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Homework Help: Law of universal gravitation and Kepler's law

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the rate of change of the free-fall acceleration with vertical position near the Earth's surface is
    [tex] \frac{d}{dr} [g] = -\frac{2GM_E}{R_E^3} [/tex]

    Assuming h is small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is
    [tex] \Delta g = \frac{(2GM_E h)}{R_E^3} [/tex]

    Show that the minimum period for a satellite in orbit around a spherical planet of uniform density ρ is
    [tex] T_{min} = \sqrt{ \frac{3\pi}{Gp}} [/tex]

    2. Relevant equations
    [tex] g=\frac{GM_E}{R_E^2} [/tex]
    [tex] T^2=Kr^3 [/tex]



    3. The attempt at a solution
    I did the first one easily
    [tex] \frac{d}{dr} [\frac{GM_E}{R_E^2}] [/tex]
    [tex] \frac{d}{dr} [GM_E R_E^{-2}] [/tex]
    simple power rule
    The second and third not so much, I substituted one R with (R+h) as that is what it seems like is going on and find the difference. That did not work out for me
    The third I thought about p= Mass/Volume and solving for R and substitute but I'm not sure what to do with K and Kepler's 3rd observation is the only equation that I have the deals with the period
     
    Last edited: Apr 24, 2016
  2. jcsd
  3. Apr 24, 2016 #2

    PeroK

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    For part 1, ##R_E## is constant so you can't differentiate your formula with respect to ##r##.
     
  4. Apr 24, 2016 #3
    yes, in the situation described using Earth it would be a constant. I think they only did that to make it relatable but still treat it as a variable as it could be used for other bodies in space
     
  5. Apr 24, 2016 #4

    PeroK

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    You could differentiate the formula for ##r## then replace ##r## with ##R_E##. That would be valid. What you have is not valid.

    What went wrong for part 2?
     
  6. Apr 24, 2016 #5
    when trying to simplify I get a bigger mess that I don't know what to do with

    [tex] \frac{-2GM_E}{R_E^3} - \frac{-2GM_E}{(R_E+h)^3} [/tex]

    expanding (R+h)^3 gets me [tex] R^3 +3R^2h+3Rh^2+h^3 [/tex]
    simplifying from there makes it worse
     
  7. Apr 24, 2016 #6

    PeroK

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    That's ##\Delta(dg/dr)## you are trying to calculate there.

    The point of part was to show that ##dg/dr## is approximately constant.

    It's ##\Delta g## you need to calculate.
     
  8. Apr 24, 2016 #7
    I thought so, I tried it with [tex] \frac{GM_E}{R_E^2} [/tex] but ran into the same issue at the same point, so I tried both
     
  9. Apr 24, 2016 #8

    PeroK

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    That should have worked. You need to use the binomial expansion doing it that way.

    Instead, can you see how to use part 1? It's very simple. Compare the answers to parts 1 and 2.
     
  10. Apr 24, 2016 #9
    I expanded (R+h)^2 to
    [tex] R^2+2Rh+h^2 [/tex]
    multiplied to get a common denominator
    [tex] \frac {(R^2 +2Rh+h^2)GM-R^2GM}{R^2(R^2 +2Rh+h^2)} [/tex]

    I could only remove R^2GM after that
     
  11. Apr 24, 2016 #10

    PeroK

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    The point of part 1 was to avoid this. You don't need to expand ##(R+h)^2##. Instead take out a factor of ##R^2## and use the binomial expansion.

    Better still, use part 1. what is the change in a function whose derivative is constant?
     
  12. Apr 24, 2016 #11
    I don't fully understand what you are saying.
    What I am getting is
    [tex] \frac{GM}{R^2} - \frac {GM}{(R+h)^2} [/tex]
    then factor a ##R^2## from each term?
     
  13. Apr 24, 2016 #12

    PeroK

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    yes, why not?
     
  14. Apr 24, 2016 #13
    so....

    [tex] \frac{1}{R^2}(GM- \frac{GMR^2}{(R+h)^2} )[/tex]

    I haven't done anything like this so I'm not too confident in what I am doing.
     
  15. Apr 24, 2016 #14

    PeroK

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    what about

    ##\frac{GM}{R^2}(1 - \frac{1}{(1 +h/R)^2})##
     
  16. Apr 24, 2016 #15
    oh yea, I could have also taken out ##GM## as well.
    The ##\frac{1}{(1+h/R)^2}## is really confusing me
     
  17. Apr 24, 2016 #16

    PeroK

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    ##R^2(1+h/R)^2 = (R+h)^2##
     
  18. Apr 24, 2016 #17
    okay so
    [tex] R^2(1+h/R)^2 = R(1+h/R)R(1+h/R) = (R+hR/R)^2= (R+h)^2 [/tex]
    that is really cool how that works out
     
  19. Apr 24, 2016 #18
    [tex] \frac{GM}{R^2}(1-\frac{1}{(1+h/R)^2}) [/tex] can be rewritten as
    [tex] \frac{GM}{R^2}(\frac{(1+h/R)^2-1}{(1+h/R)^2}) [/tex]

    Then do the "binomial expansion" as you said?
     
  20. Apr 24, 2016 #19

    PeroK

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    simply apply the binomial expansion to ##(1 + h/R)^{-2}##
     
  21. Apr 24, 2016 #20
    so
    [tex] (1+h/R)^{-1} (1+h/R)^{-1} = (1+2(h/R) +h^2/R^2)^{-1} [/tex]

    [tex] \frac{GM}{R^2} (1-(1+2(h/R) +h^2/R^2)^{-1}) [/tex]
     
  22. Apr 24, 2016 #21

    PeroK

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    Thats not right at all. you need to brush up your maths. I'm going off line now.
     
  23. Apr 24, 2016 #22
    okay.

    even dumming it down to (a+b)^2 = 1a+2ab+b^2
    a=1 b= h/R so ## 1^2 +2*(1)(h/R)+ (h/R)^2##

    something must be going right over my head if I am missing something so simple
     
  24. Apr 24, 2016 #23
    I'm still not getting what is wrong here. Am I not getting what you mean by binomial expansion?
     
  25. Apr 25, 2016 #24

    ehild

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    Yes, factor out GM/R2 from each term.
    You have problem with the second term then, ##\left ( \frac{1}{1+\frac{h}{R}} \right )^2 ##
    (h/R) is small, denote q=-h/R. So you have the expression ##\frac{1}{1-q}##. That should remind you on something: the sum of infinite geometric series.If you have a geometric progression 1, q, q2, q3......qn, what is the sum of the terms? What is the limit of the sum when n goes to infinity?
     
  26. Apr 25, 2016 #25
    I see the resemblance to the geometric series as ##h/R## would be less than one and would converge if it was a ratio in a geometric series but I don't see how that relates to the equation.
    As a side note. My progression through calculus is that I am taking a test on diverging and converging infinite series and Taylor polynomials tomorrow morning.
     
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