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Molniya Orbit: calc angular displacement of ascending node

  1. Nov 2, 2013 #1
    Hello there :),

    1. The problem statement, all variables and given/known data
    We have a molniya orbit with an eccentricity of [itex]e = 0.72[/itex] and a semi-major axis [itex]a = 26554 [km][/itex].

    The task:
    Calculate the angular displacement of the ascending node Ω after 100 sidereal days

    2. Relevant equations
    Moreover I know from Wikipedia that a molniya orbit has an inclination of [itex]i = 63.4 [deg][/itex].

    3. The attempt at a solution
    I considered using following formula
    [tex]\Delta\Omega = - \frac{3\pi J_2 R_E^2}{a^2 (1-e^2)^2} * cos i [/tex]
    But I am pretty sure it is the wrong formula, because time doesn't matter in it. (Even it has an Ω at the beginning ;)).

    Hence I hope on insightful input provided by you :). Thank you for your help!

    Paul
     
  2. jcsd
  3. Nov 3, 2013 #2
    That formula gives you ## \Delta \Omega ## per some period of time. What is that period of time?
     
  4. Nov 3, 2013 #3
    That's a simple but clever answer. I will give it a shot as soon as possible :). Thank you ...
     
  5. Nov 3, 2013 #4
    Using the hint voko provided I get the following answer

    [tex]\Delta\Omega = - \frac{3\pi J_2 R_E^2}{a^2 (1-e^2)^2} \cdot \cos i
    = \frac{3\pi 1082.7 \cdot 10^{-6} 6378^2}{26544^2 (1-0.72^2)^2} \cdot \cos 63.4
    = -2.139 \cdot 10^{-3} [\frac{rad}{rev}][/tex]

    One revolution takes
    [tex]1 [rev] = T = 2\pi \frac{a^3}{\mu_{Earth}}^{1/2}= 11.962h[/tex]

    Hence the angular displacement of the ascending node after 100 sederial days (1day = 23.934h) is:
    [tex]2.139 \cdot 10^{-3}\frac{23.934}{11.962}\cdot 100= 0.42798 [rad] = 24.5214[deg][/tex]

    Is this a plausible result?

    PS: I don't know what to do with the minus in front of [itex]-2.139 \cdot 10^{-3}[/itex] :)?

    Thanks for your help!
     
  6. Nov 3, 2013 #5
    I did not check the numbers, but the logic is sound.
     
  7. Nov 3, 2013 #6
    Thank you very much voko. The numbers are not as important as the logic is :).

    The minus in front of [itex]-2.139*10^{-3}[/itex] still bugs me. How can I get rid of it? Can I neglect it?
     
    Last edited: Nov 3, 2013
  8. Nov 3, 2013 #7
    The minus gives you the direction of rotation.
     
  9. Nov 3, 2013 #8
    Thank you, I got it :).
     
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