# Showing that a near earth orbit is P = C(1 + 3h/2R_E)

1. Oct 7, 2012

### jeanbeanie

1. The problem statement, all variables and given/known data
Show that a satellite in low-Earth orbit is approximately P = C(1 + 3h/2R_E) where h is the height of the satellite, C is a constant, and R_E is the radius of the earth)

2. Relevant equations

p=C(1+3H/2R_E)
p2 = (4∏2)/(GM) * a3. (G - gravitational constant, M - mass of the earth (in this case) and a = semi-major axis)

3. The attempt at a solution
I've been googling this and I seen one person say use Taylor series. We have not touched that so Im pretty sure it isnt how I should be finding this

so the a is equal to the radui of earth, and its high
a=h+R_E
and i can write that as

a=R_E(1+h/R_E)

SO p2 = (4∏2)/(GM) * R_E(1+h/R_E)3.

I just saw in my noes thats if (4∏2)/(GM) can be cancled out if all meaurments are in Au's, years and solars mass.
Then Id get
p2 = R_E(1+h/R_E)3
and p= R_E(1+h/R_E)3/2

Now what?

Last edited: Oct 7, 2012
2. Oct 8, 2012

### voko

Now you have to recall that $\frac {df(x) } {dx} = \lim_{\Delta x \rightarrow 0} \frac {f(x + \Delta x) - f(x)} {\Delta x}$ which for some small $\Delta x$ means that $f(x + \Delta x) - f(x) \approx \frac {df(x) } {dx} \Delta x$.

3. Oct 8, 2012

### jeanbeanie

we have never done anything like this, Im not even sure were to start with it

4. Oct 8, 2012

### voko

You do not know what the derivative of a function is?

5. Oct 8, 2012

### RockenNS42

we we have done that, but I didnt get the second part
but now that Im more awake and looking at it again, does it mean

6. Oct 8, 2012

### voko

The second part means that if you know the value of a function and its derivative at some point x, then you can approximate the value of the function at some close point, via the expression given. Note that you have a function which you want to compute at (1 + h/R). The function (and the derivative) at 1 are trivial, and (1 + h/R) is very close to 1 because h is much smaller than R.

7. Oct 8, 2012

### RockenNS42

So do you mean that p= R_E(1+h/R_E)3/2 is approximately
p= R_E(1*3/2
then
p= R_E
?
Then the derivative would be p=0
Or am I really wrong?

8. Oct 8, 2012

### voko

As shown above, $f(x + \Delta x) - f(x) \approx \frac {df(x) } {dx} \Delta x$. What would $x$ and $\Delta x$ be in this case?

9. Oct 8, 2012

### RockenNS42

isnt x the function?
p= R_E(1+h/R_E)3/2

and found Δx be the h/R because it is going to zero?

10. Oct 8, 2012

### voko

x is the independent variable. What is the function in your problem?

11. Oct 8, 2012

### RockenNS42

Ok, I really thought that the p= R_E(1+h/R_E)3/2 was the function
as in
p(h)= R_E(1+h/R_E)3/2

12. Oct 8, 2012

### voko

That is one possible choice of the function. Now you need to choose x and consequently Δx so that h = x + Δx. These are usually chosen so that f(x) and f'(x) are very simple.

13. Oct 8, 2012

### RockenNS42

like any x? like x and 1?
Or something to do with the problem like
R_E and 0?

Or would it have nothing do to with the eq?

Im really sorry, we never went over any of this is in class

14. Oct 8, 2012

### voko

Consider this very simple example. f(x) = x^2. Compute f(1.000001) using the derivative (the formula above).

15. Oct 8, 2012

### RockenNS42

right ok, woudlnt I just plug in the value for x as 1.000001?
Like

f(1.000001) = 1.000001^2
and
f'(x)=2x
f'(1.000001)=2(1.000001)

16. Oct 8, 2012

### voko

That did not use the f'(x)Δx formula.

17. Oct 8, 2012

### RockenNS42

ok

f(x+Δx)−f(x)≈df(x)/dx Δx

so 2x*Δx ? I do not understand what the Δx even is in this case

18. Oct 8, 2012

### voko

Do you see that if x = 1, then both f(x) and f'(x) are very simple? What is then Δx and what is f(x + Δx)?

19. Oct 8, 2012

### RockenNS42

I do see that f(x) = x^2 it would be 1
and in f'(x) it would be 2

f(x+Δx)−f(x)≈df(x)/dx Δx
f(x+Δx)−1≈2 Δx
f(x+Δx)≈2 Δx +1
Like I dont understand how to use the formula maybe? I understand what a derive is and how to plug and chug numbers, just not this forumla or even how it applies to my problem. Im sorry

20. Oct 8, 2012

### voko

f(x+Δx)≈2 Δx +1 is not the complete answer. You were requested to compute f(1.000001) using the formula. You have the formula, but not the numeric result.