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Error Propogation - E of Gravity due to r

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    The acceleration due to gravity is given by Newton's universal law of gravitation as:
    g=(GM)/r^2

    Derive an algebraic expression for the error in gravity (Eg) due to the uncertainty in r. You may assume that the errors in G & M = 0; (EG & EM = 0)

    Let E=error in for internet purposes
    2. Relevant equations

    g=(GM)/r^2

    3. The attempt at a solution

    r^2=(GM)/g
    r=(GM/g)^(1/2)
    r=[(GM)^(1/2)]/[g^1/2]

    This is calculus, i can't see a classical algebraic way of doing this..

    any help plz?
     
  2. jcsd
  3. Sep 16, 2007 #2
    Suppose f is a function of r. Then:

    [tex]f + \delta f \approx f(r) + \delta r f^{\prime}(r)[/tex]

    so [tex]\delta f \approx \delta r f^{\prime}(r)[/tex]

    Can you see why this is the case?
     
  4. Sep 16, 2007 #3
    in the second half of that you are taking the derivative of the function, as far as i am convinced this course is an algebra based course and i'm trying to figure out an algebraic response to this question, without taking the derivative at any point.... would it be
    r^2+Er^2=GM/g+Eg
     
  5. Sep 16, 2007 #4
    Okay, then:

    [tex]g + \delta g = \frac{GM}{(r + \delta r)^2}[/tex]

    Knowing that [tex]g = \frac{GM}{r^2}[/tex]

    Apply binomial expansions as necessary.
     
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