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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Accuracy of a gravitational force approximation

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