Accuracy of a gravitational force approximation

1. Jan 16, 2005

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

Last edited: Jan 16, 2005
2. Jan 16, 2005

Gokul43201

Staff Emeritus
(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 :

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

3. Jan 16, 2005

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

4. Jan 16, 2005

Gokul43201

Staff Emeritus
This one : (the other is dimensionally incorrect)

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

5. Jan 16, 2005

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.

Last edited: Jan 16, 2005
6. Jan 18, 2005

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.

Last edited: Jan 18, 2005