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Acceleration due to Gravity problem.

  1. May 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Acceleration due to gravity, g, is considered to be 9.8(m/s^2). However, changes in latitude alter g according to the following model:
    g = 9.78049(1 + 0.005288sin^2x-0.000006sin^2(2x))(m/s^2), where x is latitude measured in degrees.
    [Side note= Hey guys i didn't know how to type in the little zero with the line across it which is the variable for degrees so i just used x, so x = zero-with-slash]

    a) Rewrite g in terms of powers of sinx only
    b)Find the latitude of Portland, Maine, and determine g.
    c)Find the percent of change between the value calculated in part b above and the standard value of 9.8.


    2. Relevant equations
    I seriously can't think of any! I don't even know where to begin! :(


    3. The attempt at a solution
    Please excuse my sorry attempt:
    a)1.g = 9.78049(1 + 0.005288sin^2x-0.000006sin^2(2x))(m/s^2)
    2.g = 9.78049(sin0x +sin^2x-sin^2x)
    3.g = 9.78049(sin^3x)
    b)Using a calc to graph
    portland = 5.4 degrees
    g = 95
    (im probably way off)
    c)5.4/9.8 = .548
     
  2. jcsd
  3. May 5, 2008 #2

    Dick

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    To write this as a power of sin(x) only, you just need to get rid of the sin(2x) term. Do you know a trig identity that will get you an expression for sin(2x) in terms of sin(x) and cos(x)? Then to write the cos(x) in terms of sin(x) remember 1=sin^2(x)+cos^2(x).
     
  4. May 5, 2008 #3
    Dick is right. And for that 'zero with a slash in it', you mean the Greek letter 'phi' or 'theta' ;)
     
  5. May 5, 2008 #4
    Thanks Dick, you helped me out alot i just have one small question though.
    Sin^2(x) * sin^2(x) = sin^4(x) ...right? :)

    Or anyone else, in case Dick isn't here could answer.
     
  6. May 5, 2008 #5

    Dick

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    That's a really small question. Sure. x^2*x^2=x^4. So sin^2(x)*sin^2(x)=sin^4(x). How could you doubt it? Were you working on the same question as sirajoman?
     
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