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Applicaton of taylor series

  1. Apr 28, 2010 #1
    so F = mgR2/(R+h)2

    where R is the radius of the earth. consider the situation where h is much smaller than R.

    a) show that F is approximately equal to mg

    b)express F as mg multiplied by a series in h/R

    so i need help on getting started.

    would showing that F is approximately equal to mg be the same as expressing F as mg multiplied by a series in h/R since taylor series are used as approximations?

    i can pull an R2 out of the denominator and get rid of the R2 on the top and be left with mg(1+h/R2)-2 and in that case i could use the binomial expansion of a taylor series. would that solve a or b?

    any other help would be appreciated
     
  2. jcsd
  3. Apr 28, 2010 #2

    Dick

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    Yes, expanding F as a taylor series around h=0 is exactly what they are asking for.
     
  4. Apr 28, 2010 #3
    therefore a and b would be the same answer? just an expansion of the taylor series around h=0?
     
  5. Apr 28, 2010 #4

    Dick

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    You consider h to be small, so, yes, when h is near zero F is near mg. You don't need taylor series for the first one, if h is small compared with R you can just just put h=0 into the expression to find approximately what F is when h is small. You would use the taylor series to find out how good the approximation is. I'm not really sure what your question is. What they want in b is pretty clear, isn't it?
     
  6. Apr 28, 2010 #5
    there is a part c to this question asking how far above the surface of the earth you can go before the first order correction changes the estimate f ~ mg by more than 10% (and assume that R=6400km)

    so for that i just plug in .10 for h and 6400km for R into the first order approximation, i did so and got .9996875mg. is that saying that you can go 99.99% away from the surface before the first order estimate is changed by more than 10%?
     
  7. Apr 28, 2010 #6

    Dick

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    No. You want to calculate h from the series. If F changes by 10%, then mg changes to 0.9*mg. Figure out the value of the first order correction that will change mg (at h=0) to 0.9*mg.
     
  8. Apr 28, 2010 #7

    Dick

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    BTW mg*(1+h/R^2)^(-2) should be mg*(1+h/R)^(-2). Check the algebra.
     
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