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Height differences from relative uncertainty of gravimeters

  1. May 27, 2017 #1
    1. The problem statement, all variables and given/known data
    The best relative gravimeters have a relative uncertainty of 10-12, that corresponds to a height difference of 3 µm.

    2. Relevant equations
    g∝(1/r2)
    The local gravitational acceleration g outside the Earth is proportional to 1/r2, which means (Δg)/g = -2 (Δr)/r. With (Δg)/g = 10-12 we get Δr = 0.5 * 10-12 r where r is the radius of Earth.

    3. The attempt at a solution
    I've tried to get to the equation I highlighted in red, but I fail. I know it has derived from taking derivatives, but my derivatives result in different equations.
    g∝1/r2 → g=k(1/r2) → ?
     
  2. jcsd
  3. May 27, 2017 #2

    BvU

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    The rule is derived from ##f(x+\Delta x) = f(x) + f'(x)\Delta x\ + {\mathcal O} (\Delta x)^2## so you get $$
    \Delta g = g(x+\Delta x ) - g(x) \approx g'(x) \Delta x \Rightarrow \\ = {-2\over x^3} \Delta x \Rightarrow {\Delta g\over g} = -2 {\Delta x\over x}$$

    We usually report the standard deviation which is positive.
     
  4. May 27, 2017 #3
    Thanks. What's the name of that rule and the equation you wrote?
    I've totally forgotten it, so I'm asking for a reference to learn and review.
     
  5. May 27, 2017 #4

    BvU

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