Accuracy of a gravitational force approximation

[DivGradCurl]

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.

Problem

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

F=\frac{mgR^2}{(R+h)^2}

where R is the radius of Earth and g is the acceleration due to gravity.

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

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

My work

(a)

(R+h) ^{-2} = \frac{1}{R^2} \left( 1 + \frac{h}{R} \right) ^{-2}
(R+h) ^{-2} = \frac{1}{R^2} \sum _{n=0} ^{\infty} \binom{-2}{n} \left( \frac{h}{R} \right) ^n
(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]
(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]
(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

(b)

The value of mg 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

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

I've tried this with:

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

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

Thank you

 

[Gokul43201]

(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 :

|S_{\infty} - S_n| \leq t_{n+1}

 

[DivGradCurl]

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

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

but not really in

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

Which of the following is correct?

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

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

I'm just a bit confused.

Thanks

 

[Gokul43201]

This one : (the other is dimensionally incorrect)

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

 

[DivGradCurl]

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.

 

[DivGradCurl]

I've just seen my problem in a different way, and it seems to work.

From part (a), we have:

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

Applying the Alternating Series Estimation Theorem gives:

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

which can be visually verified with the aid of a plot.