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Accuracy of a gravitational force approximation

  1. Jan 16, 2005 #1
    It seems to me that I've got part (a) right, but I'm not so sure about what I have in part (b). I just need to know whether or not I am on the right direction. Any help is highly appreciated. :smile:

    Problem

    The force due to gravity on an object with mass [tex]m[/tex] at a height [tex]h[/tex] above the surface of Earth is

    [tex]F=\frac{mgR^2}{(R+h)^2}[/tex]

    where [tex]R[/tex] is the radius of Earth and [tex]g[/tex] is the acceleration due to gravity.

    (a) Express [tex]F[/tex] as a series of powers of [tex]h/r[/tex].

    (b) Observe that if we approximate [tex]F[/tex] by the first term in the series, we get the expression [tex]F\approx mg[/tex] that is usually used when [tex]h[/tex] is much smaller than [tex]R[/tex]. Use the Alternating Series Estimation Theorem to estimate the range of values of [tex]h[/tex] for which the approximation [tex]F \approx mg[/tex] is accurate within 1%. (Use [tex]R = 6400 \mbox{ km}[/tex]).

    My work

    (a)

    [tex] (R+h) ^{-2} = \frac{1}{R^2} \left( 1 + \frac{h}{R} \right) ^{-2} [/tex]
    [tex] (R+h) ^{-2} = \frac{1}{R^2} \sum _{n=0} ^{\infty} \binom{-2}{n} \left( \frac{h}{R} \right) ^n [/tex]
    [tex] (R+h) ^{-2} = \frac{1}{R^2} \left[ 1 + \frac{(-2)}{1!} \frac{h}{R} + \frac{(-2)(-3)}{2!} \left( \frac{h}{R} \right) ^2 + \dotsb \right] [/tex]
    [tex] (R+h) ^{-2} = \frac{1}{R^2} \left[ 1 -2\frac{h}{R} + 3\left( \frac{h}{R} \right) ^2 - 4\left( \frac{h}{R} \right) ^3 + \dotsb \right] [/tex]
    [tex] (R+h) ^{-2} = \frac{1}{R^2} \sum _{n=0} ^{\infty} (n+1) \left( -\frac{h}{R} \right) ^n \Longrightarrow F = mg \sum _{n=0} ^{\infty} (n+1) \left( -\frac{h}{R} \right) ^n [/tex]

    (b)

    The value of [tex]mg[/tex] doesn't seem to matter here. So, I consider just the terms in the series. It seems to me that the easiest way to go is to plot

    [tex] \left| 1 - \sum _{n=0} ^{N} (n+1) \left( -\frac{h}{R} \right) ^n \right| [/tex]

    I've tried this with:

    [tex]N=1000[/tex]
    Domain: [tex][0,37 \times 10^3][/tex]
    Range: [tex][0,0.01][/tex].

    It gives a reasonable picture to estimate that [tex]h \leq 37 \times 10^3 \mbox{ m}[/tex]. Is there a better way to find this? Am I on the right path at all?

    Thank you
     
    Last edited: Jan 16, 2005
  2. jcsd
  3. Jan 16, 2005 #2

    Gokul43201

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    (a) looks fine.
    For (b) you are required to use the Estimation Theorem for Alternating Series. For an alternating series with decreasing terms that converge to 0 for large n :

    [tex]|S_{\infty} - S_n| \leq t_{n+1} [/tex]
     
  4. Jan 16, 2005 #3
    I have a question. I can't find an example in my textbook of the Alternating Series Estimation Theorem in a similar context. I know how to apply it in

    [tex]e = \sum _{n=0} ^{\infty} \frac{1}{n!} \Longrightarrow t_{n+1} =\frac{1}{(n+1)!} < \left| \mbox{error} \right|[/tex]

    but not really in

    [tex]e^x = \sum _{n=0} ^{\infty} \frac{x^n}{n!} \Longrightarrow ? < \left| \mbox{error} \right|[/tex]

    Which of the following is correct?

    [tex]t_{n+1} = (n+2) \left( \frac{h}{R} \right) ^{n+1} < 10^{-2} = \left| \mbox{error} \right| \quad (1)[/tex]

    [tex] t_{n+1} = (n+2) \left( \frac{1}{R} \right) ^{n+1} < 10^{-2} = \left| \mbox{error} \right| \quad (2)[/tex]

    I'm just a bit confused. :smile:

    Thanks
     
  5. Jan 16, 2005 #4

    Gokul43201

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    This one : (the other is dimensionally incorrect)

    [tex]t_{n+1} = (n+2) \left( \frac{h}{R} \right) ^{n+1} < 10^{-2} = \left| \mbox{error} \right| \quad (1)[/tex]
     
  6. Jan 16, 2005 #5
    Ok. Now, we have an inequality and two unknowns---namely, n and h. What kind of trick is then required so that I am able to estimate h properly? I think it involves plotting, but I'm not sure how---for the very same reason. :confused:
     
    Last edited: Jan 16, 2005
  7. Jan 18, 2005 #6
    I've just seen my problem in a different way, and it seems to work.

    From part (a), we have:

    [tex] F = mg \left[ 1 - 2\left( \frac{h}{R} \right) + 3\left( \frac{h}{R} \right) ^2 - \dotsb \right] [/tex]

    Applying the Alternating Series Estimation Theorem gives:

    [tex]b_{n+1} = 2 \frac{h}{R} \leq 10 ^{-2} \Longrightarrow h \leq 32 \mbox{ km} [/tex]

    which can be visually verified with the aid of a plot.
     
    Last edited: Jan 18, 2005
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