# 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