1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted