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Gravity/potential energy

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    A sensitive gravimeter is carried to the top of Chicago’s Willis (formerly Sears) Tower, where its reading for the acceleration of gravity is 1.36 mm/s^2 lower than at street level.

    Find the building's height, h=?

    2. Relevant equations

    3. The attempt at a solution

    I started this problem with this:

    g - g' = 0.00136 m/s^2

    now find the change in potential energy of gravity:

    GMm/(Radius of earth)^2 - GMm/(Radius of earth + h)^2

    using this we should be able to simplify but i am not getting too far from here, it is a bit messy and i am not getting too much to cancel out here...why?

    can anyone help me out?

    please and thank you!
  2. jcsd
  3. Oct 31, 2011 #2

    Doc Al

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    You're finding the change in g, not gravitational PE.

    OK, except that it should be only GM in the numerator, not GMm.

    Assume that h is small enough compared to the earth's radius that you can use a binomial expansion approximation.

    Hint: (R + h)^2 = R^2(1 + h/R)^2; now take advantage of the fact that h/R << 1.
  4. Oct 31, 2011 #3


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    Look at [itex]\displaystyle \frac{1}{R^2}-\frac{1}{(R+h)^2}=\frac{(R+h)^2-R^2}{R^2(R+h)^2}[/itex]

    The numerator of the expression on the right side of the equation is the difference of squares.

    Factor that.

    R > 6300 km. I doubt that the building is more than 1 km tall (≈5/8 mile).
  5. Oct 31, 2011 #4
    why is m not included in the formula?

    going off the rest you said:

    Re = radius of earth

    GM/(Re)^2 - GM/(Re + h)^2

    Gm[(Re + h)^2 - (Re)^2] /(Re)^2 (Re + h)^2

    using this: (R + h)^2 = R^2(1 + h/R)^2

    i reduce down to :

    Gm [ h^2/Re^2 + 2h/Re] / Re^2

    does this sound right? still seems like a lot going on
  6. Oct 31, 2011 #5
    Did you cancel out GM?
  7. Oct 31, 2011 #6

    Doc Al

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    Because you want the acceleration, not the force.

    Looks good. Now get rid of higher order terms, like (h/Re)2. They are too small to worry about. (And that m should be M, the mass of the earth.)

    You're on the right track. Keep going.
  8. Oct 31, 2011 #7
    right. now i get :


    and get with an answer of 441.4 meters

    which turns out to be the right answer, sweet. Thanks everyone!
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