# Applicaton of taylor series

1. Apr 28, 2010

### apiwowar

so F = mgR2/(R+h)2

where R is the radius of the earth. consider the situation where h is much smaller than R.

a) show that F is approximately equal to mg

b)express F as mg multiplied by a series in h/R

so i need help on getting started.

would showing that F is approximately equal to mg be the same as expressing F as mg multiplied by a series in h/R since taylor series are used as approximations?

i can pull an R2 out of the denominator and get rid of the R2 on the top and be left with mg(1+h/R2)-2 and in that case i could use the binomial expansion of a taylor series. would that solve a or b?

any other help would be appreciated

2. Apr 28, 2010

### Dick

Yes, expanding F as a taylor series around h=0 is exactly what they are asking for.

3. Apr 28, 2010

### apiwowar

therefore a and b would be the same answer? just an expansion of the taylor series around h=0?

4. Apr 28, 2010

### Dick

You consider h to be small, so, yes, when h is near zero F is near mg. You don't need taylor series for the first one, if h is small compared with R you can just just put h=0 into the expression to find approximately what F is when h is small. You would use the taylor series to find out how good the approximation is. I'm not really sure what your question is. What they want in b is pretty clear, isn't it?

5. Apr 28, 2010

### apiwowar

there is a part c to this question asking how far above the surface of the earth you can go before the first order correction changes the estimate f ~ mg by more than 10% (and assume that R=6400km)

so for that i just plug in .10 for h and 6400km for R into the first order approximation, i did so and got .9996875mg. is that saying that you can go 99.99% away from the surface before the first order estimate is changed by more than 10%?

6. Apr 28, 2010

### Dick

No. You want to calculate h from the series. If F changes by 10%, then mg changes to 0.9*mg. Figure out the value of the first order correction that will change mg (at h=0) to 0.9*mg.

7. Apr 28, 2010

### Dick

BTW mg*(1+h/R^2)^(-2) should be mg*(1+h/R)^(-2). Check the algebra.