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

  1. Oct 7, 2012 #1
    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

    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
    and i can write that as


    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. jcsd
  3. Oct 8, 2012 #2
    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 ##.
  4. Oct 8, 2012 #3
    we have never done anything like this, Im not even sure were to start with it
  5. Oct 8, 2012 #4
    You do not know what the derivative of a function is?
  6. Oct 8, 2012 #5
    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
    just take the dertive of the function that I already had?
  7. Oct 8, 2012 #6
    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.
  8. Oct 8, 2012 #7

    So do you mean that p= R_E(1+h/R_E)3/2 is approximately
    p= R_E(1*3/2
    p= R_E
    Then the derivative would be p=0
    Or am I really wrong?
  9. Oct 8, 2012 #8
    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?
  10. Oct 8, 2012 #9
    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?
  11. Oct 8, 2012 #10
    x is the independent variable. What is the function in your problem?
  12. Oct 8, 2012 #11
    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
  13. Oct 8, 2012 #12
    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.
  14. Oct 8, 2012 #13
    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
  15. Oct 8, 2012 #14
    Consider this very simple example. f(x) = x^2. Compute f(1.000001) using the derivative (the formula above).
  16. Oct 8, 2012 #15

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

    f(1.000001) = 1.000001^2
  17. Oct 8, 2012 #16
    That did not use the f'(x)Δx formula.
  18. Oct 8, 2012 #17

    f(x+Δx)−f(x)≈df(x)/dx Δx

    so 2x*Δx ? I do not understand what the Δx even is in this case
  19. Oct 8, 2012 #18
    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)?
  20. Oct 8, 2012 #19
    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
  21. Oct 8, 2012 #20
    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.
  22. Oct 8, 2012 #21
    does that mean that x+Δx is the 1.000001?
    Making Δx (1.000001 -1)/2
  23. Oct 8, 2012 #22
    Yes. x + Δx must always be equal to whatever value we want to compute the function at.

    What does this mean?
  24. Oct 8, 2012 #23
    i ment Δx = (1.000001 -1)/2 which is really small? getting close to zero...
  25. Oct 8, 2012 #24
    If Δx = (1.000001 - 1)/2, then x + Δx = 1 + (1.000001 - 1)/2 = 1.0000005, which is not 1.000001.
  26. Oct 8, 2012 #25
    ok so
    x + Δx = 1.000001
    and i really have no clue.

    f(x+Δx)≈2 Δx +1
    f(1.000001)≈2 Δx +1? or am i still even using the right thing?
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