given by the potential energy function is GmM/R^2 at the surface, so this is equal to mg, so g = GM/R^2

    So the V function is then V = mgR[1-z/R]

    And here I am stuck. It appears that I am still taking the centre of the planet as my reference point. Can someone help me? I'm stuck.


    Somehow I need to redefine the reference point so that V = 0 for z = 0. How can I do this?
     
    Last edited: Oct 10, 2009
  2. jcsd
  3. Oct 10, 2009 #2
    Additional work:

    I now have something very close, but not quite.

    Since V is unchanged by adding a constant, I added GmM/R to V and swapped r = R+z to get

    V(z) = GmM/R - GmM/(R+z), which is 0 at z = 0 as it should be.

    Now I expand this around 0 and take the first three terms in the series:

    V(0) + zV'(0) + z^2 V''(0)

    to get:

    0 + zGmM/R^2 -2z^2GmM/R^3

    which can be factored and the identity g = GM/R^2 put in to get

    V = mgz[1-2z/R]

    Why do I have an extra factor of two in my answer....?

    Oops: forgot to divide by 2 factorial in the expansion...
     
    Last edited: Oct 10, 2009
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